Jul 1, 2010 - Let a, b, and c be the sides of triangle ABC, and angles α, β, and γ be the ... sides a, b, and c, i.e. the interior angles at vertices A, B, and C. An ...
Tiling a Triangle with Congruent Triangles Michael Beeson July 1, 2010 Abstract We investigate the problem of cutting a triangle ABC into N congruent triangles (the “tiles”), which may or may not be similar to ABC. We wish to characterize the numbers N for which some triangle ABC can be tiled by N tiles, or more generally to characterize the triples (N, T ) such that ABC can be N -tiled using tile T . In the first part of the paper we exhibit certain families of tilings which contain all known tilings. We conjecture that the exhibited tilings are the only possible tilings. If that is so, then for there to exist an N -tiling of any triangle ABC, N must be a square, or 2, 3, or 6 times a square, or a sum of two squares. We were able to reduce this conjecture to a special case. The case we could not solve is when tile has angles α, β, and γ with 3α = 2β, and sin(α/2) is rational. Some number-theoretic properties of N are also necessary. The triangle ABC must have angles 2α, β, and β + γ and α is not a rational multiple of π. The simplest unsolved case is N = 28, with a tile whose sides are 2, 3, and 4, and triangle ABC has sides 12, 14, and 16. In particular, there are no N -tilings for N = 7. We have earlier given a (rather long) traditional Euclid-style proof of the impossibility of a 7-tiling, but could not handle even N = 11, let alone N = 19. Now we know there are no N -tilings for N = 11, 19, and 23. The proof in this paper goes beyond Euclidean geometry, by bringing the tools of linear algebra and field theory to bear on the problem. √ The key idea for the case when the tile is similar to ABC is that the similarity factor N is an eigenvalue of a certain matrix. In the case when T is not similar to ABC, various geometrical considerations deal with all but a few special cases. When the tile has angles π/11, 3π/11, and 7π/11, or π/14, 4π/14, and 9π/14, we use the theory of cyclotomic fields, and when the tile has one angle of 120◦ and one angle of 2π/15 = 24◦ , or π/9 = 20◦ , or π/12 = 15◦ , we also use some algebraic number theory about the primes in cyclotomic fields.
1
Examples of Tilings
We consider the problem of cutting a triangle into N congruent triangles. Figures 1 through 4 show that, at least for certain triangles, this can be done with N = 3, 4, 5, 6, 9, and 16. Such a configuration is called an N -tiling. The method illustrated for N = 4 ,9, and 16 clearly generalizes to any perfect square N . While the exhibited 3-tiling, 6-tiling, and 5-tiling clearly depend on the exactly angles of the triangle, any triangle can be decomposed into n2 congruent triangles by drawing n − 1 lines, parallel to each edge and dividing the other two edges into n equal parts. Moreover, the large (tiled) triangle is similar to the small triangle (the “tile”). We call such a tiling a quadradtic tiling. It follows that if we have a tiling of a triangle ABC into N congruent triangles, and m is any integer, we can tile ABC into N m2 triangles by subdividing the first tiling, replacing each of the N triangles by m2 smaller ones. Hence the set of N for which an N -tiling of some triangle exists is closed under multiplication by squares.
1
Figure 1: Two 3-tilings
Figure 2: A 4-tiling, a 9-tiling, and a 16-tiling
Let N be of the form n2 + m2 . Let triangle T be a right triangle with perpendicular sides n and m, say with n ≥ m. Let ABD be a right triangle with base AD of length m2 , the right angle at D and altitude mn, so side BD has length mn. Then ABD can be decomposed into m triangles congruent to T , arranged with their short sides (of length m) parallel to the base AD. Now, extend AD to point C, located n2 past D. Triangle ADC can be tiled with n2 copies of T , arranged with their long sides parallel to the base. The result is a tiling of triangle ABC by n2 + m2 copies of T . The first 5-tiling exhibited in Fig. 3 is the simplest example, where n = 2 and m = 1. The case N = 13 = 32 + 22 is illustrated in Fig. 5. We call these tilings “biquadratic.” More generally, a biquadratic tiling of triangle ABC is one in which ABC has a right angle at C, and can be divided by an altitude from C to AB into two triangles, each similar to ABC, which can be tiled respectively by n2 and m2 copies of a triangle similar to ABC. The second 5-tiling shows that this can be sometimes be done more generally than by combining two quadratic tilings. If the original triangle ABC is chosen to be isosceles, then each of the n2 triangles can be divided in half by an altitude; hence any isosceles triangle can be decomposed into 2n2 congruent triangles. If the original triangle is equilateral, then it can be first decomposed into n2 equilateral triangles, and then these triangles can be decomposed into 3 or 6 triangles each, showing that any equilateral triangle can be decomposed into 3n2 or 6n2 congruent triangles. These tilings are neither quadratic nor biquadratic. For example we can 12-tile an equilateral triangle in two different ways, starting with a 3-tiling and then subdividing each triangle into 4 triangles (“subdividing by 4”), or starting with a 4-tiling and then subdividing by 3. Examples like these led us to the following definitions: A tiling E of triangle ABC (with tile T2 is a subtiling of another tiling F of ABC (with tile T ), if T can be tiled by the tile T2 and the tiling E is obtained by tiling each copy of T in F with triangle T2 . It is not required that the same tiling be used for each copy of T . For example, we could take F to be one of the two five-tilings, and then tile each of the tiles in that tiling by one of its two five-tilings. In this way we can obtain 32 different 25-tilings, none of them quadratic. A tiling of ABC is called composite if it is a subtiling of some tiling into fewer triangles. It is called prime if it is not composite. Note that a quadratic N 2 tiling is prime if and only if N is a prime number.
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Figure 3: Three 4-tilings
Figure 4: Two 5-tilings
The examples above do not exhaust all possible tilings, even when N is a square. For example, Fig. 7 shows a 9-tiling that is not produced by those methods: There is another family of N -tilings, in which N is of the form 3m2 , and both the tile and the tiled triangle are 30-60-90 triangles. The case m = 1 is given in Fig. 1; the case m = 2 makes N = 12. There are two ways to 12-tile a 30-60-90 triangle with 30-60-90 triangle. One is to first quadratically 4-tile it, and then subtile the four triangles with the 3-tiling of Figure 1. This produces the first 12-tiling in Fig. 8. Somewhat surprisingly, there is another way to tile the same triangle with the same 12 tiles, also shown in Fig. 8; the second tiling is prime. The next member of this family is m = 3, which makes N = 27. Two 27-tilings are shown in Fig. 9; the first obtained by subtiling a quadratic tiling, and the second one prime. Similarly, there are two 48-tilings (not shown). Until October 12, 2008, no examples were known of more complicated tilings than those illustrated above. Then we found the beautiful 27-tiling shown in Fig. 10. This tiling is one of a family of 3k2 tilings (the case k = 3). The next case is a 48-tiling, made from six hexagons (each containing 6 tiles) bordered by 4 tiles on each of 3 sides. In general one can arrange 1 + 2 + . . . + k hexagons in bowling-pin fashion, and add k + 1 tiles on each of three sides, for a total number of tiles of 6(1 + 2 + . . . + k) + 3(k + 1) = 3k(k + 1) + 3(k + 1) = 3(k + 1)2 . Figure 11 shows more members of this family.
2
Previous work
The examples given in Figures 1 through 6 are well-known. They have been discussed, in particular, in connection with “rep-tiles” [6]. A “rep-tile” is a set of points X in the plane (not necessarily just a triangle) that can be dissected into N congruent sets, each of which is similar to S. An N -tiling in which the tiled triangle ABC is similar to the triangle T used as the tile is a special case of this situation. That is the case, for example, for the n2 family and the n2 + m2 family, but not for the 3-tiling, 6-tiling, or the 12-tiling exhibited above. Thus the concepts of an N -tiling and rep-tiles overlap, but neither subsumes the other. The paper [5] also contains a diagram showing the n2 family of tilings, but the problem considered there is different: one is allowed to cut N copies of the tile first, before assembling the pieces into a large figure, but the large figure must be similar to the original tile. The two books [1] and [2] have tantalizing titles, but deal with other problems. Only after completing the work in this paper did I encounter Soifer’s book [7], when the
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Figure 5: A 13-tiling
Figure 6: A 6-tiling, an 8-tiling, and a 12-tiling
second edition came out, although the first edition had been out for 19 √ years. The book contains the observation that if the tile T is similar to the tiled triangle then N is an eigenvalue of a certain matrix, so that observation is, as it turns out, not new. The book, however, does not contain any examples of tilings beyond the quadratic tilings, though it gives an indication that at least the biquadratic tilings were known, since it says that the 1989 Russian Mathematical Olympiad contained the problem to show that if N is a sum of two squares then there is a triangle that can be N -tiled. Soifer states (p. 48) the open problem solved in this paper, and says that Paul Erd¨ os offered a $25 prize for the first solution. He does not state where or when Erd¨ os mentioned these problems. The problem statement is: Find all positive integers N such that at least one triangle can be cut into N triangles congruent to each other. This is Soifer’s “Problem 6.7.” Soifer also states some related problems. His “Problem 6.5” is: For each triangle ABC, find all positive integers N such that T can be cut into N triangles congruent to each other, and the number of distinct partitions of T into N congruent triangles. Actually this is two problems, and we have solved one and a half of them in this paper. Given a triangle ABC, our results succinctly describe the pairs (N, T ) such that ABC can be N -tiled by triangle T . That is slightly better than just finding the possible N , but not as good as completely classifying and counting the tilings. Soifer says that his Problem 6.5 is “open and very difficult.” Soifer’s “Problem 6.6” is also a $25 Erd¨ os problem: Find (and classify) all triangles that can only be cut into n2 congruent triangles for any integer n. We have solved this problem. In fact, our main theorem implies the stronger statement: Given triangle ABC, a necessary and sufficient condition for ABC to have N -tilings only when N is a square is that ABC be not isosceles, and not a right triangle whose angles have rational tangents. But we have not proved (nor does Soifer claim or even conjecture) that for triangles ABC meeting these conditions, any tiling must be the quadratic tiling, although we have no counterexample to this conjecture either. Soifer claims, without publishing a proof, that if the sides and angles of ABC are integrally independent, then ABC admits only quadratic tilings. He proves that the perfect squares are exactly the N for which every triangle ABC can be N -tiled by some triangle.
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Figure 7: Another 9-tiling
Figure 8: Two 12-tilings
Dima Fan-Der-Flaas informed me that the problem of finding an N -tiling of some triangle when N = 1989 was posed on the Russian Mathematical Olympiad in 1989; it was solved by a few students, who had to discover what we call here the “biquadratic tilings”, and realize that 1989 is a sum of two squares and the relevance of that fact. I would like to thank Dima for his careful reading of parts of some drafts of this paper.
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The impossibility of certain tilings
The elementary constructions just described suffice to produce N -tilings when N has one of the forms n2 , n2 + m2 , 2n2 , 3n2 , or 6n2 . Our main theorem implies that if N is not of one of these forms, then there is no N -tiling of any triangle. The smallest N not of one of these forms is N = 7. The next numbers not of one of the listed forms are 11, 14, and 19. It follows from the main theorem of this paper is that there is no 7-tiling, 11-tiling, 14-tiling, or 19-tiling. Indeed, our theorem settles completely the question, “for which N does there exist an N -tiling of some triangle?”. In an unpublished manuscript, we have given a proof using only traditional Euclidean geometry that there is no 7-tiling (of any triangle).1 There is no practical hope of extending that argument even to the next case N = 11; the proof would run to hundreds, if not thousands, of pages, and be completely incomprehensible. But in this paper, we solve the question for arbitrary N . A little linear algebra and Galois theory goes a long ways! The problem of finding the N -tilings of ABC by tile T has three parameters. Quantifying over two of them gives a problem; for example, quantifing over ABC and T leads to the problem of characterizing the N for which some triangle can be N -tiled by some tile; that is the problem mentioned above. Every triangle can be quadratically tiled; but aside from quadratic tilings, we can ask what triangles have a non-quadratic N -tiling for some N and some tile; or we can ask what tiles can be used to N -tile some ABC non-quadratically. The examples we have given of 1 The proof uses, in addition to obviously Euclidean arguments, some simple trigonometry that in principle can be done within Euclidean geometry.
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Figure 9: Two 27-tilings
Figure 10: A prime 27-tiling
non-quadratic tilings all involve some special circumstances: either the tile is a right triangle, or ABC is an isosceles or even equilateral triangle. In fact we will prove that this is the case. Our proof splits naturally into two cases: the case in which the tile T and the tiled triangle ABC are similar, and the case in which they are not similar. The case in which they are similar is the easier one; we use linear algebra and some simple geometry to deal with it. The case in which T and ABC are not similar is much more complicated. The key to its solution lies in the “vertex splitting”. Each vertex of ABC (or at least two of them) must “split” into a sum of angles of the tile, because several tiles meet at that vertex. The vertex splitting is described by three numbers P , Q, and R that give the total number of α, β, and γ angles (respectively) occurring at the vertices of ABC. Then we have P α + Qβ + Rγ = π, since the total sum of angles at the vertices of ABC is π. This implies that the number R cannot be too large: there will usually be more α and β angles than γ angles at the vertices of ABC. But then, somewhere else in the tiling, there must be a vertex with more γ angles than α and β angles. That will give another equation nα + mβ + ℓγ = π or 2π. Together with α + β + γ = π, that is three equations for α, β, and γ (for each fixed P , Q, and R). Thus, unless the determinant of the system is zero (which is another equation to consider), the vertex-splitting numbers P , Q, and R determine the angles of the tile! But there are still infinitely many possibilities for the vertex-splitting numbers, of course. Careful counting of the distribution of the α, β, and γ angles at the other vertices enables us to rule out most of the possibilities, except for a few special cases. For example, when γ is 120◦ . When γ is 120◦ and the tile is isosceles, we get the 3k2 tilings discussed above. In the other cases when γ is 120◦ , there are no tilings at all, but this is comparatively difficult to prove. After considerable efforts, only three sub-cases remained: the case when the tile has one 120◦ angle and smallest angle α equal to either π/12, π/9, or 2π/15. In degrees, that is 15◦ , 20◦ , or 24◦ . These cases did not yield to either geometry or linear algebra; but finally we ruled them out using the Galois theory and algebraic number theory of the cyclotomic field Q(ζ), where ζ = eiα . There were also two other special tiles that did not
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Figure 11: 3m2 tilings for m = 4 and m = 5
yield to a geometric analysis: when α = π/11 and β = 3α, and when α = π/14 and β = 4α. These cases also yielded to field theory: we consider the equation that says the area of triangle ABC is N times the area of the tile, and show that it gives a polynomial equation for tan(α/2) whose degree is not consistent with the degree known from the Galois theory of the cyclotomic fields.
4
Definitions, notation, and some simple lemmas
We give a mathematically precise definition of “tiling” and fix some terminology and notation. Given a triangle T and a larger triangle ABC, a “tiling” of triangle ABC by triangle T is a list of triangles T1 , . . . , Tn congruent to T , whose interiors are disjoint, and the closure of whose union is triangle ABC. A “strict vertex” of the tiling is a vertex of one of the Ti that does not lie on the interior of an edge of another Tj . A “strict tiling” is one in which no Ti has a vertex lying on the interior of an edge of another Tj , i.e. every vertex is strict. For example, the biquadratic tilings (illustrated above for N = 5 and N = 13) are not strict, but all the other tilings shown above are strict. The letter “N ” will always be used for the number of triangles used in the tiling. An N -tiling of ABC is a tiling that uses N copies of some triangle T . Let a, b, and c be the sides of triangle ABC, and angles α, β, and γ be the angles opposite sides a, b, and c, i.e. the interior angles at vertices A, B, and C. An interior vertex in a tiling of ABC is a vertex of one of Ti that does not lie on the boundary of ABC. A boundary vertex is a vertex of one of the Ti that lies on the boundary of ABC. By the law of sines we have a b c = = sin α sin β sin γ Up to similarity then we may assume a
=
sin α
b
=
sin β
c
=
sin γ
Since γ = π − (α + β) we have sin(γ) = sin(α + β), so p sin α + q sin β + r sin(α + β) = 0.
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The meanings of all these symbols will be fixed throughout the rest of the paper. A non-strict vertex V is one that lies on an edge of Tj , with Tj on one side of the edge and (more than one) Ti having vertex V on the other side. Consider the maximal line segment S extending this edge which is contained in the union of the edges of the tiling. This is defined to be the maximal segment of V . Since there are triangles on each side of S, there are triangles on each side of S at every point of S (since S cannot extend beyond the boundary of ABC). Hence the length of S is a sum of lengths of sides of triangles Ti in two different ways (though the summands may possibly be the same numbers in a different order). Let us assume for the moment that the summands are not the same numbers. Then it follows that some linear relation of the form pa + qb + rc = 0 holds, with p, q, and r integers not all zero (one of which must of course be negative), and the sum of the absolute values of p, q, and r is less than or equal to N , since there are no more than N triangles. If S is a maximal segment containing a non-strict vertex, then there will be integers n and m such that n triangles have a side contained in S and lie on one side of S, and m triangles have a side in S and lie on the other side of S. In that case we say S is of type m : n. For example, Fig. 3 shows a 5-tiling with a maximal segment of type 1 : 2. This definition does not require that the lengths of the subdivisions of the maximal segment all be the same (as they are in Fig. 3). A quadratic tiling is one in which N is a perfect square, say N = m2 , and the tiling is produced by drawing m − 1 equally spaced lines parallel to each side, dividing each edge into m equal segments. In such a tiling, the tile T is similar to the large triangle ABC. An angle relation is an equation pα + qβ + rγ = 2π where p, q, and r are non-negative integers, not all equal. (Since we always have α + β + γ = π, we do not count that equation or its multiples as an angle relation.) A split vertex occurs when two copies of the tile in a triangle share one of the vertices of the large triangle. The following lemma is simple and fundamental: Lemma 1 If, in a tiling, P is a boundary vertex (or a non-strict interior vertex) and only one interior edge emanates from P , then both angles at P are right angles and γ = π/2. Proof. If the two angles at P are different, then their sum is less than π, since the sum of all three angles is π. Therefore the two angles are the same. But 2α ≤ α + β < π and 2β ≤ β + γ < π. Therefore both angles are γ. But then 2γ = π, so γ = π/2. The following result we call “Euler’s equation”, because it is related to Euler’s famous formula for the numbers of vertices, edges, and faces of a polygon, although we have not derived it that way here. Lemma 2 Let a tiling contain Nb boundary vertices, Ns strict interior vertices, and Nn nonstrict interior vertices. We then have N −1
=
Nb + Nn + 2Ns
(1)
Proof. We count vertices. At each strict interior vertex, the sum of the angles of the tiles sharing that vertex is 2π. At each non-strict interior vertex and at each boundary vertex (that is, vertex lying on the boundary of triangle ABC but not equal to A, B, or C), the angle sum is π. The total angle sum of all N copies of the tile is N π, which must be accounted for by the π at the vertices of ABC, plus the contributions at the other vertices. The following lemma identifies those relatively few rational multiples of π that have rational tangents or whose sine and cosine satisfy a polynomial of low degree over Q.
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Lemma 3 Let ζ = eiθ be algebraic of degree d over Q, where θ is a rational multiple of π, say θ = 2mπ/n, where m and n have no common factor. Then d = ϕ(n), where ϕ is the Euler totient function. In particular if d = 4, which is the case when tan θ is rational and sin θ is not, then n is 5, 8, 10, or 12; and if d = 8 then n is 15, 16, 20, 24, or 30. Remark.√ For example, if θ = π/6, we have√sin θ = 1/2, which is of degree 1 over Q. Since cos θ = 3/2, the number ζ = eiθ is in Q(i, 3), which is of degree 4 over Q. The number ζ is a 12-th root of unity, i.e. n in the theorem is 12 in this case; so the minimal polynomial of ζ is of degree ϕ(12) = 4. This example shows that the theorem is best possible. Remark. The hypothesis that θ is a rational multiple of π cannot be dropped. For example, x4 − 2x3 + x2 − 2x + 1 has two roots on the unit circle and two off the unit circle.
Proof. Let f be a polynomial with rational coefficients of degree d satisfied by ζ. Since ζ = ei2mπ/n , ζ is an n-th root of unity, so its minimal polynomial has degree d = ϕ(n), where ϕ is the Euler totient function. Therefore ϕ(n) ≤ d. If tan θ is rational and sin θ is not, then sin θ has degree 2 over Q, so ζ has degree 2 over Q(i), so ζ has degree 4 over Q. The stated values of n for the cases d = 4 and d = 8 follow from the well-known formula for ϕ(n). That completes the proof of (ii) assuming (i). Corollary 1 If sin θ or cos θ is rational, and θ < π is a rational multiple of π, then θ is a multiple of 2π/n where n is 5, 4, 8, 10, or 12. Proof. Let ζ = cos θ + i sin θ = eiθ . Under the stated hypotheses, the degree of Q(ζ) over Q is 2 or 4. Hence, by the lemma, θ is a multiple of 2π/n, where n = 5, 8, 10, or 12 (if the degree is 4) or n = 3 or 6 (if the degree is 3). But the cases 3 and 6 are superfluous, since then θ is already a multiple of 2π/12.
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Quadratic and non-quadratic tilings
In this section we give a simple sufficient condition for a tiling to be quadratic. Lemma 4 Suppose tile T strictly tiles triangle ABC. If the tile T is similar to the triangle ABC, and there are no angle relations, then the tiling is a quadratic tiling. Proof. Note that since there are no angle relations, the three angles α, β, and γ are pairwise unequal: for example, if α = β, then the relation α + β + γ = π implies 2α + γ = π, which is an angle relation. Since T is similar to triangle ABC, and angle A is the smallest angle of ABC, angle A = α. Then consider the copy T1 of the tile that shares vertex A. Its two sides lie on the sides of triangle ABC. We can relabel the vertices B and C if necessary so that the angle of T1 at its vertex P1 on side AB is β, and its angle at its vertex Q1 on side AC is γ. There must be exactly three copies of the tile meeting at P1 , and the three angles at P1 are (in some order) α, β, and γ, because any other vertex behavior gives rise to an angle relation. Let the tiles meeting at P1 be T1 , T2 , and T3 , numbered so that T2 and T1 share a side. That shared side is a, since it is opposite angle A in T1 . Then T2 does not have angle α at P1 , since the α vertex of T2 has to be opposite side P1 Q1 . T2 does not have angle β at P1 , since T1 has angle β there, and only one β can occur at P1 . Therefore T2 has angle γ at P1 . Therefore T3 has angle α at P1 . Since the tiling is strict, the angle of T3 at its second vertex P2 on side AB must be β; otherwise the shared sides of T2 and T3 will have different length, since the length of that side of T2 is b. But now, we are in the same situation with T2 as we originally were with T1 : the two angles along side AB are α and β (in that order). We can argue as before that the three triangles T3 , T4 , and T5 meeting at P2 have angles β,γ, and α at P2 , in that order. Continuing down side AB in this fashion, we eventually reach a tile T2m−1 that has B for a vertex; there
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will be m copies of the tile sharing a side with AB; there will be m − 1 vertices P, . . . , Pm−1 along AB, each shared by three triangles; the number of tiles used is 2m − 1. The third vertices of these triangles are points Q1 , . . . , Qm−1 , lying on a line parallel to AB, and the last point Qm−1 lies on BC. The triangle Q1 CQm−1 is thus tiled by the restriction of the original tiling to that triangle. This restricted tiling is still strict and has no angle relations. By induction, we can assume that this restricted tiling is quadratic. Since it has m − 1 tiles along side Q1 Qm−1 , we have (m − 1)2 = N − (2m − 1). Then N = (m − 1)2 + 2m − 1 = m2 . That completes the proof. Remark. The 5-tiling in Figure 1 has T similar to ABC, but it has an angle relation 2α+2β = π, and it also has a non-strict vertex. It is natural to ask if the hypotheses of the lemma can be weakened by dropping one or the other of the hypotheses. Does there exist a strict non-quadratic tiling in which T is similar to ABC? (Angle relations are OK.) Does there exist a non-quadratic tiling with no angle relations in which T is similar to ABC? (Non-strict vertices are OK, but the angles meeting there would have to be exactly one each of α, β, and γ.) We do not know the answer to either of those questions. Note that if γ is a right angle we have an angle relation 2γ = π.
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The d-matrix, and a related eigenvalue problem
Let triangle ABC be tiled by the tile T , whose sides are a, b, and c. Let the sides of ABC be X, Y , and Z. We assume the triangle is labeled so that angles A, B, and C are listed in non-decreasing order; hence also X ≤ Y ≤ Z. In case triangle ABC is similar to the tile, this implies that angle A = α, angle B = β, and angle C = γ. Each side X, Y , and Z is a linear combination of a, b, and c, the coefficients specifying how many tiles share sides of length a, b, and c with X, Y , or Z. These nine numbers are the entries of the matrix d, such that 1 1 0 0 a X @ Y A = d@ b A c Z
If the triangle ABC is similar to the tile, then we have 1 0 1 0 X a √ @ Y A= N@ b A Z c √ because each side of ABC must be N times the corresponding side of the tile T , in order that the area of ABC can be N times the area of T . Therefore 1 0 1 0 a a √ d@ b A = N @ b A. c c √ That is, N is an eigenvalue of d, and (a, b, c) is an eigenvector for that eigenvalue. If triangle T is isosceles, then d is not (yet) uniquely defined. In that case we have either a = b or b = c; our convention is to ignore b, so that when T is isosceles, the middle column of the d matrix is zero. We will not make use of the d matrix when T is equilateral, but for completeness, we define the d matrix in that case to have non-zero entries only in the first column. If T is not isosceles, then the coefficients in the d matrix are integers between 0 and N − 1, inclusive, assuming N > 2: Not all N triangles can share a side of triangle ABC, since if N > 2, there would be two adjacent vertices along that side at which only two triangles meet; but then by Lemma 1, the copy of the tile between those vertices would have two right angles. For example, consider the 5-tiling shown in Figure 1. Here the shortest side of the large triangle consists of one c, so the top row of the d matrix is 0 0 1. The middle side of the large
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triangle consists of two c’s, so the middle row of the d matrix is 0 0 2. The longest side of the large triangles consists of one a and two b’s, so the bottom row is 1 2 0. Thus the d matrix for this example is 1 0 0 0 1 @ 0 0 2 A 1 2 0 and the eigenvalue equation is 0
0 @ 0 1
0 0 2
1 10 1 0 a 1 a √ 2 A@ b A = 5@ b A c 0 c
In √ this example we have α = π/6, β = π/3, and γ = π/2, so a = sin π/6 = 1/2, b = sin π/3 = 3/2, and c = sin π/2 = 1. One can check the eigenvalue equation numerically with these values. √ Note that the d matrix for a quadratic tiling is N times the identity. We conjecture that if N is a perfect square, say m2 , and d is m times the identity, then the tiling is quadratic.
7
Tilings with T similar to ABC
In this section, we assume triangle ABC is N -tiled by triangle T similar to ABC. In case N is a square, we have the quadratic tiling of ABC; in this section we assume N is not a square. Let the sides of T be a, b, and c, in non-decreasing order; these are opposite the angles α, β, and γ of T . We start by disposing of a special case. Lemma 5 Suppose T and ABC are both equilateral, and there is an N -tiling of ABC by T . Then N is a square and the tiling is a quadratic tiling. Proof. Since all the angles of T and ABC are equal, and all the sides of T are equal, there is only one way to place tile T1 at vertex B. Along side BC there must be a certain number m of copies of T ; hence the side of ABC is mc, where c is the side of X. We prove by induction on m that such a tiling is a quadratic tiling using m2 triangles. There are m tiles that share sides with BC. Call them T1 , . . . , Tm . This sawtooth-like configuration requires the placement of m − 1 copies of T , one between each adjacent pair of triangles T1 , . . . , Tm . Now we have identified a total of 2m − 1 triangles that participate in the original tiling, and the remaining triangles tile the smaller equilateral triangle formed by deleting the tiles identified so far from ABC. The base of this triangle is smaller than the original base BC by c, the side of T . By the induction hypothesis, the tiling of this triangle is quadratic, using (m − 1)2 tiles. Combining this with the row of 2m − 1 triangles along BC, we have a quadratic tiling with a total of (m − 1)2 + 2m − 1 = m2 tiles, completing the inductive proof. Next we review the computation of eigenvectors by cofactors. To find an eigenvector of the √ √ d matrix with eigenvalue N , consider the matrix X := d − N I. An eigenvector can be found by picking any row, and then arranging the cofactors of the elements of that row as a (column) vector. If these cofactors do not all vanish, then the result is an eigenvector. (The reader may either verify this or just check directly that the particular eigenvalues produced this way below are indeed eigenvectors.) Now we take up the general case of a tiling T with ABC similar to T , when N is not a square. Lemma 6 Let triangle ABC be N -tiled by tile T similar to ABC, and suppose N is not a square. Then the diagonal entries of the d matrix are zero.
11
Proof. Since √ the area of ABC is N times the area of T , and T is similar to ABC, the sides of ABC are N times a, b, and c. Then (as discussed in a previous section) we have the eigenvalue equation 1 0 1 0 a a √ d@ b A = N @ b A. c c
The characteristic polynomial f (x) of the d matrix, the determinant √ of d − xI, is a cubic polynomial with integer coefficients, yet has for a zero the number N . This is only possible if it factors into a quadratic factor and a linear factor. Since N is not a square, the quadratic factor must be a multiple of λ2 − N . The coefficient of x3 is −1, and so for some q we have f (x) = (x2 − N )(q − x) In general the coefficient of x2 in the characteristic polynomial of any 3 by 3 matrix d is the trace of d, and the constant term is the determinant of d. Hence q is the trace of d and −N q is the determinant of d. Since the entries of d are non-negative integers, the trace is non-negative, so q ≥ 0. To avoid so many subscripts, we use separate letters for the entries in the d-matrix, writing it as 1 0 p d e d=@ g m f A h ℓ r √ √ Since the similarity factor between ABC and T is N , there √ N tiles √ cannot be more than with a sides along X, the short side of ABC. That N . More formally, a N = X = √ √ is, p ≤ √ pa + db + ec ≥ pa, so p ≤ √ N . Similarly √ m ≤ N √and r ≤ N . Since N is not rational, we have strict inequalities: p < N , r < N , and r < N . It follows that pm < N , etc. We also note that there is just one tile sharing vertex A, where ABC has its α angle. That tile must have its b and c sides along AB and AC, or along AC and AB, we don’t know which. Thus either f and ℓ are nonzero, or m and r are nonzero. Suppose, for proof by contradiction, that q, the trace of d, is not zero. √ Then q = p+m+r > 0. Since the three eigenvalues are distinct (because q is rational and N is not), the eigenspace √ corresponding to N is one-dimensional. The eigenvalue equation is 1 0 u (d − λI) @ v A = 0 w or showing the coefficients 0
√ p− N @ g h
d√ m− N ℓ
1 10 e u f√ A @ v A = 0 w r− N
√ We claim that there exists an eigenvector (u, v, w) whose components lie in Q( N ). To prove this we will use the cofactor method described above. The resulting eigenvector is (u, v, w), provided all three components are nonzero, where ˛ ˛ ˛ d√ e ˛˛ ˛ u = ˛ m− N f ˛ ˛ ˛ √ ˛ p− N e ˛ ˛ v = −˛˛ g f ˛ √ ˛ ˛ ˛ p− N d√ ˛˛ w = ˛˛ g m− N ˛
12
Although we have not given a proof of the cofactor method’s correctness, one can easily verify √ directly that the exhibited vector is indeed an eigenvector for N ; this also provides a check that no algebraic mistake has been made. The fact that all three cofactors are nonzero is really only needed to conclude directly that the eigenspace √ of (u, √ v, w) is one-dimensional; but we know that directly in our case since the eigenvalues N , − N , and q are distinct. It therefore suffices to check that one of the cofactors u, v, w is nonzero; then the others must automatically be nonzero because (u, v, w) is a nonzero multiple of (a, b, c). But we give direct proofs that all three cofactors are nonzero anyway, as it takes only one more paragraph. √ We have u = df − em + e N . If u = 0 then e = 0 and hence df = 0. If v = 0 then similarly f = 0 and eg = 0. Finally if w = 0 then p + m = 0 and hence p = m = 0, so N = dg. Assume, for proof by contradiction, that w = 0 . Then √ √ (m − N )(p − N ) = dg √ mp + N − N (p + m) = dg √ Since N is irrational this means p + m = 0, and√since p and m are nonnegative, that implies p = 0 and m =√0. Hence dg = N . But d ≤ (a/b) N , with equality implying that p = d = 0, and g ≤ (b/a) N with equality implying m = f = 0.√ Since dg = N , equality must hold in √ both inequalities. Hence d = (a/b) N and g = (b/a) N and p = e = m = f = 0. But we showed above that either m and r are both nonzero, or f and ℓ are both nonzero. That is now contradicted by m = f = 0. This contradiction shows that w 6= 0. Next we give the proof that u 6= 0; as remarked above, this is technically superfluous, but still it is interesting because the proof we give is not simply an abstract argument about projecting onto the one-dimensional eigenspace. Assume, for proof by contradiction, that u = 0. Then √ √ df − em + e N = 0. Since N is not rational, e = 0 and df − em = 0. Then df = 0, so d = 0 or f = 0. If d = 0, then √ since both d and√e are zero, side X of triangle ABC is composed of all a sides √ and X = (p − N√ )a. But√since N is√the similarity factor between T and ABC, we have X = N a. Hence p − N = N . Hence N = p/2, so N 2 = p2 /4. Hence 4N 2 = p2 and p is even, so N is a square, contradiction. This contradiction proves d 6= 0. Since d = 0 or f = 0, we have f = 0. Now assume, for proof by contradiction, that g √ = 0. Then√since f = 0, side Y is composed entirely of b sides of tiles, so Y√ = mb. But Y = b N since N is the similarity factor between T and ABC. Hence m = N , contradiction. That proves g 6= 0. As shown above, either f and ℓ are both nonzero or m and r are both nonzero. But f = 0. Hence both m and r are nonzero. Now X
=
X √ ( N − p)a
= =
b a Y
= =
Y b a
= =
b a
=
Cross-multiplying we have
pa + db √ Na db √
N −p d √ Nb ga + mb since f = 0 g √ N −m √ g N −p = √ d N −m
√ N − (m + p) N + mp
dg
=
m+p
=
0
m
=
p = 0
since e = 0
as the coefficient of
√
N must be zero
as m and p are nonnegative
13
But m was proved above to be nonzero. This contradiction completes the proof that u 6= 0. Now assume v = 0. Then √ pf − eg + f N = 0
Since N is irrational we have f = 0 and pf = eg, but since f = 0 we have eg = 0. Hence either e = 0 or g = 0. Assume, for proof by contradiction, that g = 0. side of ABC √ √ Then the middle (corresponding to the middle row) is equal to mb but also to b N , so m = N , contradiction. This contradiction proves g 6= 0. Hence e = 0. Since either f and ℓ are both nonzero or m and r are both nonzero, and we have proved f = 0, then m and r are both nonzero. Now that we have e = 0 = f , and m 6= 0, we reach a contradiction by the same computation as in the case u = 0, shown in the series of displayed equations above. Hence v 6= 0. Thus none of the three cofactors√ is zero. That completes the √ proof that there is an eigenvector (u, v, w) for the eigenvalue N with components in Q( N ). Since the eigenspace is one-dimensional, this eigenvector is a (not necessarily rational) multiple of (a, b, c). Recall that the third eigenvalue of the d matrix is the trace q = p + m + r. We can use the cofactor method to find an eigenvector for this eigenvalue as well, namely ˛! ˛ ˛ ˛ ˛ ˛ ˛ ˛ d d e ˛˛ ˛˛ p − q e ˛˛ ˛˛ p − q ˛ ˛ , V = ˛ m − q f ˛, −˛ g m−q ˛ f ˛ ˛ g 1 0 df − em + e(p + m + r) A = @ −pf + eg + f (p + m + r) pm − dg − (m + p)(m + p + r) + (m + p + r)2 1 0 df + e(p + r) A = @ eg + f (m + r) pm − dg + r(m + p + r)
Technically, it is not an eigenvector until we prove that the components are not zero, but we do not need that right now; it suffices that it satisfy the eigenvalue equation. The eigenvalue equation dV = (p + m + r)V is 1 1 10 0 0 df − e(p + r) df − e(p + r) p d e A = (p + m + r) @ eg + f (m + r) A(2) @ g m f A @ eg + f (m + r) pm − dg + r(m + p + r) pm − dg + r(m + p + r) h ℓ r
The first component of this vector equation is
p(df − e(p + r)) + d(eg + f (m + r)) + e(pm − dg + r(m + p + r)
=
(p + m + r)(df − e(p + r))
Multiplying out and cancelling like terms, and dividing by 2, we find epm + er(m + p + r) = 0. We argue by cases, according to whether e = 0 or not. We first take up the case that e 6= 0. Then pm + r(m + p + r) = 0. Since these terms are nonnegative, they are both zero. Hence pm = 0 and r(m + p + r) = 0. Hence r = 0 or m + p + r = 0. In either case r = 0. Writing out the third component of the eigenvalue equation, and setting r = 0, we have h(df − ep) + ℓ(eg + f m)
=
hdf − hep + ℓeg + ℓf m
=
h(df − ep) + ℓ(eg + f m)
=
(p + m)(pm − dg) −(p + m)dg
−pdg − mdg
14
since pm = 0
Now we write out the second component of the eigenvalue equation (2), setting r = 0: g(df − ep) + m(eg + f m) + f (pm − dg)
=
(p + m)(eg + f m)
gdf − gep + m(eg + f m) + f pm − f dg
=
peg + pf m + m(eg + f m)
peg
=
0
pg
=
0
since e 6= 0
Assume, for proof by contradiction, that m 6= 0. Then since mp = 0 we have p = 0. The equation N (p + m) = hdg + ℓpf + ℓeg becomes N m = hdg + ℓeg. The third component of the eigenvalue equation becomes, with p = 0, hdf + ℓeg + ℓf m
=
−mdg
The left side is ≥ 0 and the right side is ≤ 0. Hence both sides are equal to zero. Since m 6= 0 and e 6= 0, we have dg = 0 and hdf = 0 and ℓg = 0 and ℓf = 0. We derived above (by observing that the b and c sides of the tile at vertex A lie on the two adjacent sides of ABC) that either f and ℓ are both nonzero or m and r are both nonzero. Since r = 0 we must have f and ℓ both nonzero. Hence ℓf = 0 is a contradiction. That contradiction completes the proof that m = 0. Now assume, for proof by contradiction, that p 6= 0. Then since pg = 0 we have g = 0. Then the equation N (p + m) = hdg + ℓpf + ℓeg becomes N p = ℓpf . Canceling p we have N = ℓf . But as proved above, ℓf ≤ N , and equality holds if and only if AC is composed only of c sides of tiles and AB is composed only of b sides. Therefore we have h = 0 as well as g = m = r = 0. Then since h = 0 and r = 0, the long side AB of ABC is composed entirely of b sides of tiles. If T is isosceles, then by convention the middle column of the d-matrix is zero. Since ℓf = N , we now have ℓ 6= 0, so the middle column is not zero, and T is not isoceles. Since side AB is composed entirely of b sides of tiles, there are equally spaced vertices V0 = A, V1 , . . . , Vℓ , spaced b apart, each one of which is one side of a tile Ti . Tile T1 , which has vertices at A and V1 , has its α angle at V0 . All these tiles have their β angles in the interior of ABC, and their α and γ angles at the Vi . If γ > π/2 then there is only one possible orientation for these tiles, as two γ angles will not fit at any Vi . In that case the angle of the last tile at vertex B must be γ, contradiction, since the angle there cannot exceed β, and β 6= γ since then T would be isosceles. Hence γ ≤ π/2. In particular, the tile that shares vertex B has its b side along AB. Therefore the tile sharing vertex B and part of side AB has its α angle at B, and the angle β at vertex B splits into some number of α angles, so for some number J, we have β = Jα. Somewhere along AB there must occur a vertex Vk at which both the tile Tk and the tile Tk+1 have angle γ. There is not room at Vk for a third tile, since 2γ + α > α + β + γ = π. Hence there are exactly those two tiles at Vk , and we have γ = π/2. Since γ is a right angle, we must have a2 + b2 = c2 . Since (u, v, w) is a multiple of (a, b, c) we also have u2 + v 2 = w2 . We now compute these expressions from the formulas for (u, v, w). In view of m = g = 0 we have √ u = ef − e N √ v = f N − fp √ w = N −p N Squaring these equations we have u2
=
v2
=
2
=
w
√ e2 (f 2 + N − 2f N ) √ f 2 (N + p2 − 2p N ) √ N 2 + p2 N − 2pN N
15
Setting u2 + v 2 = w2 we find √ e2 f 2 + e2 N + f 2 N + f 2 p2 − 2(e2 f − f 2 p) N Equating the coefficients of
√
=
√ N 2 + p2 N − 2pN N
N and equating the rational parts, we have
2 2
e2 f − f 2 p
2
2
2 2
e f +e N +f N +f p
=
pN
=
N 2 + p2 N
Since γ is a right angle, α + β = π/2. Since β = Jα, we have α =√π/(2(J + 1)), so α is a rational multiple of 2. We have tan α = b/a = v/u, which belongs to Q( N ). We have u u2 + v 2 √ √ which is also in Q(√ N ). Similarly sin α belongs to Q( N ). Then ζ = diα is of degree 4 over Q, since Q(ζ) = Q(i, N ). By Lemma 3, 4(J + 1) is 5, 8, 10, or 12. Since 5 and 10 are not divisible by 4, we have 4(J + 1) = 8 or 10. But if 4(J + 1) = 8 then J = 1, while we have J ≥ 2 since β = Jα. The only remaining possibility is 4(J + 1) =√12, which makes J = 2. Then α = π/6 and 2β = α, so β = π/3. Then a = sin α = 1/2, b = 3/2 and c = 1. But now AC = f c = f , and cos α =
ℓb
AB 2 √ AC 3 2 √ fc 3
= = =
Now we put in c = 1 and b =
√
3/2: ℓ
√
3 2 3ℓ
= =
2 √ f 3 4f
We have N = ℓf = (4/3)f 2 , so 3N = 4f 2 , so f is divisible by 3, say f = 3k; then N = 3(2k)2 is three times a square.2 It remains to show that p = 0; in fact we claim p = 0 and e = 0, so side AC is also composed entirely of b sides of triangles. We have BC
= = = =
1 AB 2 1 ℓb 2 √ ℓ 3 4 pa + db + ec
√
Now we put in the values a = 1/2, b = 3/2, and c = 1. √ √ p 3 ℓ 3 = +d +e 4 2 2 √ This is an equation in Q( 3). Equating the rational parts we have 0 = p/2 + e. Since both p/2 and e are nonnegative, we have p = 0 and e = 0, as claimed. In particular p = 0 so the diagonal 2 Tilings of this kind actually exist, but with e = 0 (we are now in the case e 6= 0). (In the next draft I will add some figures to illustrate it.)
16
elements are nonzero, which is the conclusion of the theorem; or we could say, in particular e = 0, contradicting the assumption e 6= 0 and completing the analysis of that case. Therefore we may now assume e = 0. Remember that r = 0 was derived only under the assumption e 6= 0, so the equation r = 0 is no longer in force. The third component of the eigenvalue equation (2) is (substituting e = 0) hdf + ℓf (m + r) + r(pm − dg + r(m + p + r))
=
(p + m + r)(pm + r(m + p + r))
2
Subtracting r (m + p + r) from both sides we have hdf + ℓf m + ℓf r + rpm − rdg
hdf + ℓf m + ℓf r + rpm − rdg
=
(p + m + r)pm + (p + m)r(m + p + r)
=
(p + m + r)(pm + pr + mr)
(3)
To get rid of h and ℓ, we expand the determinant of the d matrix by cofactors on the bottom row. That determinant is −N q = −N (p + m + r), so we have (remembering e = 0) ˛ ˛ ˛ ˛ ˛ ˛ ˛ p d ˛ ˛ p e ˛ ˛ p d ˛ ˛ ˛ ˛ ˛ ˛ ˛ + r˛ − ℓ˛ −N (p + m + r) = h˛ g m ˛ g f ˛ g m ˛ =
hdf + ℓpf + rpm − rdg
Adding and subtracting ℓpf to the left side of (3) the expression for the determinant appears, and we have hdf + ℓpf + rpm − rdg + ℓf m + ℓf r − ℓpf −N (p + m + r) + ℓf m + ℓf r − ℓpf
=
(p + m + r)(pm + pr + mr)
=
(p + m + r)(pm + pr + mr)
Moving everything to the right side we have (p + m + r)(pm + pr + mr) + (N − ℓf )(m + r) + (N + ℓf )p √ √ Since ℓ is√the number of b sides of tiles on the long side N c of ABC, we have√ ℓb ≤ N c, or ℓ ≤ (c/b) N . Since f is√the number of c sides of tiles on AC, whose length is N b, we have √ f c ≤ N b, or f ≤ (b/c) N . Hence ! ! c√ b√ N N ℓf ≤ b c 0
=
ℓf
≤
N
Hence all the terms on the right of the previous equation are nonnegative. Hence each of them is zero. In particular (N + ℓf )p = 0; but N + ℓf > 0, so p = 0. Then then equation becomes (m + r)mr + (N − ℓf )(m + r) = 0 If m + r = 0 then m = 0 = r and the lemma is proved. Hence we may assume mr √ = 0 and N = ℓf . But √ if N = ℓf then we must have equality in the two inequalities ℓ ≤ (c/b) N and f ≤ (b/c) N . This implies that side AC is composed only of c sides of tiles and side AB is composed only of b sides of tiles, so g = m = h = r = 0. In particular m = r = 0. That completes the proof of the lemma. We pause to observe that the d matrix for a the form 0 0 d = @ 0 n
17
biquadratic tiling, in case N = m2 + n2 , has 0 0 m
1 n m A 0
which does satisfy the conditions above (as it must). The hypothesis that N is not a square is necessary, as shown by the 9-tiling in Figure 7. Its d matrix is 1 0 1 1 0 @ 2 2 0 A 0 0 3
and as predicted, the determinant is zero, but the trace is not zero, and the characteristic polynomial is −x(x − 3)2 . Continuing with the general case of N not a square, some further conclusions can be drawn about the d matrix. We have shown that p = m = r = 0. The determinant is then given by det d = df h + egℓ
Since the matrix entries are nonnegative, that means that each of these two terms must contain a zero factor. In particular, at most four entries in the d matrix are nonzero. The negated coefficient of λ in the characteristic equation is (since the diagonal elements are zero) the sum of paired products of off-diagonal elements: N
=
dg + eh + f l
But at least one of these three terms will be zero, as shown above. In view of the lemma, the d matrix becomes 0 1 0 d e d=@ g 0 f A h ℓ 0
(4)
(5)
and the matrix equation
1 0 1 a a √ d@ b A = N @ b A c c 0
becomes the three equations
db + ec
=
ga + f c
=
ha + ℓb
=
√
√
√
Na Nb Nc
Lemma 7 Suppose ABC is N -tiled by tile T similar to ABC, and N is not a square. Then γ is a right angle. Proof. First we note that T and ABC are not equilateral, by Lemma 5. Next we will prove that T and ABC are not isosceles with β = γ. Assume, for proof by contradiction, that β = γ. Then, by our definition of the d matrix, the middle column of the d matrix is zero, i.e. b is counted as c. Then we have d = ℓ = 0 and 1 0 0 0 e (6) d=@ g 0 f A h 0 0
That implies that the short side BC of triangle ABC has only c sides of tiles on it, and the long side AB has only a sides of tiles on it. At the vertex A, there can only be one tile, since the angle at A is the smallest angle α so there can be no vertex splitting. This tile has one side of length a opposite angle A and another along side AB. Hence a = b. Since T is not equilateral, we must have b < c and β < γ. This contradicts the assumption that β is not less than γ, and thus completes the proof by contradiction that β < γ.
18
Since the d matrix has zeroes on the diagonal, no c sides of tiles occur along the longest side AB of triangle ABC; only a and b sides occur there. There are ℓ + h tiles along AB; for simplicity of notation, let k := ℓ + h and number those tiles T1 , . . . , Tk starting at vertex A. Let V1 , . . . , Vk−1 be the vertices of those tiles on AB. Tile T1 must have its α angle at vertex A, and since its c side is not on AB, it must have its γ angle at vertex V1 . Tile Tk cannot have a γ angle at B, since β < γ. Hence it has either its α or its β angle at B, and its γ angle at Vk−1 (since its c side does not lie on AB). Continuing towards C from B with T2 , T3 , . . ., and continuing towards B from C similarly, we must encounter an index j between 1 and k − 1 such that Tj and Tj+1 both have their γ angles at Vj . At that point we know γ ≤ π/2. Assume, for proof by contradiction, that γ 6= π/2. Then there is at least one additional copy T ′ of the tile between Tj and Tj+1 , sharing vertex Vj , by Lemma 1. If T ′ has its γ angle at Vj then there are exactly those three tiles meeting at Vj (else γ < π/3) and we have γ = π/3, and hence T is equilateral, which as noted above is impossible. Hence none of the additional tiles T ′ meeting at Vj have a γ angle at Vj . None of the tiles T ′ can contribute a β angle at Vj either, since 2γ + β > π. Hence there is an angle relation 2γ + pα = π, where p additional tiles contribute α each to the angle sum at Vj , and p > 0. But 2γ + pα > γ + β + α = π, since γ > β. This contradiction completes the proof of the lemma. Lemma 8 Suppose ABC is N -tiled by tile T similar to ABC, and N is not a square. Then f = d12 is not zero. Proof. Suppose, for proof by contradiction, that f = 0. Then the middle row of the d matrix is (g, 0, 0), which means that all the tiles along side AC of triangle ABC share their a sides with AC. At vertex A, where ABC has its smallest angle α, there is exactly one tile T1 , with angle α at a. Hence both the side of T1 opposite that angle, and the side shared with AC, are equal to a. Thus T is isosceles. In that case, by convention we have agreed to write the d matrix with zeroes in the second column, so the d matrix has the form 1 0 0 0 e d=@ g 0 0 A (7) h 0 0
Now the bottom row is (h, 0, 0), which means that all the tiles along side AB share their a sides with AB. In particular the tile at vertex A has an a side along AB. But we have already seen that its other two sides are a. Hence the tile is equilateral, contradicting Lemma 5, since N is not a square. That completes the proof. There are six letters for coefficients in the d matrix, but for any specific tiling, at most four of those coefficients are nonzero. We will analyze some special cases. The case corresponding to the biquadratic tilings is d = 0 and g = 0. We call that the “biquadratic case”. In the biquadratic case the d matrix has the form 1 0 0 0 e (8) d=@ 0 0 f A h ℓ 0
Equation (4) now becomes
eh + ℓf
=
N
(9)
We compute the eigenvector in the biquadratic case, using the cofactor method described above. Let 0 √ 1 − N 0 e √ A X=@ 0 − N f √ h ℓ − N
19
Taking the cofactors of the bottom row (notice the minus sign in the second component, which comes from the definition of “cofactor”) we find the eigenvector 0 √ 1 ˛! ˛ ˛ √ ˛ ˛ ˛ √ e √N ˛ ˛ ˛ ˛ − N e ˛ ˛ − N 0 e 0 ˛, ˛ ˛ √ ˛ ˛ √ ˛ = @ f N A ˛ − N f ˛, −˛ 0 f ˛ ˛ 0 − N ˛ N
Note that e 6= 0 and f 6= 0, since the first two sides of ABC are given by ec and f c. Hence the cofactors do not vanish. √ √ √ We claim that the bottom two rows of d − N I, namely (0, − N , f ) and (h, ℓ, −√ N ), are linearly √ independent. Indeed, suppose that for some constants p and q we have p(0, − N , f ) + q(h, ℓ, − N ) = 0. From the first component we√see that qh = 0. From the third component √ we see that pf = q N . If q is not zero, then N = pf /q, √ contradicting the irrationality of √ N . Hence q = 0. Hence from the √ 0. This proves √ second component, p N = 0. Hence p = that the bottom two rows of d − N I are linearly independent. Hence d − N I has rank 2; √ hence the eigenspace associated with the eigenvalue N is one-dimensional. It follows that the eigenvector computed above is a multiple of (a, b, c). That is, for some constant µ we have 1 0 0 √ 1 a e √N @ b A = µ@ f N A (10) c N
The constant µ is an arbitrary scale factor; changing µ just changes the size of the tile T and the triangle ABC by the√same factor. We are therefore free to choose µ to suit our convenience. We choose to take µ = N ; then we have 1 0 1 0 e a A @ b A = @ f (11) √ c N Lemma 9 Let triangle ABC be N -tiled by T , and suppose N is not a square and T is similar to ABC, and d = g = 0 (the biquadratic case). Then N is a sum of squares, specifically N = e2 +f 2 where e and f are as above, and tan α = e/f . In particular tan α is rational.
Proof. By Lemma 7, we have γ = π/2. By the Pythagorean theorem and (11) we see that γ = π/2 if and only if e2 + f 2 = N . Since γ = π/2, we have tan α = a/b, and by (10), tan α = e/f . That completes the proof of the lemma. Lemma 10 Suppose ABC is N -tiled by tile T similar to ABC, and N is not a square, and d = g = 0 (the biquadratic case). Then the right angle of ABC is split by the tiling, and the tangents of the other angles of ABC are rational. Proof. We suppose, as always, that the γ angle of ABC is at C, the β angle at B, and the α angle at A. Since the d matrix has the form given in (8), all the tiles along side BC share their c sides with BC (there are e of them) and all the tiles along side AC share their c sides with AC (there are f of them). Suppose, for proof by contradiction, that the vertex at C is not split. Then a single tile shares vertex √ C, so the tile has two c sides, and hence is isosceles with b = c. But by (11), we have c/b = N /e. Hence if b = c we have N = e2 , contradicting the hypothesis that N is not a square. Hence the vertex C is split as claimed. The tangents of the other two angles are e/f and f /e, which are rational. This completes the proof of the lemma. We now turn to another important case, when e = 0. We call this the “triple-square case”, because it will turn out that in this case N must be three times a square. The following lemma and its proof give a complete analysis of this case. Lemma 11 Suppose ABC is not equilateral and is N -tiled by tile T similar to ABC, and N is not a square, and e = 0 (the triple-square case). Then α = π/6, β = π/3, and N = 3d2 is three times a square.
20
Remark. There do exist tilings for each N of the form 3d2 that fall under the triple square case, as we showed in Figures 9 and 10. Proof. Under the hypotheses of the lemma we have 1 0 0 d 0 d = @ g 0 f A. h ℓ 0
In this matrix, d and h + ℓ are not zero, since they represent the number of tiles along BC and AB, respectively. By Lemma 8 we have f 6= 0. We have 1 0 √ d 0 − N √ √ A X = d − NI = @ g − N f √ h ℓ − N We will prove that the bottom two rows of X are linearly independent. If they are linearly dependent, then for some p and q, we have √ √ 0 = p(g, − N , f ) + q(h, ℓ, N ) √ = pg + pf + qℓ + qh + N (q − p) √ Since N is not a square, the coefficient of N is zero, so q = p, and 0
=
pg + pf + qℓ + qh
=
pg + pf + pℓ + ph
=
p(g + f + ℓ + h)
Since the entries of the d matrix are non-negative, and h + ℓ is strictly positive, we conclude p = q = 0. That proves that√the bottom two rows of X are linearly independent, so X has rank 2 and the eigenspace of N is one-dimensional. We then compute the eigenvector by the cofactor method. Taking the cofactors of the bottom row, we find the eigenvector 0 1 ˛ ˛ √ ˛! ˛ ˛ ˛ √ df√ ˛ ˛ ˛ ˛ − N 0 ˛ ˛ − N 0 e d ˛, ˛ ˛ √ ˛ ˛ A √ ˛ = @ f N ˛ − N f ˛, −˛ g f ˛ ˛ g − N ˛ N − dg
Since (a, b, c) is an eigenvector and the eigenspace is one dimensional, (a, b, c) is a multiple of this computed eigenvector. By scaling the triangle appropriately we can assume (a, b, c) is actually equal to the computed eigenvector: 1 0 0 1 df√ a @ b A=@ f N A c N − dg √ It follows that sin α = a/c = f d/(N − dg) is rational and tan α = a/b = d/ N is of degree 2 over Q. According to the first row of the d matrix, the tiles along BC have only b sides on BC. Assume, for proof by contradiction, that vertex B does not split. Then there is a single tile T1 at vertex B, which therefore shares one side with AB and one side with BC. Triangle T is not isosceles, since then by definition the d matrix would have zeroes in the middle column. Hence the unique b side of T1 must be opposite angle B; but it must also lie on BC, which is a contradiction. Hence vertex B does split. Therefore for some integer P we have β = P α. Since by Lemma 7, γ = π/2,we have π = α+β 2 = α + Pα =
(P + 1)α
21
Therefore α=
π 2π = . 2(P + 1) 4(P + 1)
By Lemma 3 we conclude that 4(P + 1) is one of the numbers n = 3, 4, 5, 8, 10, or 12 for which φ(n) = 4. Of these numbers, only 4, 1, and 12 are divisible by 4, which implies P = 2, since the values P = 0 and P = 1 do not correspond to vertex splitting. √ Hence P√= 2 and √ we have β = 2α, so α + β = π/2 = 3α, and α = π/6. Hence tan α = df /(f N ) = d/ N = 1 3. Hence N = 3d2 . That completes the proof of the lemma. Now we have dealt with the biquadratic case (when d = g = 0) and the triple-square case (when e = 0). It remains to show that these are the only two possible cases, when N is not a square and T is similar to ABC. Recall that df h = 0 and egℓ = 0 since det d = 0; that leaves only a few possibilities to consider. We begin by showing that if d = 0 then we are already in the biquadratic case. Lemma 12 Assume triangle ABC is N -tiled by T , that N is not a square, that T is similar to ABC, and that d and g are two entries in the d matrix of the tiling, in the notation used above (the ones that are zero in the biquadratic case). Then d = 0 implies g = 0, i.e. we are in the biquadratic case as soon as d = 0. Proof. We have X =d−
√
√ − N NI = @ g h 0
d √
− N l
1 e A f √ − N
By the cofactor method described above we compute the eigenvector √ 1 1 0 0 a df√+ e N @ b A = @ f N + eg A c N − dg
Suppose, for proof by contradiction, that d = 0 and g 6= 0. Then 1 0 √ 0 1 a e √N @ b A = @ f N + eg A c N
By Lemma 7, γ is a right angle, so by the Pythagorean theorem, we have a2 + b2 = c2 . That is, √ e2 N + (f N + eg)2 = N 2 √ e2 N + f 2 N + 2egf N + e2 g 2 = N 2 √ Since N is not a square, the coefficient of N is zero; that is, egf = 0. By hypothesis, g 6= 0, so ef = 0. The first row of the d matrix is (0, d, e) = (0, 0, e), so e 6= 0 because there must be some triangles on the first side of ABC. Therefore f = 0. Then the d matrix is 1 0 0 0 e @ g 0 0 A h ℓ 0
Hence all the tiles on the middle side AC of ABC have their a side on AC, and all the tiles on the hypotenuse AB do not have their c side on AB. Consider the tile T1 sharing vertex A (there is only one, since ABC has angle α there). It has its a side on AC and does not have its c side on AB. Hence its b side is on AB and its c side opposite angle A, which is α. Hence a = c and triangles T and ABC are equilateral, which is a contradiction since γ = π/2. This contradiction shows that the assumption d = 0 and g 6= 0 is untenable, which completes the proof of the lemma.
22
Lemma 13 Assume triangle ABC is N -tiled by T , that N is not a square, that T is similar to ABC, and that d 6= 0. Then e = 0, i.e. we are in the triple-square case as soon as d 6= 0. Proof. We have as in the proof of the previous lemma 1 0 0 d e d=@ g 0 f A h ℓ 0 0 √ 1 − N d e √ √ A X = d − NI = @ g − N f √ h ℓ − N √ 1 0 0 1 a df + e √N @ b A = @ eg + f N A c N − dg
By Lemma 7 and the Pythagorean theorem we have 0
= = =
c2 − a2 − b2
√ √ (N − dg)2 − (df + e N )2 − (eg + f N )2 √ −2(def + egf ) N + rational
Since N is not a square and the entries of the d matrix are nonnegative integers, we have def = 0 and egf = 0. Since f 6= 0 by Lemma 8, and d 6= 0 by hypothesis, we have e = 0. That completes the proof of the lemma. The following theorem completely answers the question, “for which N does there exist an N -tiling in which the tile is similar to the tiled triangle?” Theorem 1 Suppose ABC is N -tiled by tile T similar to ABC. Then either N is a square, or a sum of two squares, or three times a square. Proof. Suppose N is not a square. Then by Lemma 7, γ is a right angle. Now consider the d matrix. By Lemma 6 the diagonal entries are zero, so as stated in (5) the d matrix has the form 1 0 0 d e d=@ g 0 f A h ℓ 0
By Lemma 12, if d = 0 then also g = 0, i.e. we are in the “biquadratic case”. Then by Lemma 9, N is a sum of squares. If e = 0 then by Lemma 11, N is three times a square and T is a 30-60-90 triangle. Finally, Lemma 13 shows that the cases d = 0 and e = 0 are exhaustive. That completes the proof of the theorem. As for characterizing all N -tilings with the tile similar to the tiled triangle, we have only the following partial results: Theorem 2 Suppose ABC is N -tiled by tile T similar to ABC. If N is not a square, then T and ABC are right triangles. Then either N is three times a square then T is a 30-60-90 triangle, or N is a sum of squares, the right angle of ABC is split by the tiling, and the acute angles of ABC have rational tangents, and these two alternatives are mutually exclusive. Proof. First we observe that N cannot be both a sum of squares and three times a square, since the equation x2 + y 2 = 3z 2 has no integer solutions. To see that, we can assume without loss of generality that x, y, and z are not all even. Note that squares are always congruent to 0 or 1 mod 4, so the left side is 0, 1, or 2 mod 4. Then z 2 must be congruent to 0 mod 4, since if not, the right side is congruent to 3 mod 4. Hence z is even. But x and y must also be even to make the left side congruent to 0 mod 4, contradiction. Hence the equation has no solutions. Thus
23
the alternatives in the theorem are mutually exclusive, as claimed. The rest of the theorem follows from Lemma 7 and Lemma 11. Note that the 9-tiling in Figure 7 shows that not every m2 -tiling is a quadratic tiling, so we have not even classified all the m2 -tilings. Briefly we conjectured that a tiling in which the d-matrix is m times the identity should be a quadratic tiling, but even that is not true. One can extend the 9-tiling in Figure 7 by adding more triangles to the right and below, producing a 25-tiling in which the d-matrix is 5 times the identity.
8
Tilings of an isoceles triangle by a right triangle
Let us first review the known examples of tilings of an isosceles triangle ABC by a right triangle T . There is always the “double quadratic” tiling, in which one divides ABC into two halves by the altitude and then quadratically tiles each half. One might conjecture that any tiling contains the altitude, i.e. is a subtiling of the 2-tiling defined by the altitude. But this is easily refuted: consider the second tiling in Fig. 6. In that case ABC is equilateral; take the base BC as one of the slanting sides in Fig. 6. This is not a very satisfying counterexample since the triangle can be rotated so that, with a different vertex and base, the tiling does contain the altitude. If the triangle ABC is required to be not equilateral (so there is a unique choice of base), we have no counterexample. If we could prove that the altitude must be included when ABC is not equilateral, the theorem of this section would be a consequence of our other results. In fact, we have no counterexample to the conjecture that, whether ABC is equilateral or just isosceles, if the vertex angle A splits then the altitude AM is included in the tiling, i.e. does not meet the interior of any tile. So far are we from a complete classification of tilings that we cannot even prove that the number of tiles required to tile an isosceles ABC using a triangle similar to half of ABC must be even, although no tiling with N odd is known. (Figure 10 shows a tiling with N odd, but the tile is not similar to half of ABC.) Whenever there is an N -tiling of the right triangle ABM , there is a 2N -tiling of the isosoceles triangle ABC. Using the biquadratic tilings (see Figs. 4 and 5) and the triquadratic tilings (see Figs. 8 and 9) we can produce 2N -tilings when N is a sum of squares or three times a sum of squares. We call these tilings “double biquadratic” and “double triquadratic”. For example, one has two 10-tilings and two 26-tilings, obtained by reflecting Figs. 4 and 5 about either of the sides of the triangles shown in those figures; and one has 24-tilings and 54-tilings obtained from Figs. 8 and 9. Note that in the latter two cases, ABC is equilateral. In the case when the sides of the tile T form a Pythogorean triple n2 + m2 + k2 = N/2, then we can tile one half ABM with a quadratic tiling and the other half with a biquadratic tiling. The smallest example is when the tile has sides 3, 4, and 5, and N = 50. One half is 25-tiled quadratically, and the other half is divided into two smaller right triangles which are 9-tiled and 16-tiled quadratically. This shows that the tiling of ABC does not have to be symmetric about the altitude AM . One looks for other ways to tile the two halves of ABC differently. We cannot use two different biquadratic tilings, even if N/2 can be written as a sum of squares in two different ways, because the tiles would have to be different shapes. We cannot use a biquadratic tiling and a triquadratic tiling, since m2 + n2 = 3k2 has no solutions. The quadratic-biquadratic tilings corresponding to Pythagorean triples are the only known examples. In this section, our normal convention that A is the smallest angle of ABC is temporarily suspended. Instead, A is the vertex of the isosceles triangle and BC is the base. Similarly, we also suspend the convention that α < β. The following lemma shows that the base angles of ABC are either α or β; we will make the convention for this section that β is the base angle. The area of triangle ABC must be N times the area of the tile T . By hypothesis, the triangle T is similar to half the triangle ABC; the similarity q factor is the square root of the ratio of the half the area of ABC to the area of T , namely
24
N . 2
The sides of T are a = sin α, b = cos α,
and c = 1. Let AB be the length of AB. Since AB is opposite the right angle of one of the halves of ABC that is similar to T , we have r r N N AB = sin γ = since γ = π/2 2 2 r BC N = sin α 2 2 On the other hand, AB and BC must be integer linear combinations of a, b, and c. So we have, q for some non-negative integers p, q, and r, that AB = pa + qb + rc. Since c = 1 and AB = N2 , we have r N (12) p sin α + q cos α + r = pa + qb + r = 2 The notation in this equation will be used throughout this section. Lemma 14 Let ABC be an isosceles triangle with base BC, N -tiled by a right triangle q T similar to half of ABC. Suppose N/2 is not a rational square. Then a and b belong to Q(
N 2
).
Proof. Squaring both sides and simplifying as a polynomial in a we have (pa + qb + r)2
=
N 2
N 2
=
0
p2 a2 + 2pa(qb + r) + (qb + r)2 −
Since a = sin α and b = cos α we have a2 + b2 − 1 = 0. Replacing a2 by 1 − b2 we find 2ap(qb + r) + (qb + r)2 + p2 (1 − b2 ) −
N =0 2
which shows that a belongs to Q(b). Similarly we find that b belongs to Q(a), so Q(a) = Q(b) = Q(a, b). Starting again from (12) we have r N qb = − pa − r 2 r N q 2 b2 = ( − pa − r)2 2 r N 2 2 q (1 − a ) = ( − pa − r)2 2 “r N ” “r N ”2 = p2 a2 − 2ap −r + −r 2 2
Writing it as a polynomial in a we have 2
2
2
0 = a (p + q ) − 2ap
“r N 2
”
−r +
“r N 2
−r
”2
− q2 ,
(13)
q q which shows that a has degree 2 over Q( N2 ) or is in Q( N2 ). In case p = q = 0, (13) implies q N = r, contradicting our assumption that N/2 is not a rational square; so we may assume 2 q 6= 0. It follows that the quadratic term in (13) does not vanish. Solving (13) by the quadratic formula we have r q q q N ( N2 − r)2 a2 p2 + (p2 + q 2 )(q 2 − ( N2 − r)2 ) 2ap( 2 − r) ± a = p2 + q 2 p2 + q 2
25
Define
s r
r
N − r)2 ) 2 q q N For proof by contradiction, assume that a (and λ) do not belong to Q( N2 ). Then 1, , λ, 2 q q N N and λ 2 constitute a basis for Q(a, 2 ) over Q, as shown by the equation for a above. Let q q q N σ be the automorphism of Q( N2 ) that takes to − N2 . We extend σ to be defined on 2 q Q(a, N2 ) and fix λ. Therefore λ = λσ. We have λ :=
λσ =
s
(
(−
N − r)2 p2 + (p2 + q 2 )(q 2 − ( 2
r
N − r)2 p2 + (p2 + q 2 )(q 2 − (− 2
r
N − r)2 ) 2
Therefore λ
=
λσ
2
=
(λσ)2
λ “r N 2
−r
”2
p2 + (p2 + q 2 )(q 2 − (
r
N − r)2 ) 2
=
“
−
r
”2 N − r p2 + (p2 + q 2 )(q 2 − (− 2
r
N − r)2 ) 2
Subtracting the right hand side from both sides we have “ rN ”2 “ r N ”2 ” “ ”2 ”2 ” ““r N ”“ “r N 0 = p2 −r − − −r + p2 + q 2 −r )+ − −r − 2 2 2 2 ”2 “ r N ”2 ” ““r N −r − − −r 0 = −q 2 2 2 r r r “ N ”“r N ” N N 0 = −q 2 − r − (− − r) − r + (− −r 2 2 2 2 r N 0 = −q 2 4r 2 r = 0 since q 6= 0 (14) But with r = 0 the expression for λ simplifies considerably: s r “ “r N ”2 ” N λ = ( − r)2 p2 + (p2 + q 2 ) q 2 − −r 2 2 r ” “ N N = p2 + (p2 + q 2 ) q 2 − 2 2 r N = p2 q 2 + q 4 − q 2 2 r N = q p2 + q 2 − 2 Hence
r
N 2 Then λσ − λ = qN/2. Since q 6= 0 this contradicts λ = λσ. This contradiction completes the q proof by contradiction that a belongs to Q( N2 ), and hence the proof of the lemma. λσ = q
p2 + q 2 +
26
Lemma 15 Let ABC be an isosceles triangle with base BC, tiled by triangle T similar to half of ABC, in which angle α is not a rational multiple of π. Then the base angles do not split, and the vertex angle A is shared by exactly two tiles with the same angle at A, i.e.the vertex angle at A splits into exactly two equal angles. Proof. We have α + β = π/2. We also have an equation arising from the vertex splitting. Let P , Q, and R be the total number of α, β, and γ angles of tiles at the vertices of ABC. Then because the angles at these vertices must add to π, we have “ R” P α + Qβ = π 1 − . 2
If Q 6= P then we can solve the two equations:
α 1Q+R−2 = π 2 Q−P making α a rational multiple of π. Therefore we can assume P = Q. Now P ≥ 2 since there are two α angles at vertices of ABC. Therefore P α + Qβ ≥ 2α + 2β = π; but P α + Qβ = π(1 − R ). 2 Hence R = 0 and P = Q = 2. If neither base angle splits, then the base angles are β (by convention–as mentioned, in this section we do not assume α ≤ β), and the vertex angle splits into two α angles. The only other possibility is that one of the base angles of ABC splits into two α angles. In that case the vertex angle is β and ABC is equilateral. Hence α = π/6, which is a rational multiple of π, contrary to hypothesis. Therefore we may assume that it is the vertex angle of ABC that splits. Fix any vertex V of the tiling, and let n, m, and ℓ count the number of α, β, and γ angles at V , and let kπ be the angle sum at V , so k = 1 at a non-strict vertex or a boundary vertex, and k = 2 at a strict interior vertex. Then we have “ ℓ” nα + mβ = π k − 2 and if n 6= m, we can solve this equation together with α + β = π/2, obtaining α 1 m + ℓ − 2k = π 2 m−n Since α/π is not rational, we must have n = m at each vertex. This means that at each boundary or non-strict vertex, there are three possibilities: one each of α, β, and γ = π/2, or two right angles, or two each of α and β. Lemma 16 Suppose ABC is isosceles and tiled by triangle T similar to half of ABC, and assume α is a rational multiple of π. Then N is even and either (i) N/2 is a square, or (ii) N/2 is a twice a square (that is, N is a square) and α = π/4, or (ii) N/2 is three times a square and α = π/6. Proof. Suppose that α is a rational multiple of π. By Lemma 14, eiα has degree 2 or 4 over Q. We can therefore apply Lemma 3 to conclude that α = 2π/n, where n = 5, 8, 10, or√12. have α = π/4; p In case n = 8 we √ p hence the left hand side √ of (12) belongs to Q( 2); hence N/2 belongs to Q( 2). Then N/2 has the form u + v 2 with u and v rational. Squaring √ both sides we have N/2 = u2 + 2v 2 + 2uv 2. Hence uv = 0 In case v = 0 then N/2 is a square. In case u = 0 then N/2 is twice a square. √ In√ case n = 12,pα = π/6, so cos α =√ 3/2 and sin p α = 1/2; hence the left hand √ side belongs to Q( 3); hence N/2 belongs to Q( 3). Then N/2 has √the form u + v 3 with u and v rational. Squaring both sides we have N/2 = u2 + 3v 2 + 2uv 3. Hence uv = 0. Hence either
27
u = 0 or v = 0. In case u = 0 then N/2 is three times a square (which is possible, see Fig. 6); in case v = 0 then N/2 is a square. √ In case n = 10 we have α = π/5. Then cos α = (1/4)(1 + 5), and r √ 1 1 (5 − 5) sin α = 2 2 q √ But by Lemma 14, sin α must belong to Q(cos α); hence (5 − 5)/2 belongs to Q(cos α) = q √ √ √ Q( 5). That is, for some rational numbers u and v, we have (5 − 5)/2 = u + v 5. A bit of algebra, not reproduced here, shows that this is impossible, so the case n = 10 cannot actually arise. In case n = 5 we have α = 2π/5 and r √ 1 1 sin α = (5 + 5) 2 2 √ 1 (−1 + 5) cos α = 4 and in this case also sin α does not belong to Q(cos α), so by Lemma 14, this case cannot actually arise. That completes the proof of the lemma. Lemma 17 Suppose triangle ABC is tiled by right triangle T similar to half of ABC, and one of the acute angles of the tile is α. Suppose that a, b, and 1 are linearly dependent over Q. Let r be the number of hypotenuses of tiles lying on side AB. Then r = 0. Proof. By Lemma 14, Q(a, b) has degree 2 over Q. Hence and three members of that field are linearly dependent over Q; in particular, taking the three members 1, a, and b, we can find rational numbers u, v, and w such that ua + vb + w = 1. Note that (p, q, r) is the first row of the d matrix. The second and third rows tell how the side AC and base BC are expressed as combinations of a, b, and c. Since half of ABC is similar to the tile T we have 1 r 0 0 1 a 1 N @ 1 A d@ b A = 2 1 2a We intend to prove that r = 0. The argument splits into three cases. Case 1 is v 6= 0; case 2 is v = 0 and cos α not rational; and case 3 is v = 0 and cos α rational. First we take up case 1, v 6= 0. Define 1 0 0 0 1 M := @ u v w A 2 0 0 Then
1 1 0 1 a M@ b A=@ 1 A 2a 1 0
The inverse of M is M and we have
−1
0
0 = @ −w v 1
0 1 v
0
1 2 u − 2v
0
1 A
1 1 r 0 a a N M@ b A d@ b A = 2 1 1 0
28
Hence
1 r 0 1 a a N @ b A M −1 d @ b A = 2 1 1 0
q N is an eigenvalue of M −1 d. The characteristic polynomial f (λ) of this matrix is a Thus 2 cubic polynomial with coefficients in Q, which is of degree 2 over Q. The matrix has nonnegative coefficients, since that is true of both M −1 and d. We can assume N/2 is not a square. Then, as in the proof of Lemma 6, we conclude that the diagonal elements of M −1 d are zero. In particular, the lower right entry of M −1 d is r. That completes the proof that r = 0 in case v 6= 0. In both cases q 2 and 3, since v = 0, we have u sin α + w = 1, so sin α is rational, and cos α belongs to Q(
N ) 2
as shown by (12). Now we take up case 2, v = 0 and cos α not rational. In q q N that case {1, cos α} is a basis for Q( N2 ), so = e + f cos α for some rational e and f with 2 f 6= 0. If e 6= 0 we have N 2
=
e2 + f 2 cos2 α + 2ef cos α
=
e2 + f 2 (1 − sin2 α) + 2ef cos α
If e 6= 0 we can solve for cos α: cos α
=
−
e2 + f 2 (1 − sin2 α) 2ef
which shows that q cos α is rational when e 6= 0, contradiction. Hence e = 0. Then since pa + qb + r = N2 we have b
=
b
=
f q
N 2
f
r
N − r − pa 2 f cos α = q N 2
=
r
=
N − (r + pa) 2
N − (r + pa) 2 r
N 2
Since f and r + pa are rational, this implies r + pa = 0. But pa ≥ 0 and r ≥ 0. Hence r = 0, which is what we are trying to prove. Now we take up case 3, in which v = 0 (so sin α = a is rational) and cos α = b is also rational. Then we define 0 1 0 0 1 1 M =@ 0 b 0 A 2 0 0
Then the entries of M are rational and we have 1 0 1 0 a 1 M@ b A=@ 1 A 1 2a
29
The inverse of M is M −1 and we have
Hence, as in case 1, we have
0
0 =@ 0 1
0 b 0
1 2
1
0 A 0
1 r 1 0 a a N d@ b A = M@ b A 2 1 1 0
1 1 r 0 a a N @ b A M −1 d @ b A = 2 1 1 0
Thus
q
N 2
is an eigenvalue of M −1 d, and as in case 1, the diagonal elements of M −1 d are zero.
In particular, the lower right entry of M −1 d is r, so r = 0. That completes the proof of the lemma. Theorem 3 Suppose triangle ABC is isosceles with AB = AC. Let M be the midpoint of base BC, and suppose ABC is N -tiled by the right triangle T , and suppose T is similar to triangle ABM (half of triangle ABC). Then N is even and one of the following cases holds: (i) N/2 is a square. (ii) N is a square, and α = π/4. (iii) ABC is equilateral and N is three times a square (so α = π/6). (iv) N/2 is a sum of two squares, and α is not a rational multiple of π. Remark. All the cases mentioned in the theorem do actually occur. Proof. By Lemma 15, the vertex angles at B and C do not split. As mentioned above, we make the convention that β is the base angle; then by the lemma, 2α is the vertex angle. For this section only, we do not assume α ≤ β. If α is a rational multiple of π, then by Lemma 16, one of the first three conclusions of the theorem holds. We can therefore assume that α is not a rational multiple of π. Define α t := tan 2 a := sin α b := sin β = cos α Then as is well known we have b
=
a
=
t
=
1 − t2 1 + t2 2t 1 + t2 a 1−b = 1+b a
q By Lemma 14, a and b belong to Q( N2 ). Therefore a, b, and 1 are linearly dependent over Q, since the degree of Q(a, b) is only 2. By Lemma 17, we have r = 0. Now we will use the fact that r = 0 to complete the proof. By (12) we have r N p sin α + q cos α = 2
30
Since Q(eiα ) = Q(a, b, i) has degree 2 orq4 over Q, Q(a, b) = Q(t) q has degree 1 or 2. It follows q from (12) that Q(a, b) is contained in Q( N2 ); hence t is in Q( N2 ). We may assume that N2 q is not rational, since otherwise we are finished. Hence the degree of Q( N2 ) over Q is exactly q two, and Q(t) = Q( N2 ) = Q(a, b). We express (12) in terms of t: “ 2t ” “ 1 − t2 ” r N p + q = 1 + t2 1 + t2 2 Clearing denominators we have
2tp + q(1 − t2 ) = (1 + t2 )
r
N 2
Writing this as a quadratic in t we have r ” r ” “ “ N N t2 − q − + t(2p) + q − =0 (15) 2 2 q N is rational, in which case we are done. So we may assume If the first term vanishes, then 2 the first term does not vanish. Solving by the quadratic formula, we have r q q p2 − (−q − N2 )(q − N2 ) −p q ± q t = −q − N2 −q − N2 q p ± p2 + q 2 − N2 q = q + N2 q Since Q(t) = Q( N2 ) is of degree 2 over Q, the Galois group contains exactly one element, q q N which takes onto − N2 . The conjugate tσ of t is therefore 2 q p ± p2 + q 2 − N2 q tσ = q − N2
Elements fixed by the Galois group are in Q; in particular that applies to all the symmetric functions of t and tˆ; in particular the sum t + tˆ is rational. We calculate q q p ± p2 + q 2 − N2 p ± p2 + q 2 − N2 q q t + tσ = + q + N2 q − N2 ! ! r N 1 1 q + q = p ± p2 + q 2 − 2 q+ N q− N 2
=
p±
r
p2 + q 2 −
N 2
!
2
2q q 2 − N2
p The left hand side is rational. Hence p2 + q 2 − N/2 is rational; hence N/2 = p2 + q 2 is a sum of two squares, which is one of the cases mentioned in the conclusion of the theorem. Note that the double biquadratic tilings fall under this case. That completes the proof of the theorem.
31
9
Tilings by a non-isosceles right triangle T
We call to the reader’s attention the fact that many of the tilings exhibited in the introduction have the tile T similar to the tiled triangle ABC. Indeed that is the case for the quadratic tilings and the biquadratic tilings, but not the case for the 3-tiling of the equilateral triangle, and various composite tilings involving that tiling as a subtiling (such as the exhibited 12-tilings, and a 6-tiling obtained from the equilateral 3-tiling). There is also, of course, the 2-tiling of an equilateral triangle, which can in turn be used with a quadratic tiling of an equilateral triangle to produce various 8-tilings. The 27-tiling shown in Fig. 10 is an example of a prime tiling with N > 3 in which the tile is not similar to ABC. Eventually we shall give a complete characterization of tilings with ABC not similar to T . In this section, we deal with the special case in which the tile is a right triangle. We begin with a special case of this special case. Lemma 18 Suppose that T is a right triangle with α = π/8, and that ABC is an isosceles right triangle. Let N be a positive integer. Then there is no N -tiling of ABC by T . Proof. We have by trigonometry a
= =
b
= = =
c
=
π sin α = sin 8 r 1√ 1 − 2 2 4 sin β = sin 3α 3 sin α − 4 sin3 α 3a − 4a3
sin γ = sin
π =1 2
Assume, for proof by contradiction, that there is an N -tiling of ABC. Let X be the length of side AC. Then there are non-negative integers p, q, and r such that X = pa + qb + r. We have X2
=
(pa + qb + r)2
=
(pa + q(3a − 4a3 ) + r)2
=
(−4a3 q + (p + 3q)a + r)2
X 2 = a6 (16q 2 ) − a4 (8pq + 24q 2 ) − 8a3 qr + a2 (p + 3q)2 + 2a(pr + 3qr) + r 2
(16)
On the other hand, we will now find an explicit fourth-degree polynomial satisfied by a: a
= =
2a2
= =
1 − 2a2
=
1 8
=
a4 − a2 −
sin α r 1 − cos 2α 2 1 − cos 2α 1 1− √ since 2α = π/4 2 1 √ 2 0
Hence
1 (17) 8 Hence a is of degree 1,2, or 4 over Q. We claim that a is of degree 4 over Q. Since ζ = eiα is a primitive 8-th root of unity, its degree over Q is ϕ(8) = 4. Since Q(ζ) = Q(i, cos α, sin α), a4 = a2 +
32
its real subfield Q(cos α, sin α) has degree 4. If a = sin α were rational, this field would have degree 2 or 1; hence a has degree 2 or 4. But cos α = sin β = sin 3α = 3 sin α − 4 sin3 α, and if a = sin α were of degree 2, this polynomial of degree 3 could be reduced, modulo the minimal polynomial of sin α, to a linear expression in sin α. Hence Q(sin α, cos α) would have degree 2 over Q, contradiction. Hence a is not of degree 2. Hence a is of degree 4, as claimed. (This can also be shown by a direct calculation, without reference to cyclotomic fields.) We now use (17) to reduce the degree of equation (16): X2
a6 (16q 2 ) − a4 (8pq + 24q 2 ) − 8a3 qr + a2 (p + 3q)2 + 2a(pr + 3qr) + r 2 1 1 a2 (a2 + )(16q 2 ) − (a2 + )(8pq + 24q 2 ) − 8a3 qr + a2 (p + 3q)2 + 2a(pr + 3qr) + r 2 8 8 a4 (16q 2 ) − 8a3 qr + a2 (−22q 2 − 8pq + (p + 3q)2 ) + 2a(pr + 3qr) + pq + 3q 2 + r 2 1 (a2 + )(16q 2 ) − 8a3 qr + a2 (−22q 2 − 8pq + (p + 3q)2 ) + 2a(pr + 3qr) + pq + 3q 2 + r 2 8 −8a3 qr + a2 (−6q 2 − 8pq + (p + 3q)2 ) + 2a(pr + 3qr) + pq + 5q 2 + r 2
= = = = =
Simplifying the coefficient of the quadratic term, we obtain our final result for X 2 : X 2 = −8a3 qr + a2 (p2 − 2pq + 3q 2 ) + 2a(pr + 3qr) + pq + 5q 2 + r 2
(18)
The area AT of T is ab/2. Since b = 3a − 4a3 we have 2AT
=
ab
=
a(3a − 4a3 )
= = yielding the result
−4a4 + 3a2 “ 1” −4 a2 + + 3a2 8
by(17)
1 − a2 (19) 2 2 Since ABC is a right isosceles triangle, the area of ABC is X /2. But since ABC is N -tiled by T , the area of ABC is also equal to N AT . Using the equation (18) and (19), we find 2AT =
X2
= = =
2N AT “1 ” 2N − a2 2 3 −8a qr + a2 (p2 − 2pq + 3q 2 ) + 2a(pr + 3qr) + 5q 2 + pq + r 2
Subtracting the second to the last equation from the last one, we have 0
=
−8a3 qr + a2 (p2 − 2pq + 3q 2 + 2N ) + 2a(pr + 3qr) + 5q 2 + pq + r 2 − N
Since a has degree 4 over Q, each of the four coefficients must be zero. In particular the coefficient of a2 is zero: p2 − 2pq + 3q 2 + 2N
(p − q)2 + 2q 2 + 2N
=
0
=
0
Since N > 0 the left side is positive. That contradiction completes the proof of the lemma. Lemma 19 Suppose that T is a right triangle with α = π/8, and that ABC has angle C = 5π/8 and angle A = α. Let N be a positive integer. Then there is no N -tiling of ABC by T .
33
Proof. Let X be the length of side BC of ABC, which is opposite angle α. Let Y be the length of side AC, which is opposite angle B = π/4 = 2α. By the law of sines we have Y sin 2α
=
Y
=
X sin α sin 2α X sin α
The angle between sides X and Y is 5α. Hence the area of triangle ABC is given by 2AABC
=
XY sin 5α
=
XY cos α since sin 5π = sin 3π = cos 8 8 2 sin 2α cos α X sin α 2 sin α cos2 α X2 sin α X2 since cos2 α = cos2 (π/4) = 21
= = =
π 8
This is the same formula for the area of ABC as in the previous lemma; in other words, the area of this triangle ABC is the same as the area of a right isosceles triangle with two sides equal to X. In the proof of the previous lemma, we have derived a contradiction from the assumption that X 2 /2 = N AT ; that contradiction applies here and completes the proof of this lemma as well. The following lemma has already been proved in Lemma 16 and incorporated in the statement of Theorem 3. But it is interesting to see that it also has a completely different proof, along the lines of the lemmas immediately above, without using the theory of cyclotomic fields. Lemma 20 Suppose that T is a right triangle with α = π/8, and that ABC is an isosceles triangle with base angles α. Let N be a positive integer. Then N is twice a square. Remarks. Of course there is a tiling obtained by dividing ABC into two triangles similar to T and quadratically tiling each of the two halves of ABC. That tiling will satisfy the conclusion of the lemma. Proof. Let X be the length of side BC of ABC, which is also the length of side AC since ABC is isosceles. Then the area of triangle ABC is given by 2AABC
=
X 2 sin 6α
=
X 2 sin 2α
=
X 2 cos 2α
=
X 2 (1 − 2 sin2 α)
=
since α = π/8
X 2 (1 − 2a2 )
Assume, for proof by contradiction, that an N -tiling of ABC by tile T exists. As in the proof of the previous two lemmas, we equate 2AABC to 2N AT . We have N
“1 2
2N AT ” − a2
=
2AABC
=
X 2 (1 − 2a2 )
using (19) on the left
Multiplying both sides by 2, the factor 1 − 2a2 cancels from both sides, yielding 2N = X 2 . That completes the proof of the lemma.
34
Lemma 21 Suppose the tile T is a right triangle with α = π/10, and suppose that triangle ABC is N -tiled by T , and angle C is at least a right angle. Then either T is similar to ABC, or ABC is isoceles and T is similar to half of ABC. Proof. The right angle must occur at vertex C. The possible shapes of ABC are as follows: (i) C is a right angle and angle A = α and angle B = 4α. In that case, T is similar to ABC, so we are done. (ii) Angle C is π/2 + 3α and angles A and B are both α. In that case ABC is isosceles and T is similar to half of ABC, so we are done. (iii) Angle C is π/2 + α = 3π/5 and angles A and B are both 2α. (iv) C is a right angle and angle A = 2α and angle B = 3α. (v) Angle C is π/2 + α = 3π/5, and angle A = α and angle B = 3α. (vi) Angle C is π/2 + 2α = 7π/10, and angle A is α and angle B = 2α. The four remaining cases will be treated by similar computations, so we begin with the part that will be common to all three cases. We have α = π/10. Your favorite computer algebra system will tell you that π 10 π b = cos 10 a = sin
= =
1 √ ( 5 − 1) 4 r √ 1 1 (5 + 5) 2 2
Therefore Q(a) has degree 2 over Q and a2 can be expressed linearly in a: (4a + 1)2
=
5
16a + 8a + 1
=
5
4a2 + 2a − 1
=
0
2
and the result is
1 1 − a (20) 4 2 Similarly Q(b) has degree 4 over Q, so b4 can be expressed in terms of lower powers of b: a2 =
4b2 8b2 − 5
4
(8b2 − 5)2 2
64b − 80b + 25 16b4 − 20b2 + 5
=
√ 1 (5 + 5) 2 √ 5
=
5
=
=
5
=
0
and the result is
5 2 5 b − (21) 4 16 √ We can also see that a belongs to Q(b): a few lines above we showed 5 = 8b2 − 5, and hence b4 =
a
= =
1 √ ( 5 − 1) 4 1 2 (8b − 5 − 1) 4
with the result a = 2b2 −
35
3 2
(22)
Since a belongs to Q(b), a basis for Q(a, b) over Q is 1, b, b2 , and b3 . We will need to express other quantities in that basis. For example ab2
= = =
with the result
3 2 )b 2 3 2b4 − b2 2 “5 5” 3 2 2 − b 2 b − 4 16 2
(2b2 −
ab2 = b2 −
5 8
(23)
We also have a3
= = = = = =
with the result
a2 a 1 1 ( − a)a by (20) 4 2 1 1 a − a2 4 2 1 1“1 1 ” a− − a 4 2 4 2 1 1 a− 2 8 1“ 2 3” 1 2b − − 2 2 8
7 (24) 8 With these expressions in hand, we consider side AC of triangle ABC, whose length we call X. Let p, q, and r (respectively) be the numbers of a sides, b sides, and c sides of tiles that lie along AC in the given tiling. Since c = sin γ = 1, we have a 3 = b2 −
X = pa + qb + r 2
We next work out X in the basis 1, b, b2 , b3 . X2
= = = = =
(pa + qb + r)2 “ “ 3” p 2b2 − + qb + r)2 by (22) 2 ”” “ “ 2 3p 2pb2 + qb + r − 2 h “ “ “ 3p ”i 2 3p ” 3p ”2 2 4 3 2 4p b + 4pqb + q + 4p r − b + 2q r − b+ r− 2 2 2 h i “ 3p ” 9 2 4 3 2 2 2 2 4p b + 4pqb + q − 6p + 4pr b + 2q r − b + (r − 3pr + p2 ) 2 4
Substituting in for b4 from (21) we have i “ “5 3p ” 9 5” h 2 + q − 6p2 + 4pr b2 + 2q r − b + (r 2 − 3pr + p2 ) X 2 = 4p2 b2 − 4 16 2 4 and writing it as a polynomial in b we find
“ 3p ” X 2 = 4pqb3 + (q 2 − p2 + 4pr)b2 + 2q r − b + (r 2 − 3pr + p2 ) 2
36
(25)
We use the letters A and C to stand for the angles of ABC at vertices A and C, as well as for the vertices themselves. Let Y be the length of side BC, opposite angle A. Then by the law of sines we have X Y = sin B sin A Therefore sin A Y =X sin B The area of triangle ABC is given, according to the usual cross product formula, as AABC =
1 XY sin C 2
Substituting for Y we have
1 2 sin A sin C X 2 sin B Let AT = ab/2 be the area of the tile T . Our fundamental equation is AABC =
N AT = AABC That is
sin A sin C sin B (Since AT = ab/2, the factors of 1/2 cancel and disappear.) Now we take up the argument by cases according as the possible shapes of ABC, as enumerated at the beginning of the proof. First we take up case (iii), where angles A and B are each 2α. Then sin B and sin A are equal, and cancel out, so we have N ab = X 2 sin C N ab = X 2
Since in this case angle C = π/2 + α, we have sin C = cos α = b, so we can cancel b on the left with sin C on the right, obtaining N a = X2. Putting in on the right the expression for X 2 found in (25), and using (22) on the left, we have “ “ 3” 3p ” N 2b2 − = 4pqb3 + (q 2 − p2 + 4pr)b2 + 2q r − b + (r 2 − 3pr + p2 ) 2 2
Since both sides are expressed in the basis {1, b, b2 , b3 }, the coefficients of like powers of b are equal. Equating the coefficients of b3 we find pq = 0. If p = 0 then from the coefficients of b2 we have 2N = q 2 . Then from the coeficients of b we have qr = 0, and since q 6= 0 we have r = 0. Then the constant coefficient on the right is zero, but on the left it is −3/2. Hence p 6= 0. But since pq = 0 that implies q = 0. Then our equation becomes “ 3” N 2b2 − = (4pr − p2 )b2 + (r 2 − 3pr + p2 ) 2 Equating coefficients of b2 we have
4pr − p2
2N
=
=
r 2 − 3pr + p2
Equating the constant terms we have −
3N 2 3N
= =
2(3pr − p2 − r 2 )
2(4pr − p2 − pr − r 2 )
37
But above we derived 2N = 4pr − p2 . Using this on the right hand side we find 3N
2(2N − pr − r 2 )
=
4N − 2(pr + r 2 )
= N
=
2N
=
2(pr + r 2 ) 4pr + 4r 2
But also 2N = 4pr − p2 . Subtracting this from the last equation we have 0 = 4r 2 + p2 . Hence p = r = 0. But this contradicts q = 0, since p + q + r is the number of tiles along side AC. That contradiction completes the proof in case (iii), when ABC is isosceles with base angles 2α. Next we take up case (iv), when angle C is a right angle, angle A is 2α, and angle B is 3α. Then we have 2N AT
=
N ab
=
sin A sin C sin B 2 sin 2α X sin 3α X2
We have sin 2α = 2 sin α cos α = 2ab, so the ab on the left cancels with sin 2α on the right: N=
2X 2 sin 3α
We have sin 3α = 3 sin α cos2 α − sin2 α = 3ab2 − a3 , so N (3ab2 − a3 )
=
2X 2
We must express the left side in terms of the basis. We have already done the work; substituting the values for ab2 and a3 found in (23) and (24) we have N (3(b2 −
7 5 ) − (b2 − )) 8 8 N (2b2 − 1) 1 N (b2 − ) 2
=
2X 2
=
2X 2
=
X2
and substituting the value for X 2 found in (25) we have N (b2 −
1 ) 2
=
“ 3p ” b + (r 2 − 3pr + p2 ) 4pqb3 + (q 2 − p2 + 4pr)b2 + 2q r − 2
The right hand side is the same as in the previous case, but the left hand side is slightly different. As before, the coefficient of b3 must be zero on the right, so pq = 0, and as before, if p = 0 then we compare coefficients of b2 . This time we find N = q 2 . Comparing the coefficients of b we find qr = 0, but since q 6= 0 we have r = 0, and hence the constant term on the right is zero, while on the left it is −1/2. Therefore p 6= 0; but since pq = 0, we have q = 0. Then from equating the coefficients of b2 we have N = 4pr − p2 (26) and by equating the constant terms we find −
N 2 N 2
=
r 2 − 3pr + p2
=
3pr − p2 − r 2
=
4pr − p2 − pr − r 2
38
But above we derived N = 4pr − p2 . Using this on the right hand side we find N 2
N − pr − r 2
=
Solving for N we have
N = 2pr + 2r 2
(27)
Now we have two expressions equal to N , namely (26) and (27). Equating them, we have 2pr + 2r 2
=
2
=
2r
r 2 + (r 2 − 2pr + p2 )
=
r 2 + (r − p)2
=
4pr − p2
2pr − p2 0
0
It follows that r and r − p are both zero. Hence r and p are both zero, which contradicts q = 0 since r + p + q is the number of tiles along side AC. That completes the proof in case (iv). Next we take up case (v), in which angle C is π/2 + α, angle A is α, and angle B is 3α. Then we have N ab
sin A sin C sin B 2 sin α cos α X sin 3α ab 2 X sin 3α X2
= = =
and canceling ab we have N
X2 sin 3α
=
This equation differs from case (iv) only by a factor of 2 on the right side, but it is not exactly the same, so we must carry out the calculations. Expressing sin 3α in terms first of b we find N (2b2 − 1)
=
X2
Substituting for X 2 as before, we find N (2b2 − 1)
=
“ 3p ” 4pqb3 + (q 2 − p2 + 4pr)b2 + 2q r − b + (r 2 − 3pr + p2 ) 2
Comparing the coefficients of b3 we again find pq = 0, and if p = 0 we have from the coefficients of b that r = 0. Then the constant term is zero on the right but −1 on the left. Hence p 6= 0. But since pq = 0 that implies q = 0. Then equating the coefficients of b2 we have 2N = 4pr − p2
(28)
and from the constant terms we have N
= = =
3pr − p2 − r 2
4pr − p2 − pr − r 2
2N − pr − r 2
Solving for N we have N
=
pr + r 2
2N
=
2pr + 2r 2
39
by (28)
and equating this two expressions for 2N with the one in (28), we have 4pr − p2
0
=
2pr + 2r 2
=
p2 − 2pr + 2r 2
=
(p − r)2 + r 2
Hence r = 0 and p = 0, contradiction. That completes case (v). The last case is case (vi), in which angle C is π/2 + 2α = 7π/10, and angle A is α and angle B = 2α. Then we have sin A sin C N ab = X 2 sin B 2 sin α cos 2α = X sin 2α 2 2 2 sin α(cos α − sin α) = X 2 sin α cos α 2 a(b − a2 ) = X2 2ab Canceling a and multiplying by the denominator, we have 2N ab2
=
X 2 (b2 − a2 )
Putting in the value for a2 from (20) on the right, and the value of ab2 from (23) on the left, we find “ “ “1 5” 1 ”” 2N b2 − = X 2 b2 − − a 8 4 2 “ 1 “ 2 3 ”” 1 5 2 2 2 2b − by (22) = X b − + 2N b − 4 4 2 2 = X 2 (2b2 − 1) Substituting for X 2 from (25) we have ) ( “ 5 3p ” 2 2 2 2 3 2 2 2 2N b − b + (r − 3rp + p ) = (2b − 1) 4pqb + (q − p + 4pr)b + 2q r − 4 2 “ 3p ” 3 b − 4pqb3 = 8pqb5 + (q 2 − p2 + 4pr)2b4 + 4q r − 2 +2b2 (r 2 − 3rp + p2 ) − (q 2 − p2 + 4rp)b2 “ 3p ” −2q r − b − (r 2 − 3rp + p2 ) 2 Multiplying (21) by b we have 5 3 5 b − b 4 16 We use this to eliminate the b5 term, and also we use (21) to eliminate the b4 term: “5 “5 5 ” 5” 5 = 8pq b3 − b + (q 2 − p2 + 4rp)2 b2 − 2N b2 − 4 4 16 4 16 “ 3p ” 3 b − 4pqb3 +4q r − 2 ““ 3p ”2 5p2 ” 2 +2b r− − − (q 2 − p2 + 4rp)b2 2 4 “ 3p ” −2q r − b − (r 2 − 3rp + p2 ) 2 b5 =
40
Collecting and equating the coefficients of b3 we have “ 3p ” 0 = 10pq + 4q r − 2 0 = 4q(p + r) Hence either q = 0 or p = r = 0, since p and r are nonnegative. Suppose that p = r = 0. Then the constant coefficient on the right is zero, but on the left it is −5/4. Hence q = 0 and not both p and r are zero. Equating the constant terms and changing the sign we have 5 4
=
r 2 − 3pr + p2
But the right hand side is an integer and the left hand side is not, contradiction. That completes the proof of the lemma. Lemma 22 Suppose that the triangle ABC is N -tiled by a non-isosceles right triangle T . Then either (i) ABC is similar to T , or (ii) ABC is isosceles, and T is similar to half of ABC. Remarks. The proof shows that in case (i), we have P + Q + R = 4, which means that exactly one vertex angle splits, and it splits into exactly two pieces. Proof. Since γ = π/2 we have R ≤ 1. We begin by showing that if R = 1, then certain values of α imply conclusions (i) or (ii) of the theorem. Assume R = 1. Then by the definition of R, one tile has a right angle at a vertex of ABC; it must be at angle C, the largest angle of triangle ABC. Assume that α = π/8. If vertex C is split, and one α angle occurs at C in addition to the right angle, then angle C is 5π/8, and one of the two remaining angles is α. That case has been ruled out in Lemma 19. If the angle at vertex C is split, and two additional α angles occur there, then triangle ABC is isosceles with base angles α. So T is similar to half of ABC, which is case (ii) of the theorem. Therefore we may assume that the angle at vertex C is not split (so only the right angle occurs there). If each of the other two vertices is split into two α angles, then angles A and B are equal, and ABC is a right isosceles triangle. This case has been ruled out in Lemma 18. If the four α angles are distributed one and three, then ABC is similar to T , which is conclusion (i) of the theorem. We note, for the remark after the theorem, that the β angle is split into three α angles, and the right angle is not split, so P + Q + R = 3 + 0 + 1 = 4. That disposes of the case α = π/8. We will prove that either cases (ii) or (iii) of the theorem hold, or P + Q + R = 4. First we take up the case R = 0. Then we have P α + Qβ
=
π
α + β = π/2 We note that P = Q = 2 does provide a solution of those two equations. We shall show below that it is the only possible solution. We begin by considering the possibilities for Q. We have β < π/2 and α ≤ β, and α + β = π/2. Therefore β ≥ π/4. From P α + Qβ = π we then obtain Q ≤ 4. If Q = 4 then β = α = π/4, contradicting the assumption that T is not isosceles. Hence Q < 4. Next we consider the case Q = 3. Then subtracting the equation α + β = π/2 from the equation P α + Qβ = π we obtain (P − 1)α + 2β = π/2. If P > 1 then the left side exceeds α + β = π/2, contradiction. If P = 1 then β = π/4, but then α = π/4, contradicting α < β. Hence we must have P = 0. Then −α + 2β = π/2. Together
41
with α + β = π/2 this implies that α = π/6 and β = π/3. With R = 0 and P = 0, all the angles of ABC must be composed of three β angles, so ABC is equilateral, and there is no vertex splitting. Then case (ii) holds, since an equilateral triangle is isoceles and T is similar to half of ABC. That disposes of the case Q = 3. Next we claim that if Q = 1, then P ≥ 4, and if Q = 0 then P > 4. First suppose Q = 1. Then P α + β = π. Subtracting α + β = π/2 we have (P − 1)α = π/2. Since α < π/4, we have P − 1 > 2, i.e. P ≥ 4, as claimed. Now assume Q = 0. Then P α + Qβ = π implies α = π/P ; but since α < π/4, we have P > 4. This completes the proof of the claim. Now we consider the cases Q = 0 or Q = 1 further (still assuming R = 0). In that case P ≥ 4, so in particular P > Q. That is, there are more α angles than β angles at the vertices of ABC. Therefore, there is another vertex V of the tiling at which there are more β angles than α angles (since there are N of each altogether). Fix such a vertex V , and let k be 1 if V is a boundary or a non-strict vertex, and k = 2 if V is a strict interior vertex; so the angle sum at V is kπ. Let n, m, and ℓ be the numbers of α, β, and γ angles (respectively) at V . Then we have m > n and 1 0 1 10 0 α kπ n m ℓ @ 1 1 1 A@ β A = @ π A γ π P Q 0
Solving for γ by Cramer’s rule (or Gaussian elimination) we find
[(m − ℓ)P − Q(n − ℓ)]γ = [n(1 − Q) + m(P − 1) + k(Q − P )]π. As long as we don’t divide both sides by the coefficient of γ, the equation is valid whether or not that coefficient is zero. Remembering that γ = π/2 we have (m − ℓ)P − Q(n − ℓ)
=
2[n(1 − Q) + m(P − 1) + k(Q − P )]
(29)
Assume, for proof by contradiction, that Q = 0. Then P > 4, and we have (m − ℓ)P
=
P (−ℓ − m + 2k)
=
P (m − ℓ − 2m + 2k)
=
P (ℓ + m − 2k)
=
2[n + m(P − 1) − kP ] 2(n − m)
2(n − m) 2(m − n)
Since P > 4 when Q = 0, as shown above, the left hand side is greater than 4m. On the other hand, since m ≥ n (by our choice of V ), the right hand side is at most 2m. This contradiction shows that Q 6= 0. Now assume, for proof by contradiction, that Q = 1. Then we have (m − ℓ)P − (n − ℓ)
=
ℓ(1 − P ) + mP − n
=
m−n
=
2k − ℓ
=
(P − 1)(m − ℓ) + (m − n)
=
=
2[m(P − 1) + k(1 − P )]
2(m − k)(P − 1)
(P − 1)(2m − 2k)
(P − 1)(m − 2k + ℓ) n + (P − 2)m P −1 m−n m− P −1
The numerator m − n is positive, and the denominator P − 1 is at least 3, so P − 1 divides m − n. Since m − n is positive, we have m − n ≥ P − 1 ≥ 3, so m ≥ 4. Since β > π/4 and m ≥ 4, we have mβ > π, so the β angles at vertex V occupy more than a straight angle. Hence k 6= 1.
42
Hence k = 2, and also ℓ = 0 or ℓ = 1, since more than one right angle cannot be accommodated at V . Substituting 2 for k we have 4−ℓ
=
(P − 1)(4 − ℓ)
=
m(P − 2)
=
m
=
m−n P −1 (P − 1)m − (m − n) m−
(P − 1)(4 − ℓ) − n P −1 n (4 − ℓ) − P −2 P −2
Since P ≥ 4, the fraction n/(P − 2) is positive, and the fraction (P − 1)/(P − 2) is between 0 and 1, so m
<
n. We have nα + mβ + ℓγ “π π π π ” n +ℓ +m − 2P 2 2P 2
=
kπ
where k = 1 or k = 2
=
kπ
since α =
π 2P
Dividing by π/2 we have “ 1” n + ℓ = 2k (30) +m 1− P P Now we consider the possible values of ℓ. We have ℓ ≤ 3 since ℓγ < 2π. Since m > n there is at least one β angle at V . If ℓ = 3 then we must have m = n = 1, since T is not isosceles and α + β = π/2; but since m > n, that is ruled out. Consider the case ℓ = 2. That means π of the angle at V is used up by two γ angles. Since m > 0, there is at least one β angle at V , so the total angle at V is more than π. Therefore k = 2 and we must have α and β angles adding up to π. It cannot be two of each since m > n. Hence there are at least three β angles. Since β > π/4, there cannot be as many as four β
43
angles, as that would make more than 2π total at V . Hence there are exactly three β angles, i.e. m = 3. Equation (30) becomes “ n 1” +2=4 +3 1− P P
Solving for P we find P = 3 − n. Since P + Q ≥ 3 and Q = 0, we have P ≥ 3. Hence n = 0 and P = 3. Since R = 1 and Q = 0, we now have P + Q + R = 4, which is what we are trying to prove. That disposes of the case ℓ = 2. Now consider the case ℓ = 1. Then we have k = 2, since if k = 1 we must have α and β angles adding up to π/2, with more β than α angles, but that is not possible, since one β angle leaves room for only one α angle. With k = 2 and ℓ = 1, we must have α and β angles adding up to 3π/2, with more β than α angles. We cannot have m > 4 since 5β = 5π/2 − 5α = 5π/2 − 5π/(2P ) > 3π/2, since P ≥ 3. By (30) we have “ n 1” +1 = 4 +m 1− P P “ n 1” = 3 +m 1− P P n = (3 − m)P + m With m ≤ 3 we have n = (3 − m)P + m ≥ m, contradicting m > n. With m = 4 we have n = 4 − P . Since n ≥ 0 and P ≥ 3, the only possibilities are n = 0 and P = 4, or n = 1 and P = 3. If P = 3 then P + Q + R = 4, which is what we are trying to prove. Hence we may assume n = 0, m = 4, and P = 4. Then α = π/8 and β = 3π/8. But at the beginning of the proof we already disposed of the case R = 1 and α = π/8. Finally we consider the case ℓ = 0. First consider k = 1; then the total angle at V is π and none of it is used by right angles, so nα + mβ = π. Equation (30) becomes “ n 1” =2 +m 1− P P which implies
“ 1” m 1− ≤2 P That in turn implies m ≤ 1. But since m > n ≥ 0, we must have m = 1 and n = 0. Hence “ 1” n +m 1− 2 = P P “ 1” = 1− P < 1 This contradiction eliminates the case ℓ = 0, k = 1. That leaves the case ℓ = 0, k = 2. Then we have from (30) “ n 1” = 4 +m 1− P P m − n = (m − 4)P Since m − n > 0, we have m ≥ 5. We have 3
≤
3m − 12
≤
m ≤ 6 − n/2
44
m−n m−4 m−n
P =
Since 5 ≤ m we have only the possibilities m = 5, n = 0, 1, or 2, and m = 6, n = 0. Since P = (m − n)/(m − 4), if m = 6 we have P = 3, and in that case P + Q + R = 4 as claimed. So we may assume that m = 5, in which case we have P = 5, 4, or 3, according as n = 0, 1, or 2. Then α = π/(2P ) is π/10, π/8, or π/6. In case P = 3 we have P + Q + R = 4 as desired. In case P = 4 we have α = π/8, and R = 1, and this case was already treated at the beginning of the proof. In case m = P = 5, we have α = π/10. This case cannot occur, by Lemma 21. We have now proved that in every case in which there is vertex splitting, either P +Q+R = 4, or conclusion (i) of the lemma holds, or α = π/10. The equation P + Q + R = 4 implies that exactly one vertex angle of ABC is split, and that angle is split into only two pieces. The other two angles must therefore each be equal to an angle of the tile T . If two of them are equal to the same angle of T , then ABC is isosceles. We may assume T is not equilateral, since in that case no vertex splitting could occur. The two equal angles cannot be β since 2β + γ > α + β + γ = π. Therefore the two equal angles are α. Since an α angle cannot split, the vertex angle must split. It could split into α + β, or α + γ, or 2β, or β + γ. The latter is impossible, since then 2α + β + γ = π, which contradicts α + β + γ = π since α > 0. If the vertex angle splits into α + β, then the angle sum of ABC is π = 3α + β; subtracting α + β = π/2 we have α = π/4 = β, contradicting the hypothesis that T is not isosceles. If the vertex angle splits into α + γ, then then angle sum of ABC is 3α + γ = π. Since γ = π/2, we have α = π/6. Hence β = π/3 and α + γ = 2π/3 = 2β. Hence the vertex angle of ABC is 2β and conclusion (ii) of the lemma holds in this case. If the vertex angle splits into 2β, then conclusion (ii) of the lemma holds immediately. Hence, if the two unsplit angles of ABC are equal, conclusion (ii) of the theorem holds. Therefore we may assume that the two unsplit angles of ABC are not equal; hence they are equal to two different angles of T . Therefore triangle ABC is similar to T , which is conclusion (i) of the lemma. That completes the proof of the lemma.
10
An isosceles tile T with base angle π/6 or π/5
In this section, we investigate the tilings that can be constructed from the tile T with two π/6 angles and one 2π/3 angle, in which T is not similar to ABC. One can find examples of such tilings by first tiling an equilateral triangle with an equilateral tile, and then 3-tiling that tile; the 12-tiling shown in the introduction is of that form. But this does not exhaust the possibilities, as we can also 3-tile an equilateral triangle ABC and then quadratically tile the three resulting triangles. In this way we can produce, for example, a (different) 12-tiling. These tilings are not prime, however, and we once thought that the equilateral 3-tiling might be the only prime tiling constructed from T . That is incorrect, however, as the 27-tiling shown in the introduction proves. Lemma 23 Let triangle T be isosceles with a vertex angle of 2π/3. Suppose triangle ABC is N -tiled by T and not similar to T . Then ABC is equilateral, and N has the form 3m2 for some integer m. Remark. For each integer N of the form mentioned in the lemma, there does exist an N -tiling of an equilateral triangle by the tile T of the lemma, as discussed in the introduction. Proof. Let α = π/6 and γ = 2π/3 be the angles of T . We note that no tile has its γ angle at a vertex of ABC, since that would force the other two vertices of ABC to have angle α and hence ABC would be similar to T , contrary to hypothesis. Hence the vertex angles of ABC are composed (altogether) of six α angles. The only two ways of writing 6 as a sum of smaller positive integers are 6 = 1 + 2 + 3 and 6 = 2 + 2 + 2. Hence, ABC is either equilateral, or it is a 30-60-90 right triangle. √ We are free to choose the size of T ; let√us suppose that the sides√of T are a = 1 and c = 3, so the altitude of T is 1/2 and its area is 3/4 (because its base is 3 and its height is 1/2). If
45
X, Y , and Z are the sides of ABC, then the fundamental tiling equation is 1 1 0 0 1 X A=@ Y A d@ 1 √ Z 3
Our equations will be slightly simpler if we recall the convention that when the tile is isosceles, as it is in this proof, so that b = a, that we take the middle column of the d matrix to be zero (counting short sides as a rather than b, since it is arbitrary). We begin by proving that it is not the case that along one side of ABC, all the tiles sharing that side of ABC have their a sides on the boundary of ABC. Assume, for proof by contradiction, that all the tiles on side BC have an a side on BC. Let the copies of the tile along BC be T1 , . . . , Tm (for some m), with T1 sharing vertex B. Since T has only a and c sides, that means that the vertices V1 , . . . , Vm−1 of the tiling on side BC are equally spaced along BC, with spacing a. Since T1 must lie within triangle ABC, and the angle of ABC at B is π/3, only one orientation of T1 is possible, namely with its α angle at B and its γ angle at the next vertex V1 on BC. Then T2 has its α angle at V2 . By induction on k, for k < m each Tk has its α angle at Vk and its γ angle at Vk+1 . But then Tm , the last tile on BC, has its α angle at Vm−1 , and hence must have its γ angle at C; but since the angle of ABC at C is less than π/2 − α, this is a contradiction. We first take up the case when ABC√is equilateral. Let X be√ the side of equilateral triangle 2 ABC. Then the altitude of ABC is X 3/2, and its area is X 3/4. But √ this area is also N √ √ times the √ area of T . Since the area of T is 3/4, we have N 3/4 = X 2 3/4. Hence N = X 2 , and X = N . The tiling equations are 0 1 0 1 1 1 √ A = N @ 1 A. d@ 1 √ 1 3 We can assume that the components of this vector equation correspond to edges AB, AC, and BC respectively. That is, for i = 0, 1, and 2 we have (remembering that di1 = 0) √ √ N = di0 + di2 3 √ N = (di0 + di2 3)2 √ = d2i0 + 3d2i2 + 2(di0 di2 ) 3
That implies that di0 = 0 or di2 = 0. First assume di0 = 0. That is, no a sides of the tile occur on the side of ABC corresponding to the first row of the tiling equation (say for convenience it is AB). Then AB is entirely composed Since each √ of c edges (as it indeed is in the 27-tiling). √ side of ABC has length X and c = 3, the number of tiles on each edge is X/ 3, which must be an integer, say m. Then X 2 = 3m2 , so N = X 2 = 3m2 . That completes the proof in case ABC is equilateral and di0 = 0. Now assume di2 = 0. Then only a edges occur along AB. But we already proved that to be impossible. That contradiction completes the proof in case ABC is equilateral. We may therefore assume that ABC is a 30-60-90 triangle. To complete the proof, we must show this assumption leads to a contradiction. Let the π/6 angle of ABC be at A, the right √ angle at C and the√π/3 angle at B. Let the sides of ABC be X, X 3, and 2X. √ Then the 2 area of ABC is Xp 3/2, which must be equal to the area of N tiles, namely N 3/4. Hence X 2 = N2 , or X = N/2. Then we have 0 1 1 r 0 1 1 √ N @ 3 A. A= d@ 1 √ 2 3 2
46
First we show that have
q
N 2
is not rational. From the first row of the matrix tiling equation, we
r √ N . d00 + d02 3 = 2 q √ Assume, for proof by contradiction, that N2 is rational. Since 3 is irrational, it follows that d02 = 0, i.e., no c sides of tiles occur on the short side BC of triangle ABC.qBut we already
proved this to be impossible. This completes the proof by contradiction that N2 is irrational. q √ N is not a rational multiple of 3. From the middle row of the tiling Next we show that 2 equation, we have r √ N√ 3. d10 + d12 3 = 2 q √ Assume, for proof by contradiction, that N2 is a rational multiple of 3. Then the right hand side is rational, so d12 = 0. That means that only a sides of tiles occur on the long side AC of proved this to be impossible. That completes the proof that q triangle ABC. But we already √ N is not a rational multiple of 3. 2 q √ By the first row of the vector equation above, we have d00 + d02 3 = N2 . Squaring this equation we have N 2 N
=
√ (d00 + d02 3)2
√ d200 + 3d202 + 2d00 d02 3 √ √ Since 3 is irrational, it follows that the coefficient of 3 on the right is zero. Thus d02 = 0 or d00 = 0. Assume, for proof by contradiction, that d02 = 0; that means that no c sides of tiles occur on side BC of triangle ABC. But we already proved this to be impossible; this contradiction proves that d02 6= 0. Therefore d00 = 0. That is, only c sides of tiles occur√on BC. Hence, for some integer m (the number of tiles on BC), we have X = BC = mc = m 3. Hence =
AC √ 3 = X 2 √ 3 √ m 3 = 2 3m = 2 √ √ Hence the area of triangle ABC is XY /2 = 3m2 3/2. Since 3 is the area of T and the area of ABC is N times the area of T , it follows that N = 3m2 /2. By the second row of the matrix equation, we have r √ N√ d10 + d12 3 = 3 2 p √ = 3m2 /2 3 Y
=
=
√
3m/2
Since 3 is irrational, it follows that d12 = 0. That is, no c sides of tiles occur on AC. Hence, only a sides of tiles occur on AC. But we have already proved this to be impossible. That completes the proof of the lemma.
47
Lemma 24 Let α = π/5. Let T be an isosceles triangle with two angles α and one angle 3α. Let ABC be an isosceles triangle with base angles 2α and vertex angle α. Then there is no tiling of ABC by tile T . Proof. Let a = sin α. We will work in the field Q(a) so we begin with some identities in that field. We have q √ 1 a = sin α = 10 − 2 5 4 √ 1 a2 = (10 − 2 5) 16 √ 16a2 − 10 = −2 5 √ 5 = 5 − 8a2 (8a2 − 5)2
4
64a − 80a2 + 25
16a4 − 20a2 + 5
=
5
=
5
=
0
This is the minimal polynomial of a over Q. It is irreducible by Eisenstein’s criterion, since 20 and 5 are divisible by 5, but 16 is not, and 5 is not divisible by 52 . So Q(a) has degree 4 over Q and {1, a, a2 , a2 } is a basis for Q(a) over Q. From the minimal polynomial we see a4 = 20a2 − 5
(31)
We have b
=
= =
cos α √ 1+ 5 4 1 + (5 − 8a2 ) 4 3 − 4a2 2
Hence b= We have sin 2α = 2ab =
2a( 32
since
√
5 − 5 − 8a2
3 − 2a2 2
(32)
2
− 2a . Hence sin 2α = 3a − 4a3
(33)
Recall (or prove) the trig identity (which is true for any α) sin 3α = 3 − 4 sin α sin α Expressing this more briefly using a we have sin 3α = 3 − 4a sin α
(34)
With these algebraic preliminaries in hand, we turn to the proof of the lemma. The sides of T are sin α and sin(3α). The altitude of T is sin2 α, so the area of T is 12 sin(3α) sin2 α. On the other hand, for some number λ, the sides of ABC are λ sin 2α and λ sin α. The altitude of ABC
48
is 12 λ sin α sin(2α), so its area is area of T , we have
1 2 λ 2
sin α sin2 2α. Since the area of ABC must be N times the
1 sin(3α) sin2 α 2 N sin(3α) sin α p N sin α sin(3α) λ= sin(2α)
=
N
=
1 2 λ sin α sin2 2α 2 λ2 sin2 2α
Let d be the d matrix of the tiling. Since the tile is isosceles, we only have sides a and c, not b, and the middle column of the d matrix is zero. Then we have 1 1 0 0 sin α sin α d @ sin α A = λ @ sin 2α A sin 2α sin 3α 1 0 p sin α N sin α sin(3α) @ = sin 2α A sin(2α) sin 2α Writing the second of these three equations out we have p d10 sin α + d12 sin 3α = N sin α sin(3α) Dividing both sides of the second equation by sin α we have r sin 3α sin 3α d10 + d12 = N sin α sin α Expressing this in terms of a = sin α, we have d10 + d12 (3 − 4a)
=
Squaring both sides we have
p
N (3 − a)
(d10 + d12 (3 − 4a))2
=
d210 + 6d10 d12 + 9d212 + a(−8d10 d12 − 24d212 ) + a (16d212 )
=
d210
+ 2d10 d12 (3 − 4a) +
d212 (3 2
2
− 4a )
=
N (3 − a)
N (3 − 4a)
3N − 4aN
Since {1, a, a2 , a3 } is a basis for Q(a) over Q, the coefficients of like powers of a must be equal on both sides. Therefore the coefficient of a2 on the left is zero; that is, d12 = 0. Then the equation becomes d210 = 3N − 4aN
But now there is a nonzero coefficient of a on the right, and no a term on the left. This contradiction completes the proof of the lemma.
11
A non-isosceles tile T with largest angle 2π/3
As it turns out, one of the most difficult cases to analyze is the case of a tiling in which T is not similar to ABC, and T has γ = 2π/3 and α < β. In analyzing this case, we alternated several times between trying to construct such a tiling and trying to prove none exist. Although there are some quite interesting ways of fitting together tiles of this shape, one never seems to be able to make a triangle, and indeed we prove in this section that it is impossible to do so.
49
Lemma 25 Suppose ABC is N -tiled by a triangle T whose largest angle γ = 2π/3. Suppose ABC is not similar to T , and T is not isosceles, i.e. is not the tile used in the equilateral 3-tiling. Then either α = π/9, or α = π/12, or α = 2π/15. Moreover, let P and Q be the total number of α and β angles at the vertices of ABC. Then P 6= Q and α = π3 (Q − 3)/(Q − P ). When P < Q and α = π/9 we have (P, Q) = (1, 4), and if P < Q and α = π/12 we have (P, Q) = (0, 4). In case α = 2π/15, triangle ABC is isosceles with base angles either β or 2β, and (P, Q) = (0, 5). Proof. Since ABC is not similar to T , we have vertex splitting. Since T is not isosceles, α < β. Let P and Q be the total number of α angles and β angles, respectively, at vertices of ABC. The number R of γ angles is either zero or 1, since 2γ = 4π/3 > π. If R = 0 we have P α + Qβ = π, and if R = 1 we have P α + Qβ = π/3. Because ABC is not similar to T , at least two angles of ABC are split, so P + Q + R ≥ 5. Since R ≤ 1, we have P + Q ≥ 4. Fix any vertex V of the tiling, and let n, m, and ℓ be the number of α, β, and γ angles at V . If ℓ = 2 then either m = n = 1 or m = 0 and n > 2. If ℓ = 1 then at least two more angles are required. Thus, unless ℓ = 3, there are at least as many α and β angles at V as γ angles. Since at the vertices of ABC, there are more α and β angles than γ angles, this cannot be the case at every vertex, since altogether there must be N of each. This proves that at some vertex V we have ℓ = 3, i.e. three γ angles meet. We have 10 0 1 0 1 α 0 0 3 2π @ 1 1 1 A@ β A = @ π A (35) 2π P Q R π 3
The determinant of the matrix on the left is 3(Q − P ). Assume, for proof by contradiction, that P = Q. Then R = 0, since if R = 1 we have π/3 = P α + Qβ = P (α + β) = P π/3, so P = Q = 1, contradicting P + Q ≥ 4. Now that we know R = 0 we have π = P α + Qβ = P (α + β) = P (π/3) since α + β = π/3. Hence π = P (π/3); hence P = 3. Since P = Q we have P = Q = 3. Hence the matrix equation above becomes 10 0 1 0 1 α 0 0 3 2π @ 1 1 1 A@ β A = @ π A 2π 3 3 0 π 3 The matrix has rank 2, since the first and third rows are linearly independent. The equation is solvable if and only if (2π, π, π) lies in the span of the first and third rows. That is, if and only if the dot product of (2π, π, π) with the cross product of (0, 0, 3) and (3, 3, 0) is zero. If there is a tiling as hypothesized in the lemma, the equation is solvable. Hence ˛ 1 ˛ 0 ˛ ˆi ˆj k ˆ ˛ 2π ˛ ˛ 0 = @ π A · ˛˛ 0 0 3 ˛˛ ˛ 3 3 0 ˛ π 1 0 1 0 2 −9 = π@ 1 A·@ 9 A 1 0 = −9π
Since 0 6= 9π, we have reached a contradiction. That completes the proof by contradiction that P 6= Q. Therefore the determinant of the matrix equation (35) is not zero, and we can solve for α by Cramer’s rule: 1 0 2 0 3 π @ 1 1 1 A α = 3(Q − P ) 1 Q R π 2R + Q − 3 (36) = 3 Q−P
50
Since we have shown P 6= Q, there are only two possibilities, Q < P and Q > P . Consider the case Q < P . We will show that α must be π/9 or π/12. The meaning of Q < P is that there are fewer β angles than α angles at the vertices of ABC (taken together). Since each copy of the tile has one β and one α vertex, there must exist a vertex of the tiling at which there occur more β angles than α angles. Let V be such a vertex. Then at least one β angle occurs at V . The angle sum at V is either π or 2π. Suppose first that it is π. If a γ angle occurs at V , then there is room for only one β and one α angle in addition, contradicting the fact that more β angles than α angles occur at V . Therefore no γ angles occur at V , and at least two β angles. There cannot be exactly two β angles at V , since π − 2β = π/3 + 2α > 2α, so more than two α angles must occur at V , contradiction. If exactly three β angles occur, then there is room for only three α angles (since α + β = π/3); but that contradicts the fact that more β than α angles occur at V . If four β angles occur, then let k < 4 be the number of α angles at V ; then we have 4β + kα
=
π
(4 − k)β + k(α + β) π (4 − k)β + k 3 π (4 − k)β = (3 − k) 3
=
π
−=
π
since α + β = π/3
Let us check the possibilities for k. With k = 3 we have β = 0, a contradiction. With k = 2 we have α = β = π/6, contradicting the hypothesis that T is not isosceles. With k = 1 we have β = 2π/9 and α = π/9,as desired. With k = 0 we have β = π/4 and α = π/12, as desired. contradicting the assumption that α 6= π/12. Hence, if the angle sum at V is π, the conclusion of the lemma is verified. If, on the other hand, the vertex V has angle sum 2π, and has two γ angles, that leaves 2π/3 to be composed of α and β angles. That could be done with two α and two β angles, but that is not more α than β angles; it cannot be done with fewer than two β angles; and if three β angles occur, with zero α angles, then their sum is 2π/3 = 3β = π − 3α, which implies α = π/9. If three β angles occur with one α angle, then 3β + α
=
2π/3
3(π/3 − α) + α
=
2π/3
=
π/6
α
since α + β = π/3 contradiction, since α < π/6 < β
If three β angles occur with two α angles, then 3β + 2α 3(π/3 − α) + 2α
α
=
2π/3
=
2π/3
=
π/3
since α + β = π/3 contradiction, since α < π/6 < β
But since there are more β angles than α angles at V , this exhausts the possibilities. That completes the proof of the claim that if Q < P then α = π/9 or α = π/12. Turning to the case Q > P , since α < π/6, we have by (36) that π 2R + Q − 3 π < . 3 Q−P 6 Hence Q − P > 2(2R + Q − 3). Hence 4R + P + Q < 6. Since P + Q ≥ 4, the case R = 1 is ruled out, and we can assume R = 0. Then P + Q < 6. Hence P + Q = 4 or P + Q = 5. Since α > 0 we have Q > 3 by (36), so the possibilities for (P, Q) are (0, 4), (1, 4), (0, 5). The corresponding
51
values of α can be computed from (36), and β can then be found from α + β = π/3. The results are shown in the following table: P 0 1 0
Q 4 4 5
α π/12 π/9 2π/15
α in degrees 15 20 24
β π/4 2π/9 π/5
β in degrees 45 40 36
Note that in case (P, Q) = (0, 4), the triangle ABC must be a right isosceles triangle, since with four β angles to make three vertices of ABC, two of them must be β and the other one 2β = π/2. In case (P, Q) = (0, 5), five β angles are distributed among three vertices of ABC, so the triangle ABC must also be isosceles, but there are two possibilities: either the base angles are β and the vertex angle is 3β, or the base angles are 2β and the vertex angle is β. That is, the base angles are either 36◦ or 72◦ . That completes the proof of the lemma. Now the reader may expect us to rule out the three remaining values of α and be done with this section. It is not quite simple, because there are many possible shapes of ABC to consider. Strangely, we have to divide the argument into cases not only by the value of α, but by whether ABC is or is not isosceles. Lemma 26 Suppose that triangle ABC is isosceles. Then ABC cannot be N -tiled by a triangle T whose largest angle is γ = 2π/3 and smallest angle is α = π/12. Proof. Let a = sin α. We will work in the field Q(a), so it will be necessary to establish some identities that determine arithmetic in Q(a). One can calculate a as follows: a
= = = = =
π 12 π/6 sin 2 r 1 − cos π6 2 s √ 1 3 − 2 4 √ 3−1 √ 2 2 sin
The √ last step can be verified by squaring both sides and simplifying. Squaring and solving for 3 we have √ 3 = 2 − 4a2 (37)
Squaring and subtracting 3 we find 16a4 − 16a2 + 1 = 0. Hence
1 (38) 16 √ √ From the equation for a it √ now follows that both 2√and 3 belong to Q(a), which therefore has degree 4 over Q, since 3 does not belong to Q( 2). Hence the fourth-degree polynomial above is the minimal polynomial of a. √ In triangle T , we have β = π/3 − α = π/4. Hence b = sin β = 1/ 2 already belongs to Q(a). We need to find an explicit expression for b in powers of a. √ 3−1 √ a = 2 2 a4 = a2 −
52
= = b
= = = = =
√ 1 3−1 √ 2 2 √ 3−1 b 2 2a √ 3−1 √ 2a 3+1 √ √ 3−1 3+1 √ a( 3 + 1) a(2 − 4a2 + 1)
by (37)
2
a(3 − 4a )
with the final result
b = 3a − 4a3 (39) √ 2 The third side of the tile is c = sin(2π/3) = 3/2 = 1 − 2a . With these algebraic preliminaries in order, we now consider an isosceles triangle ABC, with base BC and vertex angle at A. The possibilities for the base angles of ABC are kα, for 1 ≤ k ≤ 5, and the corresponding possibilities for the vertex angle A are the even multiples of α up to 10α. In other words, since π/2 = 6α, the vertex angle A is either π/2, or π/2 ± 2α, or π/2 ± 4α. That is, the angle at A is π/2 + 2Jα, for −2 ≤ J ≤ 2. Fix this number J. Let X be the length of sides AC and BC. Then the area of triangle ABC is given by 1 2 X sin(π/2 + 2Jα) 2 1 2 = X cos(2Jα) 2 Next we calculate the area of the tile T . We have a = sin α, b = sin β, and by the cross-product formula for the area of a triangle we then have 1 AT = sin α sin β sin γ 2 1 = 12 = ab since sin γ = sin 2π 2 4 1 = a(3a − 4a3 ) by(39) 2 3 2 = a − 2a4 2 3 2 1 = a − 2(a2 − ) by (38) 2 16 1 3 = − a2 8 2 Since ABC is tiled by N copies of the tile T , we have AABC
=
N AT = AABC 3 ” 1 2 − a2 X cos(2Jα) = N 8 2 2 Let p, q, and r be the numbers of a sides, b sides, and c sides of tiles on side AB. In other words, p, q, r is the row of the d matrix corresponding to side AB. Then X = pa + qb + rc. Using the expressions derived in (39) and (37) we have “1
X
=
pa + qb + rc
=
pa + q(3a − 4a3 ) + r(1 − 2a2 )
=
r + (p + 3q)a − 2ra2 − 4qa3
53
Squaring both sides we have X2
=
2
=
X
(r + (p + 3q)a − 2ra2 − 4qa3 )2
r 2 + 2r(p + 3q)a + a2 ((p + 3q)2 − 4r 2 ) − a3 (8qr + 4r(p + 3q)) + a4 (4r 2 − 8q(p + 3q))
This expression for X 2 does not depend on the shape of ABC, but only the shape of T . Now we bring the shape of ABC into the equations. Using (38) to eliminate a4 we find X2
= = =
r 2 + 2r(p + 3q)a + a2 ((p + 3q)2 − 4r 2 ) − a3 (8qr + 4r(p + 3q)) + (a2 − 3 2 r + 4 3 2 r + 4
1 )(4r 2 − 8q(p + 3q)) 16
1 q(p + 3q) + 2r(p + 3q)a + a2 ((p + 3q)2 − 8q(p + 3q)) − a3 (8qr + 4r(p + 3q)) 2 1 q(p + 3q) + 2r(p + 3q)a + a2 (p + 3q)(p − 5q) − a3 (8qr + 4r(p + 3q)) 2
Assume, for proof by contradiction, that J = 0, i.e ABC has a right angle at A. Then 2N AT 3 ” 2N − a2 8 2 N − 3N a2 4 “1
=
2AABC
=
X2
= =
3 2 r + 4 3 2 r + 4
1 q(p + 3q) + 2r(p + 3q)a + a2 (p + 3q)(p − 5q) − a3 (8qr + 4r(p + 3q)) 2 1 q(p + 3q) + 2r(p + 3q)a + a2 (p + 3q)(p − 5q) − a3 (8qr + 4rp + 12rq) 2
Since {1, a, a2 , a3 } is a basis for Q(a), the coefficients of like powers of a are equal on both sides of this equation. The linear term tells us that r(p + 3q) = 0. Hence either r = 0 or both p and q are zero, since p and q are nonnegative. If both p and q are zero, then the quadratic term is zero on the right, but is −3N on the left, contradiction. Hence r = 0. Equating the coefficients of the quadratic term on both sides we have 3N = (p + 3q)(5q − p) Equating coefficients of the constant terms we have N 1 = q(p + 3q) 4 2 Multiplying by 12 we have 3N = 6q(p + 3q) Equating the two expressions we have derived for 3N we have (p + 3q)(5q − p) = 6q(p + 3q) We have already proved p + 3q 6= 0, so we can cancel it, obtaining 5q − p = 6q, which is impossible, since p ≥ 0. This contradiction shows that the assumption J = 0 is untenable, i.e. no such tiling exists with a right angle at vertex A. Assume, for proof by contradiction, that J = ±1, which corresponds to vertex angle π/2±2α. Then π 2N AT = X 2 sin( ± 2α) 2 N 2 2 − 3N a = X cos 2α 4 π = X 2 cos 6 √ 3 = X2 2 = X 2 (1 − 2a2 ) by (37)
54
Now we put in for X 2 the expression calculated above: n3 N 1 − 3N a2 = r 2 + q(p + 3q) + 2r(p + 3q)a + a2 (p + 3q)(p − 5q) 4 4 2 o −a3 (8qr + 4r(p + 3q)) (1 − 2a2 )
3 3 2 1 r + q(p + 3q) + 2r(p + 3q)a + a2 ((p + 3q)(p − 5q) − r 2 − q(p + 3q)) 4 2 2 −a3 ((8qr + 4r(p + 3q)) + 4r(p + 3q)) − 2a4 (p + 3q)(p − 5q) + 2a5 (8qr + 4r(p + 3q))
=
Now we use (38) to eliminate a4 and a5 N − 3N a2 4
=
3 2 1 3 r + q(p + 3q) + 2r(p + 3q)a + a2 ((p + 3q)(p − 5q) − r 2 − q(p + 3q)) 4 2 2 1 −a3 ((8qr + 4r(p + 3q)) + 4r(p + 3q)) − 2(a2 − )(p + 3q)(p − 5q) 16 a +2(a3 − )(8qr + 4r(p + 3q)) 16
Equating coefficients of the linear term we have 0
1 (8qr + 4r(p + 3q)) 8 1 2r(p + 3q) − qr − r(p + 3q) 2 3 r(p + q) 2 2r(p + 3q) −
= = =
Hence either r = 0 or both p and q are zero. Assume, for proof by contradiction, that both p and q are zero. Then equating the constant coefficients, we have N = 3r 2 . Equating the coefficients of a2 we have 3N
=
3 − r2 2
Substituting N = 3r 2 we have −9r 2 = −(3/2)r 2 ; hence r = 0. But we cannot have r = 0 when p = q = 0, since r + p + q is the number of tiles along one side of ABC, and hence is positive. This contradiction proves that not both p and q are zero. Therefore r = 0. Substituting r = 0 and collecting like powers of a we have N − 3N a2 4
= = =
1 1 q(p + 3q) + a2 ((p + 3q)(p − 5q) − q(p + 3q)) − 2(a2 − )(p + 3q)(p − 5q) 2 16 1 1 q(p + 3q) + (p + 3q)(p − 5q) + a2 ((p + 3q)(p − 5q) − 2(p + 3q)(p − 5q)) 2 8 1 (p + 3q)(p − q) − a2 (p + 3q)(p − 5q) 8
Multiplying by 2 and equating the coefficients of like powers of a we have 2N
=
6N
=
(p + 3q)(p − q)
(p + 3q)(p − 5q)
Subtracting the second equation from three times the first, we have 0
= =
(p + 3q)(3(p − q) − (p − 5q))
2(p + 3q)(p + q)
But since p and q are nonnegative, and not both zero, this is a contradiction. That completes the proof in case J = 1.
55
Therefore the one remaining possibility, J = 2, must be the case, and we have π ± 4α) 2
2N AT
=
X 2 sin(
N − 3N a2 4
=
X 2 cos 4α
=
X 2 cos
=
1 2 X 2
=
X2
N − 6N a2 2
π 3
The right hand side is the same as in case J = 0, and we reach a contradiction in the same way. The factor of 2 on the left side makes no difference, as the two expressions we find for 3N are both multiplied by 2, and hence can still be equated. That completes the proof of the lemma. Next comes the case α = 2π/15. If we argue in the same way as in the preceding lemma, we will be working in the field generated over Q by a = sin(2π/15) and b = sin(8π/15. The arithmetic in this field is more cumbersome, and the number of possible vertex angles is greater. Instead, we use another method, involving primes in the cyclotomic field. It requires, for the first time in this paper, concepts going beyond elementary field theory. Lemma 27 Suppose that isosceles triangle ABC is N -tiled by a triangle T whose largest angle γ = 2π/3. Then (i) α 6= 2π/15
(ii) If α = π/9 then the vertex angle of ABC cannot be 3α = π/3 or 5α.
Proof. Define ζ
:=
e2πi/30 qquad when α = 2π/15
ζ
:=
e2πi/18 qquad when α = π/9
ρ
:=
eiπ/3
Then we have eiα = ζ 2 in case (i) and eiα = ζ in case (ii). Let K = 5 when α = 2π/15 and K = 3 when α = 2π/9. Then we have ρ = ζ K . Since α + β + 2π/3 = π, we have β = π/3 − α. We calculate sin α and sin β in terms of ζ. First, in case α = 2π/15 we have 2i sin α
=
2i sin β
= = =
ζ 2 − ζ −2
ei(π/3−α) − e−i(π/3−α) ρζ −2 − ρ−1 ζ 2 ζ K−2 − ζ 2−K
In case α = π/9 we have instead 2i sin α
=
2i sin β
= = =
ζ − ζ −1
ei(π/3−α) − e−i(π/3−α) ρζ −1 − ρ−1 ζ
ζ K−1 − ζ 1−K
These expressions belong to Z[ζ] and hence are algebraic integers of the field Q[ζ]. Let X, Y , and Z be the sides of triangle ABC in non-decreasing order. Then 1 1 0 1 0 0 sin α a X @ Y A = d @ b A = d @ sin β A , sin γ c Z
56
so 2iX, 2iY , and 2iZ are algebraic integers of Q(ζ). The area AT of the tile can be computed as a cross product AT
= =
1 bc sin α 2 1 sin α sin β sin γ 2
Since the factor on the right are algebraic integers divided by 2i, it follows that 16i3 AT belongs to Z[ζ]. The area of triangle ABC is N times the area of the tile. That is, N AT
=
AABC
The area of triangle ABC is given by 12 U 2 sin θ, where U is the length of the two equal sides of ABC, and θ is the vertex angle. Inserting the forms of these expressions given above, and multiplying by 2, we have N sin α sin β sin γ = U 2 sin θ (40) Example. To fix the ideas, we pause to give an example. Of course, the point of the theorem is that no example exists, so to give an example we must choose α = β, which is not covered by the theorem, but is not contradicted by the arguments so far. Consider the 3r 2 tiling, in which ABC is equilateral and T has α = β = π/6. Then only c sides share the boundary of ABC, so in the formula for AABC , the coefficients of sin α and sin β are zero, and the coefficients of sin γ are 3r, r, r, and r in the four terms, respectively. Hence √ AABC = (r 2 sin2 γ/4) 3 √ r 2 ( 3/2)2 √ 3 = 4 √ 2 3 3r 6 = 1 √ We have AT √ = sin α sin β sin γ/2 = 3/16. Since N = 3r 2 we check our equation: indeed 2 N AT = (3r ) 3/16 = AABC . Now we return to the consideration of the area equation (40). Except for a factor of 16i, both sides of (40) are algebraic integers √ of the field Q(ζ), because they belong to Z[ζ]. Dividing by U 2 sin α sin β and using sin γ = 3/2 we have √ sin θ 3N = (41) 2U 2 sin α sin β Recall that 2iU is an algebraic integer of Q(ζ), since U is an integral linear combination of sides of the tile, which are sin α, sin β, and sin γ, and each of these is an algebraic integer over 2i. We claim that 2i sin θ is also an algebraic integer. When α = 2π/15, have shown in Lemma 25 that triangle ABC is isosceles with vertex angle θ = 3β = 3π/5, or θ = β = π/5. We have already shown that 2i sin β is an algebraic integer, and 2i sin(3β) = 2i(ζ 3K−6 − ζ 6−3K ). On the other hand, when α = π/9, Lemma 25 shows that the vertex splitting of ABC involves one α angle and four β angles, and here β = π/3 − α = 2π/9 = 2α. If ABC is isosceles, then the vertex angle θ is either α or α + β = π/3, and in either case 2i sin θ is an algebraic integer. Now suppose that the right hand side of (41) has norm a (positive or negative) power of 2; or more generally, at least the norm contains only an even power of 3, not an odd power. Suppose, in addition, that the rational prime 3 does √ not split in Q(ζ), but ramifies with ramification degree 2. Then there is just one prime P = 3 lying over the rational prime 3, and P divides N to an even power (twice the power of 3 in N ), and P divides U 2 to an even power (twice
57
√ the power of P in U ); and it divides 3 just once; hence the power of P in the left-hand side is odd. Hence the norm of the left-hand side contains a (positive or negative) power of 3. But that contradicts the assumption that the right-hand side contains only an even power of 3. A similar argument also works in case the prime p = 3 ramifies as Pe , where e is congruent to 2 mod 4. (We only need the case e = 6). Then again P divides N to an even power (twice the power N ), and it divides U 2 to an even power (twice the power of Pi in U ). Now √ √ eof 3 in e/2 3 = P = ±P , and since e is congruent to 2 mod 4, this is an odd power of P. We need to know how the rational primes split and ramify in Q(ζ). Turning to number theory textbooks for the solution to this problem, we found that many textbooks are content to state the facts only for primes p relatively prime to m, where ζ is a primitive m-th root of unity. But exercise 14, p. 337 of [3], states the facts in full generality: Let p be a rational prime, and let m = pk m′ where p does not divide m′ , then p factors into (P1 . . . Pg )e in Q(ζ), where f g = ϕ(m′ ) and f is the smallest integer such that pf ∼ = 1 mod m′ , the norm of Pi is pf , and the ramification degree e is ϕ(pk ). To prove (i), we apply this result when ζ = e2πi/30 , so m = 30, and p = 3. Then m′ = 10 and k = 1, and f is the smallest integer such that 3f is congruent to 1 mod 10. Hence f = 4. Since ϕ(m′ ) = √ ϕ(10) = 4, we have g = 1. The ramification degree e is ϕ(3) = 2. Hence there is only one Pi = 3, and its norm is 34 . We now show that these assumptions do in fact hold when α = 2π/15. Then β = π/5, and we have shown in Lemma 25 that triangle ABC is isosceles with vertex angle θ = 3β = 3π/5, or θ = β = π/5. We need to compute the norms of sin(2π/15), sin(π/5), and sin(3π/5). We use the definition of the norm of x: it is the product of the images of x under all members of the Galois group. Since the Galois group of Q(ζ) consists of the maps σj that take ζ onto ζ j for j relatively prime to 15, we have the following formulas for the norms: Y 2i sin(4jπ/30) = 1 N (2i sin(2π/15)) = (j,30)=1
N (2i sin(π/5))
=
N (2i sin(3π/5))
=
Y
2i sin(6jπ/30) = 25
(j,30)=1
Y
2i sin(18jπ/30) = 25
(j,30)=1
Since we know in advance that the norm of an algebraic integer is an integer, it suffices to compute these numbers to a few decimal places using an ordinary scientific calculator (or a computer algebra system or a short computer program), and round to the nearest integer. This computation reveals that 2i sin(2π/15) = 2i sin α has norm 1, and hence is a unit in Z[ζ], while 2i sin(π/5) = 2i sin β and 2i sin(3π/5) = 2i sin(3β) each have norm 25. Since the vertex angle θ is either β or 3β, in either case the norm of 2i sin θ is 25. Hence the ratio sin θ/ sin β has norm 1, so it is a unit, and since ϕ(30) = 8, the norm of sin α = sin(2π/15) is 1/28 = 1/256. Hence the norm of the right hand side of (41) is a power of 2, as claimed. We have already shown above that the prime p = 3 ramifies with degree 2 and does not split in this Q(ζ). That completes the proof of (i), i.e. the case α = 2π/15. Now we take up the case of α = π/9. Then 3K = m = 18 and β = π/3 − π/9 = π/6 = 2α. First we consider how the prime p = 3 splits and ramifies. With p = 3 we have m′ = 2 and k = 2, and f is the smallest integer such that 3f is congruent to 1 mod 2. Hence f = 1. Since f g = ϕ(m′ ) = ϕ(2) = 1, we have g = 1. The ramification degree e is ϕ(pk ) = ϕ(32 ) = 6. Hence 3 ramifies as P6 , where P has norm 3. We have shown above that this is sufficient for the proof, provided the norm of the right hand side of (41) contains either no power of 3, or only an even power.
58
It remains only to compute the norm of the right hand side of (41). We compute the norms of 2i sin α and 2i sin β and the two possibilities sin(3α) and sin(5α) for sin θ: Y sin(jπ/9) = −3 N (2i sin(π/9)) = 64 (j,18)=1
N (2i sin(π/6))
=
N (2i sin(π/3))
=
N (2i sin(5π/9))
=
Y
64
sin(jπ/6) = 1
(j,18)=1
Y
64
(j,18)=1
64
Y
(j,18)=1
sin(jπ/3) = −27 sin(5jπ/9) = −3
These values show that the right hand side of (41) has norm 1 or 9, depending on whether the vertex angle θ is 5π/9 or π/3. Since both are even powers of 3, the proof of (ii) is comnplete. That completes the proof of the lemma. Lemma 28 Suppose triangle ABC is N -tiled by tile T , with γ = 2π/3 and α = π/9 or √ α= π/12. Then the area of the tile is not a rational number, and not a rational multiple of 3. Proof. We begin by proving that the area of the tile is not a rational number. Since the sides of the tile are given by a = sin α, b = sin β, and c = sin γ, we can find the area by taking the cross product of two of the sides: AT =
1 sin α sin β sin γ 2
Since 2i sin α = ζ − ζ −1 , and similarly for 2i sin β and 2i sin γ, the area AT is i times an element of the field Q(ζ), and if AT were rational, or even in Q(ζ), then i would belong to Q(ζ). When α = π/9, i does not belong to Q(ζ) (see, for example, p. 145 of [4]), so we are done when α = π/9. Therefore we assume α = π/12. If AT were rational, then AT would be fixed by the Galois group of Q(ζ). The elements of that Galois group are the σj , for j relatively prime to 24, where ζ = e2πi/24 is a primitive 24-th root of unity, and σj takes ζ onto ζ j . Choose j = 5, which is relatively prime to 24. Consider σ5 . Since i = ζ 6 , σ5 takes i to ζ 30 = ζ 6 = i; in other words, σ5 fixes i. It takes i sin α = ζ − ζ −1 to i sin(5α) and, since β = 3α = π/4, it take i sin β to i sin(5β) = −i sin(β). It takes i sin(γ) to i sin(5γ) = −i sin(γ). Thus σ5 changes the signs of sin β and of sin γ; hence it fixes sin β sin γ. Assume, for proof by contradiction, that the area of the tile is rational. Then AT = sin α sin β sin γ is fixed by σ5 ; since sin β sin γ is fixed, sin α must also be fixed. Since i is fixed that means i sin α is also fixed. But as we have already observed, i sin α goes to i sin(5α) = i cos(α), which is not equal to i sin α. Hence the area of the tile is not rational, also in case α = π/12. √ Next we will show more: the area of √ the tile is not a rational multiple of √3. We have proved above that σ5 changes the sign of i 3 and fixes i; so it√changes the sign of 3. Assume, for proof by contradiction, that AT is a rational multiple of 3. Then σ5 changes the sign of AT = sin α sin β sin γ. But since it fixes sin β sin γ, it changes the sign of sin α. But on the contrary, it takes sin α to cos α 6= − sin α. That completes the proof of the lemma. Lemma 29 Suppose triangle ABC is N -tiled by a triangle T whose largest angle γ = 2π/3, and suppose ABC is not similar to T . Then α 6= π/12, where α is the smallest angle of T . Proof. Suppose, for proof by contradiction, that triangle ABC is N -tiled by the tile T mentioned in the lemma, and T is not similar to ABC. We start by calculating sin α and cos α: sin α
=
sin
π 12
59
= = = = cos α
= = = =
“π
π” 3 4 π π sin cos − cos 3 4 √ √ √ 1 2 3 2 − 2 2 2 2 √ √ 1 ( 6 − 2) 4 π cos 12 π π cos cos + sin 3 4 √ √ √ 1 2 3 2 + 2 2 2 2 √ √ 6+ 2 4 sin
−
π π sin ) 3 4
π π sin 3 4
We calculate the area of the tile AT
=
sin α sin β sin γ/2
=
sin(π/12) sin(π/4) sin(2π/3) √ √ √ 6− 2 1 3 √ 4 2 2 √ 3 3 − 8 8
= =
√ Hence AT belongs to Q( 3), √ a proper subfield of Q(ζ). 3), it is fixed by all elements σ of the Galois group of Q(ζ) that Since A belongs to Q( T √ fix 3. The elements of the Galois group are σj for j relatively prime to 24. In particular j = 13 will be of interest. The element σj takes ζ to ζ j . Since 2i sin(jα) = ζ j − ζ −j , σj takes i sin(α) to i sin(jα). Since β and γ are multiples of α, we also have that σj takes i sin(β) to i sin(jβ) and i sin γ to i sin(jγ). With j = 13 then we find that σ13 fixes i sin γ, σ13 fixes i, σ13 takes i sin sin β, and i sin α to −i sin α. Since σ13 fixes ρ = eiπ/3 = ζ 4 and σ13 fixes i, √ β to −i√ it fixes 3, since 3 = 4i(ρ − ρ−1 ). Hence σ13 fixes AT . Since 13 is congruent to 1 mod 12, sin(13jα) = ± sin(jα) for every integer j, so σ13 takes i sin(jα) to i sin(jα) for j even and to −i sin(jα) for j odd. Since it fixes i, it takes sin(jα) to ± sin(jα), according as j is even or odd. Fix two sides U and V of ABC and let θ be the angle between those sides. Since all the angles of ABC are composed of angles α, β, and γ (although we know γ cannot occur, we do not need that fact here), θ is an integer multiple of α, θ = jα for some j. We then have from the area equation N AT = U V sin θ/2 We have, for some nonnegative integers p, q, and r, that U
=
pa + qb + rc
=
p sin α + q sin β + r sin γ
where not both p and q are zero, because all the sides of ABC have degree 2. For notational simplicity we abbreviate σ13 to just σ. We now apply σ to U : Uσ
=
p(sin α)σ + q(sin β)σ + r(sin γ)σ
=
−p sin α − q sin β + r sin γ
60
Since p and q are nonnegative, we have |U σ| ≤ U . By the same reasoning we have |V σ| ≤ V . Since θ = jα for some j, we have (sin θ)σ = ± sin θ. Then AT
=
AT σ
=
(U V sin θ)σ
=
(U σ)(V σ)(sin θ)σ
=
±(U σ)(V σ) sin θ
since (sin θ)σ = ± sin θ
Taking absolute values we have AT
= = = ≤
=
|AT |
|(U σ)(V σ) sin θ|
|(U σ)||(V σ)|| sin θ|
U V sin θ
since sin θ > 0
AT
Therefore equality holds throughout. Hence |Uσ | = U and |Vσ | = V . That implies that, for each of U and V , either p = q = 0 (the side is composed of only c edges) or r = 0 (there are no c edges). If r = 0 then U σ = −p sin α − q sin β is negative, while if p = q = 0 then U σ is positive. Let J be the integer such that θ = Jα. Then sin θ changes sign under σ if and only if J is odd. Say that a side of triangle ABC is “of type c” if it is composed entirely of c sides of tiles, i.e. p = q = 0 for that side. Suppose J is even. Then sin θ does not change sign under σ, so either both of U and V change sign (i.e. neither is of type c), or neither does (i.e. both are of type c). If J is odd, then under σ, exactly one of U and V must change sign under σ, i.e. exactly one is of type c. On the other hand, if J is even, then either both the adjacent sides are of type c, or neither is of type c. For example, if ABC is similar to T and the tiling is a quadratic tiling, then the long side BC is of type c, and the other two sides are not of type c; one of them has p = 0 and the other has q = 0. Now we consider the effect of σ7 . This automorphism changes the sign of i, and takes 2i sin α = ζ − ζ −1 to ζ 7 − ζ −7 = 2i sin 7α = −2i cos α. Hence σ7 takes sin α to cos α. In general σ7 takes sin Jα to − sin((7J mod 24)α, so it fixes sin β ( where J = 3) and changes the sign of sin γ (where J = 8.) We have 2N AT
N sin α sin β sin γ
= =
AABC
U V sin Jα
Suppose, for proof by contradiction, that angle A = α. Then the tile at A has its b side on one side of ABC and its c side on the other. Let U be the side on which there is a c side, and V the one on which there is a b side. Then U is of type c, and since σ = σ7 changes the sign of sin γ = c, U σ = −U . On the other hand if V = pa + qb, then V σ = p cos α + qb ≥ V , with equality holding if and only if p = 0. Canceling sin α from both sides of the area equation, we have N sin β sin γ = U V Applying σ to both sides, the left hand side changes sign, and we have −N sin β sin γ = −U (Vσ ) Adding these two equations we find V = Vσ . Hence p = 0, i.e. only b sides of tiles occur on side U . Suppose that T1 is the copy of T with its b side along U and one end at an endpoint of U where ABC has an angle less than γ; Let V1 and V2 be the endpoints of this b side and suppose
61
the γ angle of T1 is at V2 . Then the b side of T1 ends at V2 , and since only b sides of tiles occur on DE, there is another tile T2 on DE sharing vertex V2 , with its b side on DE. Therefore its β angle is not at V2 . Its γ angle cannot be at V2 since the γ angle of T1 is there and 2γ > π. Therefore T2 has its α angle at V2 . Hence its γ angle is at the other end of its b side on U ; call that vertex V3 . Now we are in the same position with respect to T2 and vertex V3 as we formerly were with respect to T1 and V2 , and by induction there must be a sequence of such pairs of tiles, reaching all the way to the endpoint of U . Therefore the angle at that endpoint is at least γ, so that endpoint is vertex C of ABC. It cannot be the case that angle C = γ, since that would make ABC similar to T . It cannot be the case that angle C = γ + 2α, since if it were, then the third angle of ABC would be α, making ABC isosceles, which contradicts Lemma 25. Therefore angle C = γ + α and the third angle is 2α. The length of side AC (which is of type c) is qb = q sin β for some integer q. By the law of sines, since 2α is the angle opposite side AC, the length of side AB is sin(γ + α)/ sin 2α times the length of AC. But since AB is of type c, there is an integer r such that AB = rc. Equating the two expressions for AB, we have rc r sin γ sin 2α q √ r 3 sin 2α q 2 r√ 3(sin α cos α) q
=
sin(γ + α) q sin β sin 2α
=
sin(γ + α) sin β
=
sin β(sin γ cos α + cos γ sin α) √ “√ ” √ 1 2 3 cos α + sin α since sin β = 1/ 2 2 2 2
=
since c = sin γ
Putting in the values calculated at the beginning of the proof for sin α and cos α we have √ “√ √ √ √ √ √ √ √ √ √ 1 6 − 2” 2 3 6+ 2 r 3 6− 2 6+ 2 = + q 2 4 4 2 2 4 2 4
Multiplying by 16 and simplifying, we have √ √ √ √ r√ 8 3 = 2(2 3 + 2 6 − 2) q √ √ = 2 6+4 3−2 √ √ √ √ √ This is an equation in Q( 3, 2), which is of degree 4 with basis { 2, 3, 6}. Hence the constant coefficients on each side must be equal; but on the left the constant coefficient is zero and on the right it is −2. This contradiction shows the impossibility of A = α. Now suppose the smallest angle of ABC is 3α = β = π/4. Since ABC is not isosceles by Lemma 25, the only possibility for the other angles is B = 4α = π/3 and C = 5α = γ − α. Either side AB or side AC is of type c, since 3 is odd. If side AC, which is opposite angle B = 4α, is of type c, then for some integer r we have AC = rc. Then side AB, which is opposite angle 5α, has length rc sin(5α)/ sin(4α). The area equation becomes N sin α sin β sin γ
= =
N sin α sin γ sin 4α √
3 N sin α 2 √ √ √ 6− 2 3 N 4 2
= = = =
sin A AC AB sin 5α r 2 c2 sin β sin 4α r 2 c2 sin 5α r 2 c2 (sin α cos γ + cos α sin γ) ” “ √3 1 r 2 c2 sin α + cos α 2 2 √ “√ √ √ √ 1 6 + 2” 3 6− 2 + 2 4 2 4
62
Multiplying by 8 and simplifying, we have √ √ √ N (2 3 − 6) = 4r 2 c2 2 √ √ √ √ √ This √ is an equation in Q( 3, 2), for which {1, 2, 3, 6} is a basis. Hence the coeffiencts of 6 on both sides must be equal, but on the left we have −N and on the right 0. This contradiction shows that it is impossible that angle A is 3α. The smallest angle of ABC cannot be more than 4α = π/3. By Lemma 25, it cannot be equal to π/3 either, since that would make ABC equilateral. Therefore the smallest angle A is 2α, since α and 3α have already been ruled out. Either sides AC and AB are both of type c, or neither is. First suppose they both are of type c. Then there are two integers r and s such that the lengths of AC and AB are rc and sc, respectively. Then the area equation becomes N sin α sin β sin γ
=
rsc2 sin 2α
=
2rsc2 sin α cos α
Canceling sin α and putting in the values of the other trig expressions, we have √ √ √ √ 2 3 6+ 2 N = 2rsc2 2 2 4 √ √ rsc2 √ 6 = ( 6 + 2) N 4 2 √ √ √ This is an equation in Q( 2, 3), and the coefficients of 2 are not equal on the two sides of the equation. This contradiction shows that not both AB and AC can be of type c. Hence neither AB nor AC is of type c. Therefore both AB and AC are composed entirely of a and b tile edges. Let σ = σ7 be the automorphism of Q(ζ) defined above. Then σ7 fixes sin β and sin 2α but takes sin α to cos α, and changes the sign of γ. Let U be the length of AB and V the length of AC. The area equation is N sin α sin β sin γ = U V sin 2α Applying σ we have −N cos α sin β sin γ
−U σ V σ sin 2α
=
−U σ V σ2 sin α cos α
=
Multiplying both sides by − sin α/ cos α we have N sin α sin β sin γ
=
U σ V σ2 sin2 α
Dividing this equation by (42) we have 1
= = =
Hence
U σ V σ 2 sin2 α U V sin 2α U σ V σ 2 sin2 α U V 2 sin α cos α U σ V σ sin α U V cos α
cos α Uσ V σ = UV sin α
63
(42)
On the other hand, U = pa + qb for some nonnegative integers p and q, and V = na + mb for some nonnegative integers n and m. We have U
=
p sin α + q sin β
Uσ Uσ U
=
p cos α + q sin β p cos α + q sin β p sin α + q sin β
=
with equality holding if and only if q = 0. To make the equations shorter we introduce d = cos α and and use the already-defined a = sin α, so we have d Uσ V σ (pd + q)(nd + m) = = a UV (pa + q)(na + m)
(43)
This is possible if n = 0 = q or p = 0 = m. We claim those are the only conditions under which this equation is solvable. To prove this we cross-multiply and subtract, obtaining a(pd + q)(nd + m) − d(pa + q)(na + m) = 0 Expressing a and d in terms of ζ = eiπ/12 and expanding and simplifying (using a computer algebra system–we used Sage), one finds mqζ −1 + npζ 3 + npζ = 0 Multiplying by ζ we have
mq + npζ 2 + npζ 4 = 0
The field Q(ζ) has degree ϕ(24) = 8; hence all three coefficients are zero. That is, mq = 0 and np = 0. Hence either n = 0 or p = 0, and either m = 0 or q = 0. If p = 0 then by (43) we have nd + m d = a na + m which implies m = 0. Similarly if q = 0 we find n = 0. That establishes our claim about the conditions for solvability of (43). We have proved that one of the two sides U , V is composed entirely of a sides of tiles, and the other is composed entirely of b sides. We may assume that U is the one composed entirely of a sides, i.e. U = pa = p sin α and V = nb = n sin β. The area equation (42) then becomes N sin α sin β sin γ
=
pn sin α sin β sin 2α
Canceling sin α sin β we have N sin γ
=
pn sin 2α
=
2pn sin α cos α
and putting in the values for sin γ and sin 2α, we find √ √ √ √ √ 3 6− 2 6+ 2 N = pn 2 4 4 pn = 4 √ contradicting the irrationality of 3. This contradiction shows that angle A cannot be 2α, and since that was the last remaining possibility, that completes the proof of the lemma. Lemma 30 Suppose ABC is N -tiled by a triangle T whose largest angle γ = 2π/3, and ABC is not similar to T . Then α 6= π/9, where α is the smallest angle of T .
64
Proof. Note that ζ = e2πi/18 has degree ϕ(18) = 6 over Q. Since Q(ζ) does not contain i (because 4 does not divide 18), the field Q(sin α, cos α, i) has degree 12 over Q. Hence its real subfield Q(sin α, cos α) has degree 6. Hence sin α and cos α are both irrational, as if either were rational then the degree would be 4. The automorphism σj of Q(ζ) takes ζ to ζ j , where j is relatively prime to 18, namely j = 1, 5, 7, 11, 13, 17. Let a = sin α, b = sin β, c = sin γ, and d = sin 4α = sin 5α. Note that i does not belong to Q(ζ), but 2i sin mα = ζ m − ζ −m does. There is no automorphism that fixes exactly two of a, b, and c. This complicates the algebra. We will work with σ5 , which takes ib to −ia, and ic to −ic, and ia to id. Suppose, for proof by contradiction, that angle A = α. Let U and V be the lengths of the sides AB and AC. Then the area equation is N sin α sin β sin γ = U V sin α Canceling sin α we have N bc = U V
(44)
Let U = pa + qb + rc and V = ma + nb + ℓc. Then (iU )σ
ipd − iqa − irc
=
(iV )σ
=
(U V )σ
=
imd − ina − iℓc
(pd − qa − rc)(md − na − ℓc)
Applying σ to (44) (after inserting two factors of i on each side) we have N ac
=
(pd − qa − rc)(md − na − ℓc)
Dividing this equation by (44) we have b a
=
(pa + qb + rc)(ma + nb + ℓc) (pd − qa − rc)(md − na − ℓc)
This equation is solvable when V is composed entirely of c sides (so n = m = 0) and U is composed entirely of b sides (so p = r = 0), as in the quadratic tiling; and vice-versa when V is composed entirely of c sides and U entirely of b sides. We will show that these are the only possible solutions. Cross multiplying we have a(pa + qb + rc)(ma + nb + ℓc) − b(pd − qa − rc)(md − na − ℓc) = 0 Then ζ 14 times this expression is a polynomial in ζ. Taking its remainder under division by the minimal polynomial of ζ, which is ζ 6 − ζ 3 + 1, we find (using the PolynomialRemainder function in Mathematica) 0
=
−2mp + np + mq − 2nq − 3ℓr
+ζ(2mp − np − mq + 2nq + 3ℓr)
+ζ 2 (−6ℓp + 4mp − 2np − 2mq − nq − 6mr) +ζ 3 (4mp − 2np − 3q + 4nq + 6ℓr)
+ζ 4 (−4m + 2np + 3mq − 4nq − 6r)
+ζ 5 (3ℓp − 2mp + np + mq + nq + 3mr)
Each of the five coefficients is zero; but up to constant multiples there are really only two different coefficients, so we have the following two equations: 2mp + 2nq + 3ℓr
=
np + mq
4mp
=
2np + 2mq + 6mr
65
Adding twice the first equation to the second we find 6mr + 6ℓp + 2nq + 3ℓr = 0. Since all these quantities are nonnegative, this implies mr = ℓp = nq = ℓr = 0. Eliminating these terms, our two equations both become 2mp = np + mq
(45)
Suppose ℓ 6= 0. Then since ℓp = 0 and ℓr = 0 we have p = r = 0. Then q 6= 0 because p + q + r is the number of tiles on side U . Since nq = 0 we have n = 0. Then by (45) we have 0 = mq; and since q 6= 0 we have m = 0. This is the first solution mentioned above, in which only ℓ and q are nonzero; we have proved that this is the only solution with ℓ 6= 0. Now suppose ℓ = 0. Then the c side of the tile at vertex A lies on side V , since it cannot lie on side U and cannot lie opposite angle A. Hence r 6= 0. Since mr = 0 we have m = 0. Then by (45), we have np = 0. But n 6= 0 since ℓ and m are both zero. Hence p = 0. Since nq = 0 and n 6= 0, we have q = 0. Thus if ℓ = 0, it follows that only n and r are nonzero; that is the second solution mentioned above. We have thus proved our claim that one of these two solutions must be the case. Then one of the sides U , V is composed entirely of b sides of tiles. Let T1 be the tile sharing vertex A = V1 . Let V2 on side U be the vertex at the other end of the b side of T1 . Then the γ angle of T1 is at V2 . Hence the γ angle of the next tile T2 on U cannot be at V2 ; and since the b side of T2 is on U , the β angle is opposite that side, and not on U . Hence the β angle of T2 is at V2 , and the γ angle of T2 is at the next vertex V3 . Continuing in this way down side U , we find that the last tile Tk on U has its γ angle at the vertex of ABC, which must be vertex C. Therefore angle C is at least γ. Suppose, for proof by contradiction, that it exceeds γ. Then angle C must be γ + α and the third angle must be α, making ABC isosceles. But that contradictions Lemma 25. This contradiction shows that angle C is equal to γ. But then ABC is similar to T , which is contrary to hypothesis. That completes the proof by contradiction that angle A cannot be equal to α. The smallest angle A of ABC cannot be 3α, since then ABC would be equilateral, contradicting Lemma 25. Hence angle A must be 2α = β. Then angle B is at least 3α, since ABC cannot be isosceles. That means angle C is at most 4α. But since we cannot have angle B equal to angle C, the only remaining possibility is that angle B = 3α and angle C = 4α. Let U and V be the sides AB and AC (in some order). Then the area equation is N sin α sin β sin γ = sin βU V or in terms of a, b, and c, it is N abc = bU V Dividing sides by b and setting U = pa + qb + rc and V = ma + nb + rc, we have N ac = (pa + qb + rc)(ma + nb + ℓc) Applying σ we have −N dc = (pd − qa − rc)(md − na − ℓc)
Dividing these two equations we find −
a pa + qb + rc ma + nb + ℓc = d pd − qa − rc md − na − ℓc
Cross multiplying, we have a(pd − qa − rc)(md − na − ℓc) + d(pa + qb + rc)(ma + nb + ℓc) = 0.
66
After multiplying by ζ 16 , this becomes a polynomial in ζ. Reducing that polynomial modulo the cyclotomic polynomial 1 − ζ 3 + ζ 6 (using Mathematica) we find 0
=
2mp − np − mq + 2nq + 3ℓr
+ζ(mp − 2np − 3ℓq − 2mq + nq − 3nr)
+ζ 2 (−4mp + 2np + 2mq − 4nq − 6ℓr)
+ζ 3 (−4mp + 2np + 2mq − 4nq − 6ℓr)
+ζ 4 (mp − 2np − 3ℓq − 2mq + nq − 3nr)
+ζ 5 (2mp − np − mq + 2nq + 3ℓr)
Equating the coefficients to zero gives us only two different equations: 2mp − np − mq + 2nq + 3ℓr
mp − 2np − 3ℓq − 2mq + nq − 3nr
=
0
=
0
Subtracting the second equation from the first we have 3ℓq + mp + np + mq + nq + 3ℓr + 3nr = 0 Since all these quantities are non-negative this implies that each of ℓq, mp, np, mq, nq, ℓr, and nr is zero. Suppose, for proof by contradiction, that ℓ 6= 0. Then q = r = 0, so p 6= 0. Hence n = m = 0. Hence U is composed only of a sides of tiles, and V is composed only of c sides. The tile T1 at A that shares side U therefore has its a side on U and its β angle at A. Hence its γ angle is at vertex V2 , the other vertex of T1 on U . Hence the next tile T2 on side U has its β angle at V2 (since its a side is on U and its γ angle is too large to fit). Hence it has its γ angle at the next vertex V3 . Continuing down side U in this way we find that the last tile on side U has its γ angle at the other vertex of side U of ABC. But this is impossible, as the largest angle of ABC is 4α, which is less than γ. This completes the proof by contradiction that ℓ = 0. Similarly, if r 6= 0 then n = ℓ = 0, so m 6= 0. Hence p = q = 0, and by considering tiles on side V we find that the angle at the other end of side V must be at least γ, contradiction. Hence r = ℓ = 0. Then no tile at A has a c side on the boundary of ABC, so the vertex at A must be split, and there are two tiles T1 and T2 with their α angles at A, sharing their c sides and having their b sides along U and V respectively. Hence q 6= 0 and n 6= 0. But this contradicts nq = 0. That contradiction completes the proof of the lemma. Theorem 4 Suppose ABC is N -tiled by a triangle T whose largest angle γ = 2π/3. Suppose T is not isosceles, i.e. is not the tile used in the equilateral 3-tiling. Then T is similar to ABC. Proof. Assume, for proof by contradiction, that T is not similar to ABC. Then by Lemma 25, α is π/9, or π/12, or 2π/15. First suppose that α = 2π/15. By Lemma 25, ABC is isosceles with base angle β or 2β; hence the vertex angle is 3β or β. But these cases have been ruled out in Lemma 27. That disposes of the possibility α = 2π/15. The case α = π/9 is ruled out in Lemma 30, and the case α = π/12 is ruled out in Lemma 29. That completes the proof of the theorem.
12
A non-isosceles tile T with largest angle 2π/5
Lemma 31 Suppose ABC is N -tiled by a triangle T . Suppose ABC is not similar to T , and T is not isosceles. Then the largest angle γ of T cannot be 2π/5. Proof. Since ABC is not similar to T , we have vertex splitting. As before, let P , Q, and R be the total number of α, β, and γ angles, respectively, at vertices of ABC. Since T is not similar to ABC, we have P + Q + R ≥ 5. Since T is not isosceles, α < β. We have β > (π − γ)/2 = 3π/10 = (3/4)γ. The number R of γ angles is at most 2, since 3γ = 6π/5 > π.
67
Assume, for proof by contradiction, that R = 2. Then P α + Qβ = π/5, and since β > 3π/10 > π/5, we must have Q = 0. Hence α = π/(5P ) and β = π − γ − α = 3π/5 − π/(5P ). Since β < γ, we have π 2 3 π− < π 5 5P 5 3 1 2 − < 5 5P 5 3P − 1 < 2P
P 3π/10, we must have Q ≤ 1. We have two equations for α and β: P α + Qβ
=
α+β
=
3π 5 3π 5
since R = 1 since α + β + γ = π
We cannot have P = Q = 1, since P + Q ≥ 5. Hence P 6= Q, and the equations are uniquely solvable for α and β. By Cramer’s rule we have α
=
β
=
3π 5 3π 5
1−Q P −Q P −1 P −Q
Since Q ≤ 1, the equation for α shows that Q = 0, since α > 0. Hence α
=
β
=
3π 5P 3π P − 1 5 P
We have β < γ = 2π/5. Hence 3π P − 1 5 P
α. Hence the angle at the other endpoint of U is less than β, whether that endpoint is A or B. Let T1 be the tile along U at that vertex of ABC. Then T1 has its α angle in the corner of ABC. But then it cannot have its a side along U , contradiction. That completes the proof that θ 6= γ. Returning to (47), the last equation before we assumed θ = γ, we now assume, for proof by contradiction, that θ = α. Then sin θ = a and on the left we have N bc/a2 . Substituting the expressions derived above for b/a and c/a on the left, we have N (3 − 4a2 )(−64a6 + 112a3 − 56a2 + 7)
72
=
256(qℓ + rn)a8 −(640(qℓ + rn) + 64(ℓp + ℓr + mr))a6
+(16(q + 14r)(14ℓ + n) + 112(p + 3q + 7r)ℓ + 112(7ℓ + m + 3n)r − 4576ℓr)a4 +(−4(q + 14r)(7ℓ + m + 3n) − 4(14ℓ + n)(p + 3q + 7r) + 704ℓr)a2 +(p + 3q + 7r)(7ℓ + m + 3n) − 33ℓr
Expanding the left hand side we have
=
N (256a8 − 640a6 + 560a4 − 196a2 + 21)
256(qℓ + rn)a8
−(640(qℓ + rn) + 64(ℓp + ℓr + mr))a6
+(16(q + 14r)(14ℓ + n) + 112(p + 3q + 7r)ℓ + 112(7ℓ + m + 3n)r − 4576ℓr)a4 +(−4(q + 14r)(7ℓ + m + 3n) − 4(14ℓ + n)(p + 3q + 7r) + 704ℓr)a2 +(p + 3q + 7r)(7ℓ + m + 3n) − 33ℓr
Now we can equate coefficients of like powers of a. From the coefficient of a8 we see N = qℓ+rn. Then from the coefficient of a6 we have 640N
=
640(qℓ + rn) + 64(ℓp + ℓr + mr)
=
640N + 64(ℓp + ℓr + mr)
Subtracting the left side from the right we see ℓp = ℓr = mr = 0. Turning to the coefficient of a4 , we have 560N
= = =
16 · 14(qℓ + rn) + 16qn + 3 · 112(qℓ + rn) 560(qℓ + rn) + 16qn
560N + 16qn
Subtracting the left side from the right, we see qn = 0. Finally, from the constant term we have 21N
=
pm + 3pn + 3qm + 21(qℓ + rn)
=
pm + 3pn + 3qm + 21N
Subtracting the left side from the right, we see pm = 0 = pn = qm. Since we already deduced pℓ = 0 we now have p(m + n + ℓ) = 0, which implies p = 0. Similarly, having deduced mp = mq = rm = 0, we have m(p + q + r) = 0, which implies m = 0. Now, if q 6= 0 then since nq = 0 we would have n = 0; and then since m = 0 we would have ℓ 6= 0, and since ℓr = 0 we would have ℓ = 0. Thus if q 6= 0 then U is composed entirely of b sides of tiles (since p = r = 0) and V is composed entirely of c sides of tiles (since m = n = 0). On the other hand, if q = 0 then since p = 0, we would have r 6= 0, so ℓ = 0, so n 6= 0, and then U would be composed entirely of c sides of tiles and V entirely of b sides. Hence in either case, one of U and V is composed entirely of b sides; we may assume without loss of generality that it is U . Let T1 be the tile at vertex A (there is only one since the angle there is α). Let V2 be the next vertex on side U , at the other end of the b side of T1 . Then T1 has its γ angle at V2 , since it has its β angle opposite its b side on U , and its α angle at A. Let T2 be the next tile on side U . Then T2 must have its α angle at V2 , since the γ angle will not fit, and the β angle must be opposite the b side of T2 , hence in the interior of ABC. Let V3 be the next vertex on side U . Then T2 has its γ angle at V3 . Continuing in this fashion down side U , we eventually reach the other end of U , and the last tile Tk must have its γ angle at the vertex C of ABC, and U is the side AC. But angle C cannot be exactly γ, since then ABC would be similar to T . It must be γ + α, since that allows angle B to be 2α, and any larger value of angle C will force angle B to be
73
α, making ABC isosceles, contradicting Lemma 25. Let P be the point on side AB such that angle ACP is γ; since angle C is more than γ, such a point P exists. Then triangle ACP is similar to T , and the similarity factor is q, since the side opposite the β angle of ACP is qb. Hence the length of AP is qc and the area of triangle ACP is q 2 AT . Since the length of AP is qc and that is less than the length of AB, which is ℓc, we have q < ℓ. Observe that triangle CP B is similar to triangle ABC, since angle P CB is α and the two triangles share angle B = 2α. The length of P C is qa, because q is the similarity factor between T and triangle ACP . Side P C is opposite angle B in P CB; and AC, which has length qb, is opposite angle B in triangle ABC. The similarity factor between CP B and ABC is thus a/b. We now decompose triangle ABC into triangle ABP and triangle CP B (although we do not claim that segment CP belongs to the tiling). We have AABC
AACP + ACP B a AABC b
=
q 2 AT +
=
Solving for AABC we find AABC
q2 AT 1 − ab
=
On the other hand AABC = N AT because ABC is N -tiled by T . Hence N AT
q2 AT 1 − ab
=
Canceling AT and simplifying, we find q2
=
q2 b
=
“ a” N 1− b N (b − a)
Substituting b = 3a − 4a3 and canceling a from both sides we have q 2 (3 − 4a2 )
−4q 2 a2 + 3q 2
= =
N (2 − 4a2 )
−4N a2 + 2N
Comparing the coefficients of a2 we have N = q 2 . Comparing the constant coefficients we have q 2 = 2N/3. Since N > 0 this is a contradiction. This contradiction completes the proof that triangle ABC does not have an α angle. Recall that triangle ABC is not isosceles, by Lemma 25. The smallest angle of triangle ABC cannot exceed 2α, since if it is 3α or more then the second smallest angle is at least 4α, and the largest angle would exceed 5α, making the total at least 12α, while it must be only 11α = π. The smallest angle of triangle ABC must therefore be exactly 2α. Then the second smallest angle is at least 3α = β. If the next smallest angle is β then the largest angle is γ, which we have shown to be impossible. Hence the second smallest angle must be at least 4α. If it is 4α then the third angle is 5α. The second smallest cannot be 5α or more because then the sum of all three angles would exceed π = 11α. Hence there is only one possible shape of ABC remaining when angle A = 2α, angle B = 4α, and angle C = 5α. But that shape is ruled out by hypothesis. That completes the proof of the lemma. Lemma 33 Let ABC be tiled using as a tile the triangle T with α = π/14, β = 4π/14, and γ = 9π/14. Suppose angle C is equal to γ or γ + α. Then ABC is similar to T .
74
Proof. Let ζ = e2πi/28 . The degree of Q(ζ) over Q is ϕ(28) = 12. Since 28 is divisible by 4, i belongs to Q(ζ). Let σ = σ15 be the automorphism of Q(ζ) that takes ζ to ζ 15 . Then σ takes i to −i. Since 2 sin(jα) = −i(ζ j − ζ −j ), σ takes sin(jα) to − sin(15jα), so σ fixes a = sin α and also fixes c since cσ
= =
− sin(15 · 9α)
− sin 23α
=
sin 5α
=
sin 9α
=
sin γ = c
On the other hand bσ = − sin 60α = − sin 4α = −b. Let U and V be the lengths of two sides of ABC. Then let the number of a, b, and c sides of tiles on side U be p, q, and r respectively, and let m, n, and ℓ be the number of a, b, and c sides on V . Then we have U
=
pa + qb + rc
V
=
ma + nb + ℓc
Uσ
=
Vσ
=
pa − qb + rc
ma − nb + ℓc
For some integer J, the angle of ABC between sides U and V is θ = Jα. The area equation is 2N AT
=
2AABC
sin α sin β sin γ
=
U V sin θ
N abc
=
U V sin C
=
(pa + qb + rc)(ma + nb + ℓc)
Applying σ we have abc = (pa − qb + rc)(ma − nb + ℓc) sin θ
Equating the right sides of the last two equations and canceling sin C we have (pa + qb + rc)(ma + nb + ℓc) = (pa − qb + rc)(ma − nb + ℓc) That can be written as ((pa + rc) + qb)((ma + ℓc) + nb) = ((pa + rc) − qb)((ma + ℓc) − nb)) Multiplying out and dropping terms that appear on both sides, we have qb(ma + ℓc) + nb(pa + rc) = −qb(ma + ℓc) − nb(pa + rc) and subtracting the right side from both sides and dividing by 2 we have qb(ma + ℓc) + nb(pa + rc) = 0. But since all these quantities are nonnegative, we have qm, qℓ, np, and nr all equal to zero. Assume, for proof by contradiction, that q 6= 0. Then m = ℓ = 0, so n 6= 0, so p = r = 0. In that case sides AC and BC both are composed of only b sides of tiles. In that case the area equation becomes N abc = qnb2 sin In case C = γ we have c = sin C, so N a = qnb, which is impossible. In case C = γ + α, we have (finish this). That completes the proof by contradiction that q = 0. That means that no b sides
75
of tiles occur on side U . But side U could have been any side of ABC; hence no b sides of tiles occur on any side of ABC. Hence U = pa + rb and V = na + ℓc. Hence σ fixes both U and V . When we apply σ to the area equation N abc = U V sin θ we get −N abc = U V (sin θ)σ
because σ changes the sign of b but fixes a, c, U , and V . Therefore σ changes the sign of sin θ, for each angle θ of ABC (since U and V could be any two sides). All the angles of ABC are multiples of α. For which J do we have sin Jα not fixed by σ? Exactly for J odd, because (2i sin(Jα))σ
= = = = =
(ζ J − ζ −J )σ
(ζσ)J − (ζσ)−J
(−ζ)J − (−ζ)−J
(−1)J (ζ J − ζ −J ) (−1)J 2i sin(Jα)
so for J even, σ fixes i sin Jα; but since σ changes the sign of i, that means σ fixes sin Jα for J odd. Now the angles of ABC all have the signs of their sines changed by σ. That means that they are even multiples of α. By Lemma 25, ABC is not isosceles. That means that the smallest angle cannot be 4α, as then angle B would be at least 6α and angle C would be at least 8α, which is more than 14α = π. Hence the smallest angle A must be 2α. Then angle B = 4α, because two 6α angles violates Lemma 25. Hence angle C = 8α. We will need to know that a2 , ac, and c2 are linearly independent over Q, so we prove that before continuing. Suppose they are linearly dependent. Then for three rational numbers λ, µ, and ρ, we have 0
=
λa2 + µac + ρc2
=
λ(ζ − ζ −1 )2 + µ(ζ − ζ −1 )(ζ 9 − ζ −9 ) + ρ(ζ 9 − ζ −9 )2
The minimal polynomial of ζ is ψ(x) = x12 − x10 + x8 − x6 + x4 − x2 + 1 as can be found using the minpoly function of Sage or by factoring x28 − 1 in Mathematica or Maple. Replacing ζ by x in the linear-dependence relation and multiplying by x18 , and then taking the polynomial remainder on division by ψ, using Mathematica, and replacing x by ζ again, the result must be zero: 0
=
ρ + (µ − λ)ζ 2 + (2λ − µ + 2ρ)ζ 4 + (µ − λ)ζ 6 + ρζ 8
Since this is expressed in the basis of Q(ζ) consisting of powers of ζ up to the ζ 11 , each coefficient is zero. The constant coefficient is ρ, so ρ = 0. Then the coefficient of ζ 4 is 2λ − µ, so 2λ − µ = 0. But from the coefficient of ζ 2 we have λ−µ = 0. Subtracting these two equations we have λ = 0. Hence µ = 0, completing the proof of linear independence of a2 , ac, and c2 . Choose θ = 4α = β, which is angle B. Then the area equation becomes N abc
=
(pa + rc)(ma + ℓc) sin β
76
Canceling the b on the left with sin β on the right we have N ac
=
(pa + rc)(ma + ℓc)
=
pma2 + (rm + pℓ)ac + rℓc2
Hence pm = 0, rℓ = 0, and N = rm + pℓ. First assume, for proof by contradiction, that ℓ 6= 0. Then r = 0 since rℓ = 0, and then pℓ = N , so p 6= 0. But since pm = 0 we have m = 0. That is, one side, say U , of angle B is composed entirely of c sides of tiles, and the other entirely of a sides. Let T1 be the tile at vertex B with an a side on the boundary of ABC. Then T1 does not have its α angle at B. But the γ angle is larger than angle B = 4α, so T1 has its β angle at B, and fills angle B. Now let V2 be the next vertex on side U , at the other end of the a side of T1 , and let T2 be the next tile on U . Then T1 has its γ angle at V2 , since its c side is opposite this vertex, on the other side of angle B. Then T2 has its β angle at V2 , since its γ angle will not fit and its α angle is opposite the side it shares with U . Let V3 be the next vertex on U ; then we are in the same position with respect to V3 and T2 as we were with respect to V2 and T1 . Hence T2 has its γ angle at V3 . Continuing in this way down side U , we reach the other end of side U , and the last tile on side U has its γ angle at vertex C of ABC. But this is a contradiction, as that angle is only 8α, which is less than γ = 9α. That completes the proof by contradiction that ℓ = 0. Then m 6= 0 since there must be some tiles on each side; then since pm = 0 we have p = 0 and since there must be some tiles on the (pa + rc) side, we have r 6= 0. That is, one side is composed only of c sides of tiles, and the other is composed only of a sides. But we have just shown that these conditions lead to a contradiction. That completes the proof of the lemma. In Lemma 22, when investigating the possible tilings using a right triangle, we could not rule out the case α = π/8. Now we can. The reader who is only interested in checking the proof of the main theorem can skip the following lemma. Lemma 34 Suppose T is a right triangle with α = π/8, and suppose ABC is a right triangle and ABC is N -tiled by T for some N . Then ABC is similar to T . Proof. Let ζ = eiπ/8 . Then the degree of Q(ζ) over Q is ϕ(16) = 8. Let σ = σ9 be the automorphism taking ζ to ζ 9 = −ζ. Then σ fixes i, fixes sin Jα for J even and changes the sign of sin Jα for J odd, since (2i sin Jα)σ
= = = =
(ζ J − ζ −J )σ
(ζσ)J − (ζσ)−J
(−ζ)J − (−ζ)−J
(−1)J 2i sin Jα
Hence σ changes the signs of a = sin α and b = sin β = sin 3α. Since T is a right angle we have c = 1. Let X = pa + qb + r and Y = ma + nb + ℓ, where p, q, r, m, n, and ℓ give the numbers of tiles with sides a, b, and c along X and Y . The area equation N ab = XY becomes is N ab = (pa + qb + r)(ma + nb + ℓ) Applying σ we have N ab = (−pa − qb + r)(−ma − nb + ℓ)
Subtracting the two equations and dividing by 2 we have 0
=
r(ma + nb) + ℓ(pa + qb) = 0
=
rma + rnb + ℓpa + ℓqb
Since all these quantities are nonnegative we have rm, rn, ℓp and ℓq each equal to zero. Assume, for proof by contradiction, that ℓ 6= 0. Then q = p = 0. Hence r 6= 0. Hence m = n = 0. Hence
77
both X and Y are composed entirely of c sides, and hence are rational. Hence ab = XY /N is also rational. But it can be expressed as ab
=
sin α sin 3α
=
(ζ − ζ −1 )(ζ 3 − ζ −3 )
=
ζ 4 − ζ 2 − ζ −2 + ζ −4
Multiplying by ζ 4 we have 0
= =
z8 − ζ 6 + ζ 2 + 1
−ζ 6 + ζ 2
since ζ 8 = −1
But this is a contradiction, since the powers of ζ up to ζ 7 are a basis for Q(ζ). That completes the proof of the lemma.
14
The case when 3α + 2β = π or 2α + 3β = π
For this section only, we drop the assumption that α ≤ β, assuming only that α ≤ γ and β ≤ γ. We consider tilings of a triangle ABC that is not isosceles, by a tile with angles α, β, and γ, where 3α + 2β = π and α is not a rational multiple of π. Note that we must have γ = β + 2α, since γ = π − α − β = 3α + 2β − α − β = 2α + β. The possible shapes of ABC are quite limited: Lemma 35 Let ABC be N -tiled by a tile with angles α, β, and γ, and suppose 3α + 2β = π. Suppose ABC is not similar to the tile and not isosceles, and suppose that α is not a rational multiple of π. Then either ABC has angles 2α, β, and β + α, or ABC has angles α, 2α, and 2β. Proof. Note that β is not a rational multiple of α, since then the relation 3α + 2β = π would make α a rational multiple of π; similarly β is not a rational multiple of π. Suppose first that ABC does not have an α angle. If one angle (say A) is 2α, then the tiling must split that angle into two α angles, since β 6= 2α. Suppose, for proof by contradiction, that the tiing split one of the other two vertex angles of ABC into several α angles. It must be least 3 of them since ABC is not isosceles. Hence at least five α angles are involved at the vertices of ABC. Hence the equation that says the sum of the vertex angles is π is an integral relation of the form nα + mβ + rγ = π, with n ≥ 5. Since γ = β + 2α we have (n + 2r)α + (m + r)β = π. Since n + 2r ≥ 5 this relation is not a multiple of 3α + 2β = π, and hence the two equations can be solved for α and β, which will be rational multiples of π. This contradicts the hypothesis, and the contradiction shows that ABC does not have an angle that splits into α angles. As a consequence of this observation, ABC does not contain an angle larger than γ − α = β + α, since if it did, then the third angle would be less than β and so would have to split into α angles. Hence the other two angles of ABC must be at least β and at most β + α. There is only only possibility, namely the first triangle mentioned in the lemma. Now suppose that ABC does have an α angle. If the second smallest angle is 2α, then the remaining angle must be 2β; that is the second possibility mentioned in the lemma. If the second smallest angle is greater than 2α, it must be at least β, since if not it will split into at least three α angles, giving rise to an integral relation that is not a multiple of 3α + 2β = π. It cannot be exactly β as that would make ABC similar to the tile. Therefore the second angle is more than β, and the third angle is therefore less than γ. If neither of these two angles splits into α angles, then each one contains a β angle plus one or more α angles. But since ABC is not isosceles, they cannot both be β + α; hence at least four α angles are involved in the tiling at the vertices of ABC. That is contradictory, since 4α + 2β > π. Hence one of the second two angles does split into α angles. Since the second smallest angle is more than 2α, at least 3 α angles are required, and that gives rise to an integer relation that is not a multiple of 3α + 2β, contradiction. That completes the proof of the lemma.
78
Lemma 36 Suppose 3α + 2β = π and α is not a rational multiple of π. Suppose triangle ABC is N -tiled by a tile with angles α and β, and there are 5 angles of tiles altogether at the vertices of ABC. Then there is exactly one vertex at which three γ angles and one α angle meet, and all other vertices either have one each of α, β, and γ, or two each. We have N = 3 + Nπ + 2N2π where Nπ is the number of vertices with angle sum π and N2π is the number with angle sum 2π. Proof. In an N -tiling there are N triangles. Each has 3 edges so altogether there are 3N edges. Since 5 angles occur at the vertices of ABC and each has 2 edges, that accounts for 10 edges. Since α is not a rational multiple of π, it follows from 3α + 2β = π that α is not a rational multiple of β or γ, and γ = β + 2α is not a rational multiple of β. Each vertex is therefore of one of the types (1, 1, 1), (2, 2, 2), (0, 1, 3), (3, 2, 0), (6, 4, 0), and (4, 3, 1), where the three numbers in a “type” are the number of α, β, and γ angles at the vertex. Each vertex has angle sum π or 2π depending on its type. There must be one vertex (at least) of type (0, 1, 3) to “balance” the excess of α and β angles over γ angles at the vertices of ABC, since all other types have at least as many α and β angles as γ angles. If there are more vertices of type (0, 1, 3), they must be matched by vertices of types (3, 2, 0), 6, 4, 0), or (4, 3, 1) in such a way that the total number of α angles equals the total number of β angles equals the total number of γ angles. Let us count the total number of edges, which must be 3N . We get 10 at the vertices of ABC. 8 at each (0, 1, 3) vertex. 18 per (0, 1, 3) and (3, 2, 0) pair. 36 per pair of two (0, 1, 3) vertices and one (6, 4, 0) vertex. 24 per (0, 1, 3) and (4, 3, 1) pair. 6 per each (1, 1, 1) vertex. 12 per each (2, 2, 2) vertex. Now we calculate the number of edges divided by the total angle sum (in multiples of π) for each line above. We get 4 per each (0, 1, 3) vertex. 6 per each other vertex or pair listed one a line above. Let N013 be the number of (0, 1, 3) vertices. Since each of these has angle sum 2π, the total number of edges at such vertices is 8N013 . Let Nπ be the number of vertices with angle sum π, and N2π be the number of vertices not of type (0, 1, 3) with angle sum 2π. Then equating the total angle sum of all tiles to (N − 1)π, we have 2N013 + 2N2π + Nπ = N − 1 Hence 2N2π + Nπ
=
N − 1 − 2N013
(48)
Now, counting edges, the total number of edges should be 6N , since each edge counts twice, once at each end. Thus 6N
=
8N013 + 10 + 6(Nπ + N2π )
=
8N013 + 10 + 6(N − 1 − 2N013 )
by (48)
10 + 6(N − 1) − 4N013
6N
=
4N013
=
4
N013
=
1
We have deduced that there is exactly one vertex of type (0, 1, 3). It follows that there are no vertices of type (3, 2, 0), (6, 4, 0), or (4, 3, 1), since if there were, then the total number of α
79
and β angles would exceed the total number of γ angles. There are (besides the single (0, 1, 3) vertex) only standard vertices of types (1, 1, 1) and (2, 2, 2). That completes the proof of the lemma. We next introduce notation and definitions that will be used throughout this section, i.e., regardless of the shape of ABC. As always, a, b, and c are the sides of the tile, and we assume the tile and the triangle ABC are scaled so that a
=
sin α
b
=
sin β
c
=
sin γ
For use in this entire section, we introduce the following notation: z
=
eiπα/2
ζ
=
eiπβ
We note that
z 6 ζ 2 = −1
since 3α + 2β = π. Hence ζ = iz −3 . We have 2ia
=
2ib
=
z 2 − z −2 ζ − ζ −1
=
i(z 3 + z −3 )
2b
=
z 3 + z −3
2 sin 2α
=
2ic
=
−i(z 4 − z −4 )
ei(2α+β) − e−(2α+β)
z 4 ζ − z −4 ζ −1
=
iz 4 ζ −3 + iz −4 z 3
= 2c
=
i(z + z −1 )
=
z + z −1
In other words, sin γ = cos(α/2). We have sin(β + α)
= = =
sin(π − γ)
sin γ 1 (z + z −1 ) 2
We note the factorizations z 2 − z −2
(z + z −1 )(z − z −1 )
=
z 3 + z −3
(z + z −1 )(z 2 − 1 + z −2 )
=
In view of these factorizations we have a c b c
=
= i(z − z −1 )
=
z 2 − 1 + z −2
80
We introduce t s Then t = 2 − s2 , and
:=
2 cos α
=
z 2 + z −2
:=
2 sin(α/2)
=
−i(z − z−1)
a c
=
b c
=
=
= = sin 2α
=
= i(z − z −1 ) s
z 2 − 1 + z −2 t−1
1 − s2
2 sin α cos α
=
at
=
a(2 − s2 )
The d matrix equation is 1 a d@ b A c 0
=
1 X @ Y A Z 0
We introduce individual names for the components of the d matrix to avoid so many subscripts: 1 0 p d e d = @ g m f A h ℓ r
Therefore we have for the three sides X, Y , and Z of triangle ABC X
=
pa + db + ec
Y
=
ga + mb + f c
Z
=
ha + ℓb + rc
Generally our convention is that X, Y , and Z are in order of size, so X is opposite the smallest angle A, and Y is opposite the middle angle B, and Z is opposite the largest angle C. In this section, however, we are not assuming α < β, and that means that we may not have Y < Z either. In case triangle ABC has angles α, 2α, and 2β, we will assume that X is opposite the α angle at A, and the 2α angle is at B, opposite Y , and that Z is opposite the 2β angle. In case ABC has angles 2α, β, and β + α, we will assume X is opposite the 2α angle at A, and Y is opposite the β angle at B, and Z is opposite the β + α angle at C. Thus in case β < α, we may not have X, Y , Z in order of size. Lemma 37 Suppose triangle ABC is N -tiled by a tile in which γ > π/2 (as is the case when 3α + 2β = π). Suppose all the tiles along one side of ABC do not have their c sides along that side of ABC. Then there is a tile with a γ angle at one of the endpoints of that side of ABC. Proof. Let P Q be the side of ABC with no c sides of tiles along it. Then the γ angle of each of those tiles occurs at a vertex on P Q, since the angle opposite the side of the tile on P Q must be α or β. Let n be the number of tiles along P Q; then there are n − 1 vertices of these tiles on the interior of P Q. Since γ > π/2, no one vertex has more than one γ angle. By the pigeonhole principle, there is at least one tile whose γ angle is not at one of those n − 1 interior vertices; that angle must be at P or Q. That completes the proof of the lemma.
81
Lemma 38 Suppose 3α + 2β = π, and suppose triangle ABC has angles 2α, β, and β + α. Let T be a triangle with angles α, β, and γ and suppose that ABC is N -tiled by T in such a way that just five tiles share vertices of ABC. Then the following restrictions on the elements of the d matrix apply: We have e 6= 0, f 6= 0, and r 6= 0. In other words, the third element in each row is nonzero. Proof. None of the angles of ABC is large enough to accommodate the γ vertex of a tile, since those angles are 2α, β, and β + α < β + 2α = γ. The lemma then follows from Lemma 37, since the three numbers e, f , and r are the numbers of c sides of tiles on the three sides of ABC. That completes the proof of the lemma. Lemma 39 Let 3α + 2β = π, and let ABC have angles α, 2α, and 2β, and suppose there is an N -tiling of ABC by a tile with angles α, β, and γ such that exactly two tiles meet at the 2β vertex of ABC. Then we cannot have m = ℓ = 0, i.e. the bottom two entries in the second column cannot both be zero. Also e 6= 0 and f 6= 0 and r 6= 0. Proof. At vertex A, where the angle is α, there is just one tile. It must have its b side on AB and its c side on AC, or vice-versa. The number of b sides on AC is m and the number of b sides on AB is ℓ, so they cannot both be zero. In triangle ABC, no γ angles of tiles can occur at any vertex, since the α and 2α angles are too small and the 2β angle splits into two β angles. Hence the third column of the d matrix cannot contain any zero entries, by Lemma 37. Hence e, f , and r are each nonzero. That completes the proof of the lemma.
14.1
The case when ABC has angles α, 2α, and 2β
Lemma 40 Let 3α + 2β = π, and let ABC have angles α, 2α, and 2β, and let N be arbitrary. Then there is no N -tiling of ABC by a tile with angles α, β, and γ in which exactly two tiles meet at the 2β angle of ABC. (Here we do not assume α < β.) Proof. Let X, Y , and Z be the sides of triangle ABC opposite A, B, and C respectively. Let λ be the ratio of the sides to the opposite angle, which (by the law of sines) is the same for all three sides: X
=
λ sin α
Y
=
λ sin 2α
Z
=
λ sin 2β
We note that sin α
=
a
sin 2α
=
2 sin α cos α
sin 2β
=
ta
=
(2 − s2 )a
= = =
The d matrix equation is 0 p @ g h
d m ℓ
sin 3α
4 sin3 −3 sin2 α 4a3 − 3a2
1 1 0 1 0 10 a X a e f A @ b A = @ Y A = λ @ sin 2α A sin 2β Z c r
82
(49)
As noted above, we have sin 2β 0 p @ g h
= 4a3 − 3a2 10 d e m f A@ ℓ r
and sin 2α = a(2 − s2 ) we have 1 1 0 a a 2 b A = λ @ a(2 − s ) A 4a3 − 3a2 c
Taking the ratio of the second row to the first row, we have ga + mb + f c pa + db + ec
=
2 − s2
Multiplying by the denominator, we have ga + mb + f c
=
(2 − s2 )(pa + db + ec)
Dividing both sides by c and expressing a/c and b/c in terms of s we have b a +m +f c c gs + m(1 − s2 ) + f g
= =
a b + d + e) c c (2 − s2 )(ps + d(1 − s2 ) + e) (2 − s2 )(p
This is a fourth-degree polynomial equation for s over Q. Bringing the equation to polynomial form we have 0
=
ψ(s) := ds4 − ps3 + (m − 3d − e)s2 + (2p − g)s + (2d + 2e − f − m)
(50)
If ψ is identically zero, then from the first two coefficients of ψ in (50) we have d = p = 0. From the coefficient of s2 we have m − 3d − e = m − e = 0, so m = e. From the coefficient of s we have 2p − g = 0, and since p = 0 we have g = 0. From the constant coefficient we have 2d + 2e − f − m = 2e − f − m = e − f , so f = e. The d matrix then has the form 1 0 0 0 e @ 0 e e A h ℓ r This is not immediately contradictory, and we shall return to proving that ψ is not identically zero below. From the first row of the d matrix equation we have pa + db + ec = λa Dividing by c and using a/c = s and b/c = 1 − s2 we have λs = ps + d(1 − s2 ) + e Solving for λ we have e+d s The d matrix equation (49) can be written as an eigenvalue problem this way: 0 1 1 0 10 p d e a a f b gb mb @ A@ b A = λ@ b A sin 2α sin 2α sin 2α hc ℓc rc c c sin 3α sin 3α sin 3α λ = p − ds +
(51)
To find an eigenvector by the cofactor method, we need to compute the cofactors of the matrix 1 0 p−λ d e fb mb @ gb A −λ sin 2α sin 2α sin 2α hc ℓc rc −λ sin 3α sin 3α sin 3α
83
Taking the cofactors of the third row, and multiplying by sin 2α, we find a candidate for an eigenvector (it is only a candidate until we prove that its components are all nonzero): 1 1 0 0 u (df − em)b + eλ sin 2α A @ v A = @ (eg − pf )b + f λb (52) w (pm − dg)b − (mb + p sin 2α)λ + λ2 sin 2α
Now assume, for proof by contradiction, that ψ matrix then has the form 0 0 0 @ 0 e h ℓ
Then we have
is identically zero. As shown above, the d 1 e e A r
1 0 1 u −be2 + eλ sin 2α A @ v A = @ ebλ w ebλ + λ2 sin 2α 0
Since e 6= 0 (otherwise the whole first row of the d matrix is zero), the second two components are nonzero. Suppose, for proof by contradiction, that u = 0. Then since e 6= 0 we have eb
=
λ sin 2α
Since λ sin 2α = ga + mb + f c = eb + ec, we have eb = eb + ec, and hence ec = 0. Since e 6= 0 we have c = 0, a contradiction. This contradiction shows that u 6= 0 (still under the assumption that ψ is identically zero). Having proved u 6= 0, we conclude that the eigenspace has dimension 1, and (u, v, w) is a multiple of (a, b, c). Therefore v/w = b/c. Cross multiplying, we have vc = bw. Putting in v = ebλ and w = ebλ + λ2 sin 2α we obtain ebλc
=
eb2 λ + bλ2 sin 2α
=
eb2 λ + bλ2 at
since sin 2α = at)
Dividing by bcλ we have e
=
b a e + λt c c
Since p = d = 0, the first row of the d matrix equation is ec = λa. Hence λ = ec/a. Putting this value in for λ we have e
= = =
b ec a e + t c a c b e + et c e(t − 1) + et
since b/c = t − 1
Since e 6= 0 we can divide by e: 1
= =
t−1+t
2t − 1
Solving for t we find t = 1. Since t = 2 cos α we have cos α = 1/2, so α = π/3. Then β = π/2 − 3α/2 = 0, contradiction. Note that this is a contradiction even if we do not assume α ≤ β. This contradiction shows that ψ is not identically zero.
84
We next want to prove that the components of (u, v, w) are not zero, without the assumption that ψ is identically zero. First assume, for proof by contradiction, that v = 0. Then (eg − pf )b + f λb
=
0
Dividing both sides by b we have eg − pf + f λ
=
0
Putting in the value of λ from (51) we have 0
“ e + d” eg − pf + f p − ds + s (eg − pf )s + f (ps − ds2 + e + d)
= =
Collecting terms and changing the sign, we have 0
=
df s2 − egs − (e + d)f
Let H(s) = df s2 − egs − (e + d)f . Then H(0) = (e + d)f ≤ 0, and H(1) = df − eg − (e + d)f = −eg − ef ≤ 0. Since H is a quadratic polynomial with positive leading coefficient, it cannot have a zero between 0 and 1. But s is such a zero, contradiction. Hence v 6= 0. Next assume, for proof by contradiction, that u = 0. Then 0
(df − em)b + eλ sin 2α
=
(df − em)b + eλat
=
since sin 2α = at
Dividing both sides by c and using (51) we have 0
=
“ b e + d” a (df − em) + e p − ds + t c s c
Using b/c = 1 − s2 and t = 2 − s2 and a/c = s, we have “ e + d” s(2 − s2 ) 0 = (df − em)(1 − s2 ) + e p − ds + s = (df − em)(1 − s2 ) + e(ps − ds2 + e + d)(2 − s2 ) =
eds4 − eps3 + s2 (em − df − 3ed − e2 ) + 2eps + (df − em + 2e2 + 2de)
By Lemma 37, we have e 6= 0, since there cannot be a tile with a γ angle at B (where the 2α angle of ABC is), and there cannot be a tile with a γ angle at C, where the 2β angle of ABC is, since by hypothesis, there are two tiles each with a β angle at that vertex. Define “ ” “ df ” df H(s) := ds4 − ps3 + s2 m − − 3d − e + 2ps + − m + 2e + 2d e e Since e 6= 0, we can divide the previous equation by e, obtaining H(s) = 0: “ ” “ df ” df 0 = ds4 − ps3 + s2 m − − 3d − e + 2ps + − m + 2e + 2d e e
(53)
The first two terms of H are equal to the first two terms of ψ, so H − ψ is a quadratic in s: 0
= =
H(s) − ψ(s) ” “ df − 3d − e − (m − 3d − e) + s(2p − (2p − g)) s2 m − e ” “ df − m + 2e + 2d − (2d + 2e − f − m) + e
85
df + gs + (54) e e We have f 6= 0, by Lemma 39. Assume, for proof by contradiction, that d = 6 0. Then we can divide by −df /e, obtaining eg 0 = s2 − s−1 df =
−s2
“ df ”
This quadratic function is negative when s = 0, and negative when s = 1, and its leading coefficient is positive. Therefore it has no zero between 0 and 1, contradiction. This contradiction shows that d = 0. Since d = 0, equation (54) becomes gs = 0. Hence g = 0, since s 6= 0. The equation (50) (ψ(s) = 0) then becomes 0
=
−ps3 + (m − e)s2 + 2ps + (2e − f − m)
On the other hand, the equation (53) (H(s) = 0) now becomes (with d = g = 0) −ps3 + (m − e)s2 + 2ps + (2e − m) Subtracting that from the previous equation we find f = 0; but we already proved f 6= 0 using Lemma 37. That contradiction depended on the assumption u = 0, and that completes the proof by contradiction that u 6= 0. Now assume, for proof by contradiction, that w = 0. That is, 0
=
(pm − dg)b − (mb + p sin 2α)λ + λ2 sin 2α
To write this as a function of s, use sin 2α = at = cs(2 − s2 ) and b = c(1 − s2 ) and λ = p − ds + (e + d)/s, and then multiply by s/c. We find (with the aid of a computer algebra system) the following polynomial equation: 0
=
−d2 s7 + dps5 + (4d2 + 2de − dm)s4 + (dg − 3dp − ep)s3
(−5d2 − 6de − e2 + 2dm + em)s2
+(−dg + 2dp + 2ep)s + (2d2 + 4de + 2e2 − dm − em)
Computing the polynomial remainder of this on division by ψ, and changing the sign, we find 0
=
G(s) := df s2 − egs − f (d + e)
Assume, for proof by contradiction, that d = 0. Then 0 = −egs − f e. By Lemma 39, e 6= 0 and f 6= 0, so f e > 0; since egs ≥ 0 this is a contradiction. Hence d 6= 0. Since f 6= 0 as just noted, the leading coefficient df is not zero. Thus G is a quadratic function such that G(0) = −f (d + e) < 0. We have G(1) = df − eg − f (d + e) = −eg − ef < 0. Since G(0) and G(1) are both negative, and G′′ is positive, G has no zero between 0 and 1. This is a contradiction, since G(s) = 0 for s = 2 sin(α/2), which is between 0 and 1 since α < π/3. That completes the proof by contradiction that w 6= 0. Hence, the eigenspace is one-dimensional, and (u, v, w) is a multiple of the eigenvector (a, b, c). Returning to equation (52), now that we know (u, v, w) is a multiple of (a, b, c), we have we have a/c = u/w. That is, a c
=
(df − em)g + eλ sin 2α (pm − dg)b − (mb + p sin 2α)λ + λ2 sin 2α
We have sin 2α = at = a(2 − s2 ). Putting that in, and dividing numerator and denominator on the right by c, and then using b/c = 1 − s2 and a/c = s, we have s
= =
(df − em)(1 − s2 ) + esλ(2 − s2 ) (pm − dg)(1 − s2 ) − (m(1 − s2 ) + ps(1 − s2 ))λ + λ2 s(1 − s2 ) (df − em)(1 − s2 ) + esλ(2 − s2 ) (1 − s2 )(pm − dg − (m + ps)λ + λ2 s
86
Multiplying by the denominator we have s(1 − s2 )(pm − dg − (m + ps)λ + λ2 s)
=
(df − em)(1 − s2 ) + esλ(2 − s2 )
Putting in λ = p − es + (e + d)/s and bringing this to polynomial form (using Mathematica, for example, or by hand if you wish) we have 0
d2 s6 − dps5 + (dm − de − 3d2 )s4 + (2dp − dg)s3
=
+(3d2 + de − df − 2dm)s2 + (dg − dp + ep)s + (d2 − e2 − df − dm)
Now taking this polynomial mod ψ, using the PolynomialRemainder command of Mathematica (you won’t want to do that polynomial division by hand) we find a simple quadratic equation: 0
=
d2 − e2 − df − dm − (dg − dp + ep)s + (dm + de − d2 )s2
(55)
By Lemma 39, e 6= 0. Suppose, for proof by contradiction, that d = 0. Then (55) becomes eps + e2 = 0. Since e 6= 0 then ps + e = 0. Since e, p, and s are all ≥ 0 and e > 0, this is a contradiction. Hence d 6= 0. Since s = −i(z + z −1 ) = 2 sin(α/2), we have sin
α 2 s
<
0. Assume, for proof by contradiction, that p = 0. Then F (s)
d2 s4 − 2d(d + e)s2 + (d + e))2
=
(ds − (d + e))2
=
Hence the only zero of F is s = (d + e)/d ≥ 1. But (if there is an N -tiling), F has a zero between 0 and 1, contradiction. Hence p 6= 0. Now we compute the value of F (1). F (1)
= = =
F (1)
>
d2 − dp − 2d(d + e) + p(d + e) + (d + e)2
d2 − dp − 2d2 − 2de + pd + pe + d2 + 2de + e2 pe + e2
0
since p > 0 and e > 0
We have F (0) = (d + e)2 . This is positive since d > 0. Now F is a quartic that is positive at 0 and positive at 1, and its derivative at 0 is F ′ (0) = p(d + e), which is positive since p 6= 0 and d 6= 0. The derivative F ′ (s) is a cubic; one of its zeroes is negative, since for large negative s we have F ′ (s) < 0. It therefore has at most two zeroes between 0 and 1. The only way that
87
F (s) can have a zero strictly between 0 and 1 is if F first increases (as s increases from 0), then reaches a maximum and decreases, crossing the s-axis (or possibly just touching the s-axis, if F has a double zero), then reaches a minimum, then increases (crossing the s-axis again, unless it has a double zero) and reaches its positive value at s = 1. Then F ′ has two zeroes between 0 and 1, and in particular F ′ (1) must be positive. We compute F ′ (s)
=
4d2 s3 − 3dps2 − 4d(d + e)s + p(d + e)
The product of its three roots is thus −p(d + e)/4d2 , which is negative. Because F ′ (0) is positive, and F ′ (s) is negative for large negative s, one of the three roots is negative. But (if there is an N -tiling), the other two are between 0 and 1, so the product of the three roots is negative. The sum of the three roots of F ′ is 4d(d + e)/4d2 = (d + e)/d. Since one of these roots is negative and the other two are between 0 and 1, the sum is less than 2. Hence (d + e)/d < 2. Hence e < d. We proved above that F ′ (1) ≥ 0. Therefore 0
≤
= = =
F ′ (1) 4d2 − 3dp − 4d(d + e) + p(d + e) −2dp − 4de + pe p(e − 2d) − 4de
But since e < d, and e > 0 and d > 0, the right side is negative, contradiction. That completes the proof of the lemma.
14.2 The case when ABC has angles 2α, β, and β + α and sin(α/2) is irrational Recall that sin(β + α) = sin γ = c in this case, as follows from 3α + 2β = π. The 2α angle is at vertex A, the β angle at vertex B, and the β + α angle at vertex C. We do not assume α < β. The d matrix equation is 1 1 0 10 0 X a p d e @ g m f A@ b A = @ Y A Z c h ℓ r Let λ = X/ sin(2α). By the law of sines for triangle ABC we then have X
=
λ sin 2α
Y
=
λ sin β
Z
=
λ sin(β + α)
Since sin 2α = at and sin(β + α) = sin γ = c as proved above, the d matrix equation can be written 1 1 0 10 0 at a p d e @ g m f A@ b A = λ@ b A (56) c c h ℓ r Taking the ratio of the first row to the third, we have pa + db + ec ha + ℓb + rc
=
at c
Multiplying by the denominator and then dividing both sides by c2 we have ” b at “ a b a h +ℓ +r p +d +e = c c c c c
88
We now express this equation in terms of s, using a/c = s, b/c = 1 − s2 , and t = 2 − s2 : ps + d(1 − s2 ) + e
s(2 − s2 )(hs + ℓ(1 − s2 ) + r)
=
This is a polynomial equation of degree 5. Bringing it to polynomial form we have 0
=
F (s) := ℓs5 − hs4 − (3ℓ + r)s3 + (d + 2h)s2 + (2ℓ + 2r − p)s − (e + d)
If F were identically zero, then we would have ℓ = 0 from the leading coefficient, and then r = 0 from the coefficient of s3 , contradicting Lemma 38. Therefore F is not identically zero. Taking the ratio of the second row to the third, we have ga + mb + f c ha + ℓb + rc
=
(ga + mb + f c)
=
b a + m + f) c c (gs + m(1 − s2 ) + f ) (g
= =
b c b c a b b h + ℓ + r) c c c (hs + ℓ(1 − s2 ) + r)(1 − s2 ) (ha + ℓb + rc)
Subtracting the left side from the right and bringing the equation to polynomial form, we have 0
=
H(s) := ℓs4 − hs3 + (m − 2ℓ − r)s2 + (h − g)s + ℓ + r − m − f
If H were identically zero, we would have ℓ = 0 from the leading coefficient, and h = 0 from the coefficient of s3 , then r = m from the coefficient of s2 , then g = 0 from the coefficient of s, then f = 0 from the constant coefficient; but by Lemma 38, we do not have g = f = 0 contradiction. Hence H is not identically zero. We compute K = −(F mod H) (using the PolynomialRemainder command in Mathematica, for example), and we find 0
=
K(s) = (ℓ + m)s3 − (d + g + h)s2 − (f + m + ℓ + r − p)s + (d + e)
(57)
If K were identically zero, then we would have ℓ = m = 0 from the coefficient of s3 , and d = g = h = 0 from the coefficient of s2 , and then e = 0 from the constant term, but e = 0 contradicts Lemma 37. Hence K is not identically zero. Dividing the top row of (56) by t we write it as an eigenvalue equation: 1 0 0 p d e 10 1 a a t t t @ g m f A@ b A = λ@ b A c c h ℓ r Our next goal is to prove that the eigenspace of the eigenvalue λ is one-dimensional. To do that, we use the method of computing a candidate eigenvector by the cofactor method, and showing that its three components are each nonzero. The matrix whose cofactors we need is 0 p 1 d e −λ t t t @ A g m−λ f h ℓ r−λ We expand in cofactors of the elements (u, v, w) where 0 1 0 u @ v A = @ w
on the first row. We find a candidate eigenvector 1 (m − λ)(r − λ) − f ℓ A hf − g(r − λ) gℓ − h(m − λ)
89
Lemma 41 Suppose that ABC is N -tiled by tile T with 3α + 2β = π and sin(α/2) irrational, and triangle ABC has angles 2α, β, and β + α. Suppose also that not both g and h are zero. Then the eigenspace of λ is one-dimensional and (u, v, w) defined above is a multiple of (a, b, c). Proof. Our first aim is to prove v 6= 0. Since f 6= 0, if g = 0 then also h = 0, which contradicts the hypothesis that not both g and h are zero, and similarly, if h = 0 then g = 0. Hence g 6= 0 and h 6= 0. Then hf > 0. But we have grc ≤ λc from the third row of th e d matrix equation; hence gr ≤ λ. Then v
= = >
hf − g(r − λ)
hf + g(λ − r) 0
since not both g = 0 and h = 0
Next we will prove w 6= 0. From the second row of the d matrix equation, we have ga + mb ≤ λb. Since g > 0, we have m < λ, not just m ≤ λ. Then w
= = >
gℓ − h(m − λ)
gℓ + h(λ − m)
0
since not both g = 0 and h = 0
Finally we will prove u 6= 0. We have from the second and third rows of the d matrix equation mb + f c ℓb + rc
≤
≤
λb
with equality only if g = 0
λc
with equality only if h = 0
Rearranging the terms of these equations we have fc ℓb
≤
≤
(λ − m)b (λ − r)c
Multiplying these two equations and dividing by bc we have ℓf
≤
(λ − m)(λ − r)
The left hand side is u. We have proved u ≥ 0 with equality only if g = 0 and h = 0. But at present we have assumed that not both g and h are zero; so we have u > 0. We have proved all three components of the candidate eigenvector are nonzero. It follows that the eigenspace is one-dimensional; hence (u, v, w) is a multiple of (a, b, c). That completes the proof of the lemma. Lemma 42 Suppose 3α + 2β = π, and suppose triangle ABC has angles 2α, β, and β + α. Suppose that sin(α/2) is irrational. Then a, b, and c are linearly independent over Q; or what amounts to the same thing, 1, sin(α/2), and cos α are linearly independent over Q. Proof. Suppose a, b, and c are linearly dependent over Q. Then there are (positive or negative) integers n, j, and k such that na+jb+kc = 0. With s = 2 sin(α/2) we have a/c = s, b/c = 1−s2 . Therefore 0 = ns + j(1 − s2 ) + k.
If j = 0 then s = −k/n = 2 sin(α/2) so sin(α/2) is rational, contradiction. Hence j 6= 0. Solving for 1 − s2 we have ns k 1 − s2 = − − j j
90
(58)
There is an angle β + α at vertex C; the adjacent sides are X and Y and the area equation can be written as N ab = XY = (pa + db + ec)(ga + mb + f c) Dividing by c2 we have ab a b a b = (p + d + e)(g + m + f ) cc c c c c Putting everything in terms of s we have N
N s(1 − s2 ) k” ns − Ns − j j “
= =
ps + d(1 − s2 ) + e)(gs + m(1 − s2 ) + f “ ns “ ns k” k” ps + d − + e)(gs + m − f) − − j j j j
This is quadratic in s. We can then use (58) to replace s2 by a term linear in s. Hence s satisfies a linear equation over Q. Hence s is rational. But s = 2 sin(α/2), so sin(α/2) is rational. It is not necessary to write the equation out explicitly. That completes the proof of the lemma. Lemma 43 Let 3α + 2β = π and assume triangle ABC is N -tiled by a tile with angles α and β. Suppose ABC has angles 2α, β, and β + α, and suppose s = 2 sin(α/2) is not rational. Then the d matrix entries g and h are not both zero. Proof. Suppose, for proof by contradiction, that g = h = 0. The d matrix equation is 1 1 10 0 0 a at p d e @ 0 m f A@ b A = @ b A c c 0 ℓ r
Writing this in eigenvalue form we have 0 p d −λ t t @ 0 m−λ 0 ℓ
e t
1 a A@ b A f c r−λ 10
=
0
Hence the determinant of the matrix on the left is zero. Expanding it by cofactors we have “ ” p = 0 ( − λ) (m − λ)(r − λ) − f ℓ) t
There are two cases: either p = λt or not. In case p = λt then the last row of the d matrix equation is ℓb + rc
=
λc
Dividing by c and using b/c = 1 − s2 and λ = p/t we have ℓ(1 − s2 ) + r
= =
ℓ(1 − s2 )(2 − s2 )
=
p t
p 2 − s2 p
Thus s2 satisfies a quadratic equation over Q. Hence the degree of Q(s) over Q is either 2 or 4, since s is not rational by hypothesis. But this contradicts (57), which says s satisfies a cubic equation and hence has degree 3 or 1. That completes the case p = λt. Now assume p 6= λt. Then the second factor in the characteristic equation is zero: (m − λ)(r − λ) − f ℓ)
91
=
0
This is a quadratic equation for λ; hence Q(λ) has degree 2 or 1 over Q. But from the third row of the d matrix equation we have λ = ℓ(1 − s2 ) + r
which implies that the degree of Q(s) over Q(λ) is 1 or 2. Hence the degree of Q(s) over Q is 1, 2, or 4. Degree 1 contradicts the hypothesis that s is irrational; degrees 2 or 4 contradict (57). That completes the proof of the lemma.
Lemma 44 Let 3α + 2β = π and assume triangle ABC has angles 2α and β and is N -tiled by a tile with angles α and β. Suppose sin(α/2) is irrational. Then ℓ 6= 0, p 6= 0, h 6= 0, gℓ − hm 6= 0, and we have the following equations between the elements of the d matrix, where ∆ is the determinant of the d matrix: p g+h p g+h p g+h
= = =
mr − ℓf gℓ − hm −p(m + r) + eh + dg − N gℓ − hm + hf − gr ∆ + N (m + r) −N h
and in the second equation, if the denominator on the right is zero, so is the numerator. Proof. Since t = N/λ2 , t belongs to Q(λ). The third row of the d matrix equation, ha+ℓb+rc = λc, shows that λ belongs to Q(a, b, c) = Q(t). Since t belongs to Q(λ) and λ belongs to Q(t), we have Q(λ) = Q(t) = Q(a, b, c). This field has degree at least 3 over Q, by Lemma 42. We will soon see that the degree is exactly 3. We are in a position to find three different cubic equations for λ. The first equation come from the characteristic equation for λ: 0 p 1 d e −λ t t t @ A=0 g m−λ f h ℓ r−λ Multiplying the top row by t, the determinant is still zero: 1 0 p − λt d e A=0 @ g m−λ f h ℓ r−λ
Since λ2 = N/t, we have λt = N/λ. Multiplying by λ we have 1 0 d e p− N λ A=0 g m−λ f λ@ h ℓ r−λ
Expanding and collecting like powers of λ we find the following equation, in which ∆ is the determinant of the d matrix: pλ3 + (−p(m + r) + eh + dg − N )λ2 + (∆ + N (m + r))λ − N (mr − ℓf ) = 0
(59)
This is a cubic polynomial equation for λ. Hence Q(λ) has degree 3 over Q, and this is the minimal polynomial of λ. Since we have already proved that the eigenspace of λ is one-dimensional, it is not the case the all the coefficients are zero. Since the degree of Q(λ) is 3, the highest and lowest coefficients are not zero: p 6= 0 and mr 6= 0.
92
Next we compute b/c in terms of λ: b c
=
(z 2 − 1 + z 2 )
=
t−1
since t = z 2 + z −2
Since λ2 = N/t, we have t = N/λ2 , which gives us N b = 2 −1 c λ
(60)
Since the eigenvector (u, v, w) is a multiple of (a, b, c), we have b c
=
N −1 λ2
=
v w hf − g(r − λ) gℓ − h(m − λ)
Cross multiplying we have (N − λ2 )(gℓ − hm + hλ)
=
(hf − gr)λ2 + gλ3
Subtracting the left side from the right and expanding and collecting, we have 0
=
(g + h)λ3 + (gℓ − hm + hf − gr)λ2 − N hλ − N (gℓ − hm)
(61)
By Lemma 43, not both g and h are zero. Hence (since both are nonnegative) g +h > 0, and this is a non-trivial equation. Since Q(λ) = Q(a, b, c) (from the third row of the d matrix equation), and this field has degree 3, the constant term N (gℓ − hm) 6= 0 (or else the equation could be divided by λ to yield a quadratic equation for λ). Hence gℓ − hm 6= 0. Similarly, the constant term of (59), namely N (mr − ℓf ), is also not zero, so mr − ℓf 6= 0. Now (61) and (59) are two cubic equations for λ. Therefore (61) is a multiple of (59). The ratio of the coefficients of λ3 is p/(g + h). Therefore the other nonzero coefficients are also in that ratio. From the constant coefficients (neither of which is zero) we have p g+h
=
mr − ℓf gℓ − hm
(62)
It follows that ℓ 6= 0
(63)
since otherwise the numerator and denominator on the right of (62) have opposite signs, but the left side is positive. From the ratio of the quadratic terms we have (unless the numerator and denominator are both zero) p g+h
=
−p(m + r) + eh + dg − N gℓ − hm + hf − gr
(64)
From the linear terms we have (unless both numerator and denominator are zero) p g+h
=
∆ + N (m + r) −N h
(65)
These are the three equations mentioned in the lemma. It remains to prove h 6= 0. Assume, for proof by contradiction, that h = 0. Then g 6= 0, since the constant term g + h of (61) is not zero. Then from (62) we have pℓ = mr, or m = pℓ/r. When h = 0 we have ˛ ˛ ˛ p d e ˛ ˛ ˛ ∆ = ˛˛ g m f ˛˛ ˛ 0 ℓ r ˛ = pmr + egℓ − gdr − pf ℓ
93
Since both numerator and denominator on the right of (65) are zero we have 0
=
∆ + N (m + r)
=
pmr + egℓ − gdr − pf ℓ + N (m + r)
Substituting m = pℓ/r and multiplying by r we have 0
=
p2 ℓr + egℓr − gdr 2 − pf ℓr + N (pℓ + r 2 )
Bringing the last term to the left side and changing signs we have N (pℓ + r 2 )
pf ℓr + gdr 2 − egℓr − p2 ℓr
=
pf ℓr + gdr 2
≤
Note that gd ≤ λ2 , since ga ≤ λb and db ≤ λa, from the d matrix. We also have r ≤ λ and f c ≤ λb < λc, so f r < λ2 . Then N (pℓ + r 2 )
π/5. That contradiction completes the proof that h 6= 0. Hence equation (65) is valid, i.e. its denominator is not zero. That completes the proof of the lemma.4 Lemma 45 Suppose 3α + 2β = π, and triangle ABC has angles 2α and β and is not isosceles. Suppose sin(α/2) is irrational. Let N be arbitrary. Then there is no N -tiling of triangle ABC by a tile with angles α and β. Proof. Suppose, for proof by contradiction, that there is a triangle ABC and an N -tiling as in the lemma. Then the equations of the previous lemma hold. From (65) we have N (m + r) p g+h “ hp ” N m+r+ g+h
N (m + r) + N h
≤
−∆ − N h
≤
−∆
p g+h
≤ −∆ ≤
≤
dgr + pf ℓ + hme − pmr − df h − egℓ
mhe + rdg + p(ℓf − mr) − df h − egℓ
From the third row of the d matrix we have ha < Z = λc. Equality cannot hold since ℓ 6= 0 (and we have ha + ℓb < Z). From the first row we have ec ≤ λat. Multiplying these two inequalities (which is legal since all these quantities are positive) and dividing by ac we have he ≤ λ2 t < N . Similarly, we have dg ≤ λ2 t = N . Putting these results into the inequality above, we have N (m + r) + N
hp g+h
π/3, we have ℓ ≤ 5. That leaves the possibilities 3, 4, and 5 for ℓ. We will rule out each of these in turn; each one requires a detailed argument.
103
We first assume ℓ = 3. Since γ > π/3, we have k = 2, i.e. the angle sum at V is 2π; and we cannot have n = m = 0, since then γ would be π/3. Since m + n < ℓ, we have m + n ≤ 2. If n = 1 and m = 1 then 2π
=
3γ + α + β
=
2γ + (γ + α + β)
=
2γ + π
Hence γ = π/2, contradiction. Hence n = 1 and m = 1 is impossible. If n = 2 and m = 0, then 2π
=
3γ + 2α
=
γ + 2γ + 2α
2(γ + α + β)
=
γ + 2γ + 2α
2β
=
γ
Assume R = 1. Then by Lemma 54, β = (P − 1)α. Then since γ = 2β, we have π
=
α+β+γ
=
α + 3β
=
α + 3(P − 1)α α(3P − 2)
= We can then solve for all three angles: α
=
β
=
γ
=
π 3P − 2 P −1 π 3P − 2 2(P − 1) π 3P − 2
In that case the inequality γ ≤ π/2 is equivalent to 2(P − 1)/(3P − 2) ≤ 1/2, which is equivalent to P ≤ 2. Since P ≥ 4 we have γ > π/2 in this case. If R = 0 then we have π
=
P α + Qβ
2π
=
3γ + 2α
γ
=
2β
π
=
α + 3β
since ℓ = 3, n = 2, and m = 0
by subtracting 2γ + 2α since 3γ + 2α = 2π and γ = 2β
Since P + Q + R ≥ 5, and R = 0, we do not have P = 1 and Q = 3. If 3P − Q = 0 and (P, Q) 6= (1, 3), then the equations are contradictory. Hence 3P − Q 6= 0, and the equations are uniquely solvable: α
=
β
=
γ
=
3−Q π 3P − Q P −1 π 3P − Q 2P − 2 π 3P − Q
104
Since α < β, if 3P − Q < 0 we have Q − 3 < 1 − P ; hence P + Q < 4, contradiction. Hence 3P − Q > 0. We have P − 1 > 0, hence 3P > Q; hence 3 − Q > 0, hence Q < 3. If n = 1 and m = 0 then 2π
=
α + 3γ
2(α + β + γ)
=
α + 3γ
γ
=
α + 2β
α + β + (α + 2β)
=
π
2α + 3β
=
π
since ℓ = 3, n = 1, and m = 0
since α + β + γ = π
If R = 1, then by Lemma 54, β = (P − 1)α, so (2 + 3(P − 1))α = π, and we have α
=
β
=
γ
=
π 3P − 1 P −1 π 3P − 1 2P − 1 π 3P − 1
If R = 0 then we have 2α + 3β
=
π
as shown above
P α + Qβ
=
π
since R = 0
(69)
If P = 2 and Q = 3 then we have shown in Theorem 5 that case (iii) of the theorem holds. Therefore we may assume (P, Q) 6= (2, 3). It follows that 2Q − 3P 6= 0, since if 2Q − 3P = 0 and (P, Q) 6= (2, 3), subtracting the two equations (69) yields a contradiction. Hence the determinant of the system (69) is nonzero, and we have α
=
β
=
γ
=
3−Q π 3P − 2Q P −2 π 3P − 2Q 2P − Q − 1 π 3P − 2Q
Since α < β, if 3P − 2Q < 0 then we have Q − 3 < 2 − P , or P + Q < 5, contradicting P + Q + R ≥ 5. Hence 3P − 2Q > 0. Then since α > 0 we have Q ≤ 2, and since β > 0 we have P ≥ 3, and since α < β we have 3 − Q < P − 2, or P + Q > 5. If n = 0 and m = 1 then 2π
=
β + 3γ
2(α + β + γ)
=
β + 3γ
γ
=
2α + β
α + β + (2α + β)
=
π
3α + 2β
=
π
since α + β + γ = π
If R = 1, then by Lemma 54, β = (P − 1)α, so (3 + 2(P − 1))α = π, and we have α
=
β
=
γ
=
π 2P + 1 P −1 π 2P + 1 P +1 π 2P + 1
105
Since α < β we have 2P + 1 < P − 1, which is impossible. Hence this case does not occur. If R = 0 then we have 3α + 2β
=
π
P α + Qβ
=
π
If P = 3 and Q = 2 then we have shown in Theorem 5 that case (iii) of the theorem holds. Therefore we may assume (P, Q) 6= (3, 2). Therefore 3Q − 2P 6= 0 and α
=
β
=
γ
=
2−Q π 2P − 3Q P −3 π 2P − 3Q P − 2Q + 1 π 2P − 3Q
Since α < β, if 2P − 3Q < 0, we have Q − 2 < 3 − P , or P + Q < 5, contradiction. Hence 2P − 3Q ≥ 0. Then since α > 0 we have Q = 0 or Q = 1. If m = 2 and n = 0 we have 2π
=
3γ + 2β
= 3α + β
=
3(π − α − β) + 2β π
If R = 1 then β = (P − 1)α so 3α + (P − 1)α = π, and we have α
=
β
=
γ
=
π P +2 P −1 π P +2 2 π P +2
Since β < γ we have P − 1 < 2, or P < 3; since Q = 0 this makes P + Q + R < 4, contradiction; so this case cannot occur. If R = 0 then we have 3α + β
=
π
P α + Qβ
=
π
We do not have P = 3 and Q = 1 since P + Q ≥ 5. Hence α
=
β
=
1−Q π P − 3Q P −3 π P − 3Q
Since α < β, if P − 3Q < 0 we have Q − 1 < 3 − P , or P + Q < 4, contradiction. Hence P − 3Q > 0. Since α > 0 we have Q = 0. Hence α
=
β
=
γ
=
π P P −3 π P 2 π P
106
Since α < β, we have P > 4. Since β < π/2, we have P < 6. Hence P = 5. That makes α = π/5 and β = γ = 2π/5, a case which has already been ruled out. The following table summarizes the results obtained above for ℓ = 3: n 2 2 1 1 0
m 0 0 0 0 1
R 1 0 1 0 0
P
P ≥3 P ≥ 6−Q
Q 0 Q≤2 0 Q≤2 0 or 1
α/π
β/π
γ/π
1 3P −2 3−Q 3P −Q 1 3P −1 3−Q 3P −2Q 2−Q 2P −3Q
P −1 3P −2 P −1 3P −Q P −1 3P −1 P −2 3P −2Q P −3 2P −3Q
2P −2 3P −2 2P −2 3P −Q 2P −1 3P −1 2P −Q−1 3P −2Q P −2Q+1 2P −3Q
info γ = 2β γ = 2β γ = 2β + α γ = 2β + α γ = 2α + β
We have no further need of vertex V and the associated numbers n and m, so we reprint the table without the first two columns. The point is, that the vertex-splitting numbers P , Q, and R determine α, β, and γ uniquely. R 1 0 1 0 0
P
P ≥3 P ≥ 6−Q
Q 0 Q≤2 0 Q≤2 0 or 1
α/π
β/π
γ/π
1 3P −2 3−Q 3P −Q 1 3P −1 3−Q 3P −2Q 2−Q 2P −3Q
P −1 3P −2 P −1 3P −Q P −1 3P −1 P −2 3P −2Q P −3 2P −3Q
2P −2 3P −2 2P −2 3P −Q 2P −1 3P −1 2P −Q−1 3P −2Q P −2Q+1 2P −3Q
info γ = 2β γ = 2β γ = 2β + α γ = 2β + α γ = 2α + β
Now that we see how P , Q, and R determine the angles, we will show that each of these determinations leads to a contradiction. Let the vertices of the tiling, other than A, B, and C, be V1 , V2 , . . .; let the angle sum at Vi be ki π, so that ki = 1 for a non-strict or boundary vertex, and 2 for a strict interior vertex. At each vertex Vi , let ni , mi , and ℓi be the number of α, β, and γ angles at that vertex. altogether, each having one α, one P Since there P are N tilesP β, and one γ angle, we have P + ni = Q + mi = R + ni = N π. Consider the quantity qi = 2ℓi + mi . We will show below that, for the first three rows of the table, at each vertex Vi we have 3ni ≥ qi . Once that is proved, we finish the proof as follows: Adding over all vertices, we have X X 3ni ≥ (2ℓi + mi )
The sum on the left is three times the total number of α angles at the vertices Vi . This must be equal to 3(N − P ). The sum on the right is the total number of β angles plus twice the number of γ angles. This is equal to N − Q + 2(N − R) = 3N − Q − 2R. Hence 3(N − P )
≥
3N − Q − 2R
−Q − 2R
≤
−3P
3N − Q − 2R 3P
≤ ≤
3(N − P ) Q + 2R
But the table above shows that in every line of the table, we have Q + 2R ≤ 2. Hence 3P ≤ 2. Since P is a positive integer, this is a contradiction. It remains to supply the proof that 3ni ≥ qi , for the cases given in the first three rows of the table. We first claim that in case ki = 1 (i.e. the angle sum at vertex Vi is π), we can assume ℓi ≤ 1, i.e. if ℓi > 1, the conclusion 3ni ≥ qi holds. We certainly have ℓi ≤ 2, since 3γ > π. Now assume that ℓi = 2. Then mi = 0, since 2γ + β > π. Hence qi = 4. Then 3ni ≥ qi unless ni = 0 or ni = 1. If ni = 0, then the equation ni α + mi β + ℓi γ = π becomes 2γ = π, so γ = π/2; but we have proved that (we can assume that) γ is not a right angle, so ni 6= 0. Hence we may
107
assume ni = 1. Then, we have 2γ + α = π. But this, together with α + β + γ = π, implies β = γ, which contradicts the assumption that T is not isosceles. This contradiction shows that ni 6= 1, and hence disposes of the case ki = 1 and ℓi = 2. That completes that proof of the claim that we can assume ℓi ≤ 1 when ki = 1. For the first two rows of the table, we have γ = 2β. In that case π = α + β + γ = α + 3β, so 4β > π. Then 2γ + β = 5β > π as well. If ki = 1 and ℓi = 0 then qi = mi ≤ 3, since 4β > π. If ki = 1 and ℓi = 1 then qi = 2 + mi , and since γ = 2β and 4β > π, we have γ + 2β > π, so mi ≤ 1. Hence qi ≤ 3. Since we showed above that we can assume ℓi ≤ 1 when ki = 1, we have proved that (for the first two rows of the table) qi ≤ 3 when ki = 1 (or else the conclusion of the lemma holds). Now we assume, for proof by contradiction, that ki = 2 and the first row of the table applies. Then qi β + ni α
=
(2ℓi + mi )β + ni α
=
ℓi γ + mi β + ni α
qi β + ni α
=
2π
since 2β = γ (70)
We wish to bound qi . First we take up the first case (first row of the table). Then in addition to γ = 2β we have β = (P − 1)α. We have 6β + 2α
=
2π
since 6β = 2β + 2γ
qi β + ni α
=
2π
by (70)
Subtracting these two equations we have (6 − qi )β = (ni − 2)α. If qi > 6 then ni < 2. Then ni = 0 or ni = 1, and β = ((2 − ni )/(qi − 6))α. But β = (P − 1)α, so (2 − ni )/(qi − 6) = P − 1, an integer at least 4. This is impossible since the numerator 2 − ni is at most 2. This shows that in the first row of the table, we cannot have ki = 2. Now we give a bound for qi in case ki = 2 and the second row of the table applies, so γ = 2β and β = ((P − 1)/(3 − Q))α. Assume qi > 6. Then as in the previous paragraph, we have ni = 0 or ni = 1 and β = ((2 − ni )/(qi − 6))α. But now also β = ((P − 1)/(3 − Q))α. Hence 2 − ni qi − 6 (2 − ni )(3 − Q) qi − 6
=
P −1 3−Q
=
P −1
On the left side, the factor 2 − ni is either 1 or 2, and the factor 3 − Q is either 1, 2, or 3, since Q ≤ 2. Since R = 0 in the second row of the table, we have P + Q ≥ 5. If Q = 2 then we have (2 − ni )/(qi − 6) = P − 1 and P − 1 ≥ 2; but then (2 − ni )/(qi − 6) is an integer at least 2, with numerator at most 2. This is possible only if qi = 7 and ni = 0 and (P, Q) = (3, 2). But in the second row of the table, we have R = 0, so in this case, we have shown in Theorem 5 that case (iii) of the theorem holds. Therefore the case qi > 6 has led to a contradiction or to conclusion (iii) of the theorem. Therefore we may assume qi ≤ 6 in case ki = 2 and the second row of the table applies. Now we will prove qi ≤ 3ni for the first two rows of the table. Case (i) ki = 1. We have
qi
=
2ℓi + mi
definition of qi
Multiplying by β and adding ni α we have qi β + ni α
=
2ℓi β + mi β + ni α
=
ℓi γ + mi β + ni α
=
π
108
since γ = 2β
Now if ni = 0 then qi β = π. As shown above, we can assume qi ≤ 3. Then, since qi β = π, we have β ≥ π/3. Hence γ = 2β ≥ 2π/3 and β + γ ≥ π, contradiction. Hence ni 6= 0. Therefore ni ≥ 1. Since qi ≤ 3, we have qi ≤ 3ni as desired. Case (ii) ki = 2. We showed above that this is impossible for the first row of the table, so the second row applies. We have 3β + α = π and qi β + ni α = 2β + ni α = π. Therefore 2β + ni α
=
3β + α
From the second row of the table, we have β = ((P − 1)/(3 − Q))α. Hence we obtain ! P −1 2β + ni α = 2β + 1 + α 3−Q ! P −1 ni = 1+ 3−Q Since Q ≤ 2 the denominator is positive and at most 2; since P + Q ≥ 5, we have P ≥ 3 so the numerator is at least 2. It follows that ni ≥ 2. Since we proved above that qi ≤ 6 in case ki = 2 and the second row of the table applies, we have 3ni ≥ qi as desired. That completes the proof for the first two rows of the table. Next we take up the third row of the table. In this case we have γ = 2β + α instead of γ = 2β. Again we consider qi = 2ℓi + mi and try to prove 3n ≥ q. We start by observing that π = α + β + γ = 2α + 3β. Hence 2γ = 4β + 2α > π, so γ > π/2. Therefore, if ki = 1, we have ℓi ≤ 1, and if ki = 2, we have ℓi ≤ 3. We argue by (more) cases. Case (iii). ki = 1 and ℓi = 1. Then we have mi ≤ 1, since γ + 2β = 2β + α + 2β = 4β + α > 3β + 2α = π. Then π
=
3β + 2α
=
γ + mi β + ni α
3β + 2α
=
(2 + mi )β + (ni + 1)α
(1 − mi )β
(1 − mi )(P − 1)α
ni − 1
= = =
γ + mi β + ni α since γ = 2β + α
(ni − 1)α
(ni − 1)α
(1 − mi )(P − 1)
If mi = 1, so the right side is zero, then ni = 1 also, and we have ni = mi = ℓi = 1. In that case 3ni ≥ qi and we are finished. Hence (since mi ≤ 1) we may assume mi = 0. Then we have ni = P . Then q = 2ℓ + mi = 2, and 3ni = 3P > qi since P ≥ 2, since Q ≤ 2 and R ≤ 1 and P + Q + R ≥ 5. That disposes of Case (iii). Case (iv) ki = 1 and ℓi = 0 and mi > 4. This case can be ruled out, because 5β > 3β + 2α = π. Case (v) ki = 1 and ℓi = 0 and mi = 4. Then π = 4β + ni α > 3β + (ni + 1)α. Since 3β + 2α = π, this implies ni = 0 and β = π/4. Then γ = 2β + α = π/2 + α. Hence π
=
α+β+γ
=
α + π/4 + π/2 + α
α
=
π/8
γ
=
5π/8
For this special case of T , we cannot establish the desired bound. But we can rule out this case by cosidering the possibilities for P and Q, as follows. We have assumed that row 3 of the table applies. Then R = 1, so P + Q ≥ 4, and α = π/(3P − 1), and since α = π/8, we have P = 3,
109
and since P + Q ≥ 4, we have Q ≥ 1. But in row 3 of the table, we have Q = 0, contradiction. That disposes of this case. Case (vi) ki = 1 and ℓi = 0 and mi ≤ 3. Then π = mi β + ni α; and since π = 3β + 2α, we have ni ≥ 2, and so 3ni ≥ 6, and qi = 2ℓi + mi = mi ≤ 3, so 3ni > qi as required. This disposes of all cases where ki = 1 and the third row of the table applies. Now suppose ki = 2 and the third row of the table applies, so we have γ = 2β + α and π = 2α + 3β. Then we have 2π
=
ℓi γ + mi β + ni α
=
ℓi (2β + α) + mi β + ni α
=
qi β + (ℓi + ni )α
Subtracting from this 2π = 4α + 6β (which is twice π = 2α + 3β), we obtain 0
=
(qi − 6)β + (ℓi + ni − 4)α
(71)
Case (vii) The third row of the table applies, and ki = 2, and qi > 6. Then we have β = (P − 1)α, so we find 0
=
4 − ℓi − ni
=
(qi − 6)(P − 1)α + (ℓi + ni − 4)α
by (71)
(P − 1)(qi − 6)
In the third row of the table, we have Q = 0, and since P + Q + R ≥ 5 and R = 1, we have P ≥ 4, so P − 1 ≥ 3. Since qi > 6, the factor qi − 6 is positive. The left side, on the other hand, is an integer at most 4. Hence, qi − 6 must be 1, so qi = 7, and both sides are equal either to 3 or to 4. Hence, looking at the left side 4 − ℓi − ni is 3 or 4, so ℓi + ni is 0 or 1. In case it is zero, we have ℓi = ni = 0 and P = 5. Since qi = 2ℓi + mi = 7, we have mi = 7, so β
=
α
=
γ
=
2π/7 β P −1 β/4
=
π/14
=
2β + α
=
9π/14
This case is impossible, by Lemma 33. Now suppose ℓi = 1 and ni = 0. Then qi = 7 implies mi = 1, so the equation ℓi γ + mi β + ni α = 2π becomes γ + β = 2π; but since γ + β < π, that is a contradiction. Suppose ℓi = 0 and ni = 1. Then qi = 7 implies mi = 7, so the equation ℓi γ +mi β +ni α = 2π becomes 7β +α = 2π. In this case we have P = 4, so β = (P − 1)α = 3α. Hence 2π
=
7β + α
=
21α + α
=
22α π 11 3π 11 7π 11
α
=
β
=
γ
=
110
Since R = 1, the largest angle of ABC must be at least γ, but according to Lemma 32, that is not possible. Case (viii) The third row of the table applies, and ki = 2, and qi = 6. Then by (71) we have 0
=
ℓi + ni
=
(ℓi + ni − 4)α
4
Since qi = 2ℓi + mi , we have ℓi = (qi − mi )/2. Hence qi − mi + 2ni
qi
=
8
≤
2ni + mi
≤
3ni
which disposes of this case. Case (ix) The third row of the table applies, and ki = 2, and qi < 6. By (71) we have 4 − ℓi − ni
=
(P − 1)(qi − 6)
The right side is negative, since P ≥ 5 and qi < 6. Hence the left side 4 − ℓi − ni must also be negative. Since ℓi ≤ 3, we must have ni ≥ 2. But then 3ni ≥ 6 > qi , and we have disposed of this case. That completes all cases involving the third row of the table. Now we take up the fourth row of the table. Here we also have γ = 2β + α, but the formulas for β and α in terms of P and Q are different, and we have R = 0. It is simplest just to enumerate the possibilities for ℓi , mi , and ni . We have 3γ + α = 2π (since ℓ = 3, m = 0, and n = 1, from the first table); that gives the first row in the following table. The others are obtained by using the equation γ = 2β + α and the equation β = (P − 2)/(Q − 3)α to “trade in” some larger angles for smaller ones. The entries containing fractions apply only if the fraction shown is an integer. This table assumes ki = 2, so the angle sum is 2π. ℓi 3 2 2 2 1 1 0 0
mi 0 2 0 1 4 4-r 6 6-r
ni 1 2 −2) 1 + 2(P 3−Q P −2 1 + 3−Q 3 −2 3 + rP 3−Q 4 −2 3 + rP 3−Q
2ℓi + mi 6 6 4 1 4 6−r 6 6−r
2mi + ℓi 3 6 2 4 9 9 − 2r 12 12 − 2r
We note that the estimate qi = 2ℓi + mi ≤ 3ni , which we used for the first three rows of the table, fails here, e.g. in the first row. But we can replace it by ri = 2mi + ℓi ≤ 3ni , shown in the last column of the table. Remembering that the rows with fractions occur only if the fraction is an integer, and that P − 2 and Q − 3 are positive integers since Q ≤ 2 and P + Q ≥ 5, inspection of the table shows that ri ≤ 3ni in each row. This table applies only when ki = 2, so we also need to consider the case ki = 1. Since 3γ + α = 2π, we have γ > π/2, so when ki = 1, we have ℓi = 0 or 1. If ℓi = 1, then since α 6= β, we have mi = ni = 1 (in which case ri = 3 ≤ 3ni ), or mi = 0 and β is an integer multiple of α. In that case, ni is at least 3, since β > α, and ri = 1, so ri ≤ 3ni . Now suppose ki = 1 and ℓi = 0. Since π = (2β + α) + β + α = 3β + 2α, we have mi = 3 and ni = 2 as one possibility, in which case ri = 6 ≤ 3ni ; and other possibilities may arise if 3β is a multiple of α, but if that happens, then ri < 6 and ni > 2, so we still have ri ≤ 3ni .
111
Hence, for the fourth row of Table 2, we always have ri ≤ 3ni . Now we sum over all vertices of the tiling that are not vertices of ABC. We have X X ri = (2mi + ℓi ) X X = 2 mi + ℓi 2(N − Q) + N
=
since R = 0
3N − 2Q
= On the other hand we have 3
X
ni
=
3(N − P )
P P Since ri ≤ 3 ni , we have 3N − 2Q ≤ 3N − 3P . Hence 3P ≤ 2Q. But Q ≤ 2 and P ≥ 3, so 3P ≥ 9 > 4 ≥ 2Q, contradiction. That completes the proof that the fourth row of Table 2 is impossible. Now we turn to the fifth (and last) row of Table 2. Here we have γ = 2α + β. The following table shows the possible vertices. The rows containing fractions represent possible vertices only in case the fraction shown is actually an integer. Since Q ≤ 1 in the fifth row of table 2, the denominator of the fractions shown is either 1 or 2. ki 2 2 2 2 2 2 2 2 2 1 1 1 1 1
ℓi 3 3 2 2 2 1 1 0 0 1 0 0 0 0
mi 1 0 2 1 0 3 3-r 4 4-r 1 0 1 1 0
ni 0 P −3 2−Q
2 P −3 2 + 2−Q P −3 2 + 2 2−Q 6 P −3 6 + r 2−Q 8 P −3 8 + r 2−Q 1 P −3 2−Q
2 P −3 2−Q P −3 2 2−Q
For this row of Table 2, the estimates used in the previous rows do not work, because of the possibility shown in the first row of this table, namely ℓi = 3 while ni = 0. Therefore we give a different argument. Note that the first four rows of the table correspond to vertices with at least two γ angles. Since there are altogether N γ angles, there are at most N/2 such vertices. Hence (and since R = 0) there are at least N/2 vertices other than those of ABC, each contributing at most one γ angle. Now consider the difference mi − ni at the two kinds of vertex. At the vertices with two or three γ angles, mi − ni ≤ 1; indeed only the first row has mi − ni > 0, and in that row it is 1. In all the other rows of the table, we have ni −mi ≥ 1. Hence, adding ni −mi over all the vertices, we obtain at least one from each of at least N/2 vertices, and at least −1 from the remaining (smaller number of) vertices, so the sum of ni − mi over all the vertices (not counting the vertices of ABC) is non-negative. But when we add in the contributions of the vertices of ABC, we add P − Q, which is a positive number (since Q ≤ 1 and P + Q ≥ 5). That is a contradiction, since altogether there are N each of α angles and β angles, so the sum over
112
all vertices, including those of ABC, of ni − mi must be zero. That completes the proof that the fifth row of Table 2 is impossible, and it also completes the proof that ℓ = 3 is impossible. We still have the cases ℓ = 4 and ℓ = 5 to deal with. We now assume ℓ = 4. That is, there is at least one vertex V where four γ angles meet, and there are at most 3 other angles, i.e. n + m ≤ 3. Since γ > π/3, we have k = 2, i.e. the angle sum at V is 2π, not π. If n + m = 0, then γ = π/2, contradiction. If n = 0 then we have 2π
=
4γ + mβ
2(α + β + γ)
=
4γ + mβ
2α + (2 − m)β
=
2γ
Since α < γ, we must have 2 − m > 0, or m < 2. Since n = m = 0 is impossible, that leaves m = 1 as the only possibility when n = 0, and in that case 2γ = 2α + β. We will prove that in this case (ℓ = 4 and n = 0) we have R = 0. By Lemma 53, we have R ≤ 1. Assume, for proof by contradiction, that R = 1. Then by Lemma 54 we have Q = 0 and β = (P − 1)α, so 2γ = 2α + β = (P + 1)α. We then have β
3, which is not possible, since m ≥ 0. Hence the case ℓ = 5 is impossible. That was the last possibility for ℓ, so the proof of the lemma is complete.
16
Main Theorem and Open Problems
Our main result is that if there is any N -tiling, then N is either a square, or a sum of two squares, or is 2,3, or 6 times a square, or twice a sum of two squares. For example, there are no N -tilings for N = 7, 11, or 19. The following theorem gives more information about the possibilities for the shapes of the tile and the tiled triangle. Theorem 7 (Main Theorem) Suppose triangle ABC is tiled by triangle T . Then one of the following holds: (i) N has the form 3m2 or 6m2 for some integer m and ABC is equilateral; and the tile is (in case N = 3m2 ) the tile used in the equilateral 3-tiling, that is, γ = 2π/3, α = β = π/6, or (in case N = 6m2 ), it is half of that tile. (ii) N is twice a square, six times a square, or twice a sum of two squares; γ = π/2; and ABC is isosceles, having angles α, α, and 2β, or angles β, β, and 2α, so that T is a right triangle similar to half of ABC. (iii) T and ABC are both equilateral. Then N is a square and the tiling is a quadratic tiling. (iv) T and ABC are similar, but not equilateral, and N is a square. (v) T and ABC are similar, but not equilateral, and N is a sum of squares but not a square. Then ABC and T are right triangles, and we have N = e2 + f 2 where e is the number of tiles on the short side of ABC, f is the number of tiles on the other side of ABC, and only hypotenuses of tiles occur on those sides of ABC, and no hypotenuses of tiles occur on the hypotenuse of ABC. The tangent of α is rational, namely e/f . (vi) T and ABC are similar, but not equilateral, and N is three times a square. Then T is a 30-60-90 triangle.
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(vii) 3α + 2β = π, triangle ABC has angles 2α, β, and β + α, s = sin(α/2) is rational, and N ≥ 28, and N contains only even powers of primes not congruent to ±1 mod 8. Here we do not assume α < β. [It is not known whether any such tilings actually exist.] Remark. Cases (v) and (vi) are mutually exclusive, since the equation 3m2 = e2 + f 2 has no integer solutions (since it has none mod 4); and also because in case (v), tan α is rational, while in case (vi), it is irrational. Proof. By Theorem 6 for cases (i) and (ii); Lemma 5 for case (iii). Case (iv) follows from Theorem 1 and Lemma 9. Corollary 3 There are no N -tilings when N is a prime congruent to 3 mod 4, such as 7, 11, 19, and 23. Proof. There is a well-known number-theoretic criterion for determining when a number can be written as a sum of two squares, that covers the stated cases. Namely, an odd number N is a sum of two squares if and only it contains only even powers of primes congruent to 3 mod 4. Hence case (v) cannot apply when N is a prime congruent to 3 mod 4. Case (vii) can also not apply. That completes the proof. We have reduced the question, “for which N do there exist N -tilings of some triangle?” to the unsolved case (vii) of the theorem, and completely answered the question “for which N do there exist N -tilings with the tile similar to the tiled triangle?”. Even if case (vii) were solved, there would still remain open problems: (1) In case (i), must the tiling be exactly the tiling described in the paper, or are there other tilings for those values of N ? The following questions were mentioned earlier, but are listed again here for completeness: (2) Does there exist a strict non-quadratic tiling in which T is similar to ABC? (Angle relations are OK.) Note that Figure 7 exhibits a non-quadratic tiling in which T is similar to ABC, but it is not a strict tiling. (3) Does there exist a non-quadratic tiling with no angle relations in which T is similar to ABC? (Non-strict vertices are OK, but the angles meeting there would have to be exactly one each of α, β, and γ.) Note that the 9-tiling in Figure 7 does have an angle relation 2γ = π. Finally, there are a number of generalizations of the tiling problem that one could consider. Here we will mention just one: what if we allow two different triangles as tiles, and require triangle ABC to be tiled using copies of those two tiles, instead of only one tile? The reader will have no difficulty constructing a triangle that can be 7-tiled if two different tiles are allowed.
References [1] Boltyanskii, V. G., and Gohberg, I. T. The decomposition of figures into smaller parts, translated and adapted from the Russian edition by Henry Christoffers and Thomas P. Branson. [2] Boltyanskii, V. G. Equivalent and Equidecomposable Figures, Translated and adapted from the 1st Russian ed. (1956) by Alfred K. Henn and Charles E. Watts. D. C. Heath (1963). [3] Borevich, Z. I., and Shafarevich, I. R., Number Theory, Academic Press, New York and London (1966). [4] Cohen, Henri, Number Theory, Volume 1: Tools and Diophantine Equations, Springer (2007). [5] Goldberg, M., and Stewart, B. M., A dissection problem for sets of polygons, Amer. Math. Monthly 71 (1964) 1077–1095. [6] Golomb, Solomon W., Replicating figures in the plane, The Mathematical Gazette48 No. 366. (Dec., 1964), pp. 403-412. [7] Soifer, Alexander, How Does One Cut a Triangle?, second edition, Springer (2009).
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