No triangle can be decomposed into seven congruent triangles

Michael Beeson

November 27, 2008

Abstract

We investigate the problem of cutting a triangle into N congruent triangles. While this can be done for certain values of N, we prove that it cannot be done for N = 7. This result is a special case of much more general results obtained in [1], but the proof in this paper may still be of some interest, because only methods of Euclidean geometry are used (including simple trigonometry that can in principle be done by geometric arguments).

1 Introduction

We consider the problem of cutting a triangle into N congruent triangles. Figures 1 through 1 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.

Figure 1: Two 3-tilings

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 exact angles of the triangle, any triangle can be decomposed into n^{2} 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”). 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 Nm^{2} triangles by subdividing the

Figure 2: A 4-tiling, a 9-tiling, and a 16-tiling

Figure 3: Three 4-tilings

first tiling, replacing each of the N triangles by m^{2} smaller ones. Hence the set of N for which an N-tiling of some triangle exists is closed under multiplication by squares.

Let N be of the form n^{2} + m^{2}. 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 m^{2}, 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 n^{2} past D. Triangle ADC can be tiled with n^{2} copies of T, arranged with their long sides parallel to the base. The result is a tiling of triangle ABC by n^{2}+m^{2} copies of T. This is a rigid tiling. The 5-tiling exhibited in Fig. 3 is the simplest example, where n = 2 and m = 1. The case N = 13 = 3^{2} + 2^{2} is illustrated in Fig. 4.

If the original triangle ABC is chosen to be isosceles, then each of the n^{2} triangles can be divided in half by an altitude; hence any isosceles triangle can be decomposed into 2n^{2} congruent triangles. If the original triangle is equilateral, then it can be first decomposed into n^{2} equilateral triangles, and then these triangles can be decomposed into 3 or 6 triangles each, showing that any equilateral triangle can be decomposed into 3n^{2} or 6n^{2} congruent triangles. Note that these are different tilings than those obtained by the method of the first paragraph of this section. 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.

The elementary constructions just described suffice to produce Ntilings when N has one of the forms n^{2}, n^{2} + m^{2}, 2n^{2}, 3n^{2}, or 6n^{2}. The smallest N not of one of these forms is N = 7. The main theorem of this paper is that there is no 7-tiling. In [1], we have completely solved

Figure 4: A 5-tiling

Figure 5: A 6-tiling, an 8-tiling, and a 12-tiling

the problem of determining the values of N for which there exists some N-tiling. The proof given here for the special case N = 7 may still be of some interest, since it uses only elementary methods of of Euclidean geometry (including some elementary trigonometry, which could be done by geometric methods). To tackle the next interesting case, N = 11 by these methods would require hundreds, if not thousands, of pages. Luckily, we found a more abstract approach in [1].

The examples of N-tilings given above are well-known. They have been discussed, in particular, in connection with “rep-tiles” [5]. A “reptile” 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 n^{2} family and the n^{2} + m^{2} 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. As far as I have so far been able to discover, there is (until now) not a single publication mentioning the concept of an N-tiling in general. The paper [4] also contains a diagram showing the n^{2} 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 [2] and [3] have tantalizing titles, but deal with other problems.

2 Definitions and Notation

We give a mathematically precise definition of “tiling” and fix some terminology and notation. Given a triangle T and a larger triangle ABC, a

Figure 6: A 13-tiling

“tiling” of triangle ABC by triangle T is a list of triangles T_{1}, …, T_{n} 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 T_{i} that does not lie on the interior of an edge of another T_{j}. A “strict tiling” is one in which no T_{i} has a vertex lying on the interior of an edge of another T_{j}, i.e. every vertex is strict. For example, the tilings shown 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 T_{i} that does not lie on the boundary of ABC. A boundary vertex is a vertex of one of the T_{i} that lies on the boundary of ABC.

In the case of a non-strict tiling, there will be a non-strict vertex V; so V lies on an edge of T_{j}, with T_{j} on one side of the edge and T_{i} (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. 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 T_{i} 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.

By the law of sines we have

\frac{a}{sin α} = \frac{b}{sin β} = \frac{c}{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.

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).

3 2-tilings, 3-tilings and 4-tilings

In this section, we warm up by characterizing 2-tilings, 3-tilings and 4tilings. Not only will these results be used later, but the ideas introduced in the proofs will also be used later.

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 either 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.

Theorem 1 If triangle ABC is 2-tiled by T, then ABC is isosceles and the tiling divides it into two right triangles by means of an altitude.

Proof. The two triangles T_{1} and T_{2} have a total of 2π angles, of which π are accounted for by the vertices of ABC. An interior vertex (strict or not) would require there to be three triangles. Hence there is exactly one more vertex, and it is a boundary vertex. Call that vertex P. Since there are only two triangles, only one interior edge emanates from P, and its other end must be at the opposite corner of triangle ABC. Relabeling if necessary, we can assume this corner is B and P lies on AB. By the lemma, γ = π/2 and the angles at P are right angles. Then AB = BC = c since these sides are opposite the right angles of T_{1} and T_{2} respectively. Hence triangle ABP is congruent to triangle CBP and the tiling is as described in the theorem. That completes the proof.

We could have reached the conclusion that just one interior edge emanates from P in another way, which seems overly complicated in this example, but will be useful below. The triangles T_{1} and T_{2} have together six boundary segments. Four of these occur on the boundary of ABC, and the other remaining boundary segments must suffice to count each interior edge from both sides. In this case there are just two remaining (because 6 − 4 = 2) and hence there is exactly one interior edge, whose two sides account for these two boundary segments. In general in an Ntiling there are Nπ radians to account for, of which π are in the corners of triangle ABC, and the rest are distributed between boundary and interior vertices. If there are k boundary vertices then there are k + 3 boundary segments on the boundary of ABC, leaving 3N −k −3 to be accounted for by counting each side of each interior edge. In a strict tiling, the number of interior edges will thus be half of 3N −k −3, but in a non-strict tiling, a more detailed accounting must be made. In the next proof, we will apply this technique to the case N = 3.

Theorem 2 If triangle ABC is 3-tiled by T, then either (i) ABC is equilateral and the tiling consists in connecting the center of ABC to its vertices, or (ii) ABC is a 30-60-90 triangle, and there is no interior vertex of the tiling; the shared side of two of the T_{i} is perpendicular to the hypotenuse of ABC at its midpoint P, and meets side b at Q, say, and the other interior edge connects Q to the vertex B (where the angle of ABC is π/3). See Fig. 1.

Proof. Suppose ABC is 3-tiled by T_{1}, T_{2}, and T_{3}. First we suppose the tiling is strict. The total of the angles in the tiling is 3π, since there are three copies of T. The total angle accounted for by the vertices of ABC is π. Each strict interior vertex accounts for 2π and each boundary vertex for π. Thus there are only two possibilities: one interior vertex and no boundary vertices, or two boundary vertices and no interior vertex.

First assume that there is one interior vertex and no boundary vertices. Since there are no boundary vertices, three of the nine boundary segments of the T_{i} are on the boundary of ABC, and the other six are doublecounted as the two sides of three interior edges. Since at least three edges must emanate from an interior vertex, all three edges do emanate from the one interior vertex P. Since there are no boundary vertices, they must terminate in the three vertices A, B, and C. That is, at least the tiling has the topology of the tiling in (i). The three angles at P must all be γ, since any other sum of three angles chosen from α, β, and γ is at most β + 2γ, which is less than 2π because

β + 2γ = β + α + (γ − α) + γ
= π + γ − α
< 2π

Hence 3γ = 2π, so γ = 2π/3. Hence the c sides of all three T_{i} are the faces of triangle ABC, which is thus equilateral. Now let AP = a; then in triangle AP C, we have P C = b; hence in triangle CP B, we have P B = a; hence in triangle P BA we have AP = b. Hence AP is equal to both a and b, so a = b and the T_{i} are isosceles. Hence the tiling is the one described in (i) of the theorem.

Next assume that there are two boundary vertices P and Q (and hence no interior vertex). Then there are five boundary segments, i.e. sides of copies of T lying on the boundary of ABC. Since there are only three triangles, two of the triangles must account for two sides each, i.e. two of the angles of ABC are not “split”, i.e. are not shared by more than one T_{i}. Hence each T_{i} is similar to triangle ABC. Of the 9 sides of the three T_{i}, five occur on the boundary of ABC, and the other 4 occur in the interior. Since each interior side is counted twice, as a boundary of the triangles on either side, there must be exactly two interior edges. One of these interior edges must connect P and Q, because if not, then both interior edges would have to connect P or Q to the opposite vertex. But if one edge connects (say) Q to the opposite vertex, then the edge from P is blocked from reaching the opposite vertex, and vice-versa, if one edge connects P to the opposite vertex, the other edge cannot connect Q to the opposite vertex. Hence it cannot be that both interior edges connect P or Q to the opposite vertex. The only other possibility is that one of these edges connects P to Q. The other interior edge must connect one of P or Q to the opposite vertex of ABC, which must be split. That means that one of P or Q (by relabeling we can assume it is P) has only one interior edge emanating from it. That implies that γ is a right angle, by Lemma 1. Changing the labels A, B, and C if necessary, we can assume that P lies on AB, Q lies on BC, and QA and QP are the interior edges. Angles QP B and QP A are right angles, and triangle AQP is congruent to triangle QP B. Sides AQ and QB are opposite the right angle and hence are equal. Hence AP = P B and P is the midpoint of AB. The angle at B is not split. Since triangle ABC is similar to each triangle T_{i}, but its area is 3 times larger, the similarity factor is \sqrt{3}. Let T_{1} and T_{2} be the two triangles sharing side P Q, with T_{1} = QP A and T_{2} = QP B. Then T_{3} shares side AQ, which is side c in triangle T_{1}, so the third vertex C of ABC, which is also the vertex of T_{3} opposite AQ, must be the right-angled vertex of ABC. Now triangle CAB is similar to triangle P QB, since they have the same angle at B and right angles at C and P respectively. Hence AB and QB are corresponding sides. Their ratio is therefore \sqrt{3}, i.e. AB = \sqrt{3}QB. But since AB = AP + P B = 2AV, we have 2P B = \sqrt{3}QB. Hence angle B = π/6 and angle P QB = π/3, and the tiling is as described in (ii) of the theorem. That completes the proof in case of a strict tiling.

Now suppose the tiling is non-strict. Since only three triangles are involved, the only possible type of non-strict vertex is the type we shall call 2 : 1 below, where one side of (say) T_{1} is matched by two sides, one of T_{2} and one of T_{3}. There cannot be two such vertices as the three triangles will have only this one side of T_{1} in common, and if the sides of T_{2} and T_{3} that touch do not have the same length, a triangle ABC will not be formed. Hence, of the 3π in total angles, π is accounted for at the interior vertex P, and π is accounted for by the vertices of ABC, leaving π to be accounted for by a single boundary vertex Q. With one boundary vertex there are 4 boundary segments on the boundary of ABC, leaving 3 · 3 − 4 = 5 in the interior (counting each side of each interior segment). Three of those are the three sides that lie on the maximal segment of the non-strict vertex P. The other two are the two sides of one more interior segment with an endpoint at P. Since there is only one boundary vertex, the three endpoints of the interior segments must end at Q and at two corners of the triangle. The maximal segment must end at Q and one corner, which we may label A, and the other interior segment runs from P to another corner, say B. Since only two triangles share vertex Q, we have γ = π/2 by Lemma 1. But now triangle QP B has two right angles, at Q and P. That contradiction completes the proof.

Theorem 3 If triangle ABC is 4-tiled by T, then (a) there is no interior vertex, and (b) T is a 30-60-90 triangle, and the tiling is one of those shown in Fig. 3 (or a reflection of these), or T can be any triangle and the tiling is one of the n^{2} family as illustrated in Fig. 2.

Proof. First suppose the tiling is strict. The four triangles have angles totaling 4π. The vertices of ABC account for π of this, and the remaining 3π must be accounted for. There are just two possibilities: one interior vertex and one boundary vertex, or no interior vertices and three boundary vertices.

First assume there is one interior vertex P and one boundary vertex Q. Then there are four boundary segments and (12 − 4)/2 = 4 interior edges. Then these four edges must emanate from P and go to A, B, C, and P. By the lemma, the angle at Q is a right angle and γ = π/2. Hence all four angles at P must be right angles. Then triangle AP Q has two right angles, contradiction. That disposes of the case of one interior vertex and one boundary vertex.

Next assume there are three boundary vertices and no interior vertex. Then there are six boundary segments and (12 − 6)/2 = 3 interior edges.

First assume that one of the interior edges terminates in A, B, or C (splitting the angle there). Then there are not enough edges to provide two edges at each boundary vertex, so one boundary vertex has only one edge terminating there. Hence by the lemma, γ = π/2. Label the split vertex B and let Q be the interior vertex at the other end of the interior edge emanating from B. Then either triangles ABQ and CBQ are both 2-tiled, or one is 3-tiled and the other is congruent to the tile T. First assume ABQ is 3-tiled. There are two possible 3-tilings; in one case, angle AQB is π/3, and the fourth triangle T_{3} can contribute only an angle of π/6 at each vertex, not enough to make π and remove a vertex. So this case is impossible. In the other possible 3-tiling, ABQ is a 30-60-90 triangle similar to T, and the three sides of ABC are a + c, 2b, and b. Neither a + c nor 2b can be a side of T, so we must have QB = b, and angle C = π/3. We then necessarily have the third tiling show in Fig. 3 (or its reflection). Next assume that ABQ is 2-tiled. Then AQ = QB and QP is an altitude of triangle AQB. The third interior vertex R must lie on QC or on BC. First assume R lies on BC. Then the third interior edge is QR, and both angles at R are right angles by the lemma. We therefore have the first tiling shown in Fig. 3. Next assume R lies on AC. Then the third interior edge must be BR, and angle QRB must be a right angle by the lemma. Then triangle P QB is congruent to triangle RQB and we have the third tiling in Fig. 3. This disposes of the case in which one of the interior edges terminates in a vertex of ABC.

Now the three interior edges terminate only in the three boundary vertices. It follows that triangle ABC is similar to triangle T; since there are four triangles, the similarity factor is 2: triangle ABC is twice the size of T, and the same shape. If two boundary vertices lie on the same side of ABC, three interior edges cannot exist. Therefore one boundary vertex lies on each side of ABC. Label them so that P lies on AB, Q lies on BC, and R lies on AC. Then the three interior edges form triangle P QR. Let RQ = a, P Q = b, and RQ = c. Then if CQ = a, it follows that QB = a and AR = RC = b, and we have an n^{2}-family tiling. If, on the other hand, CQ ≠ a, then since RQ = c we must have CQ = b and b ≠ a. But then RC = a and hence AR = a. But by definition RP = a, and by the similarity of ABC to T, we have AB = 2c and hence AP = c. Hence a = b, contradiction. This completes the proof in the case of a strict tiling.

Now assume that there is a single non-strict vertex P. If this vertex is of type 3 : 1 then every one of the four triangles shares a side with the maximal segment S. There are no more triangles that can share the two interior vertices on S, so only one edge emanates from each of these vertices on the side bounding three triangles. By the lemma the angles at these vertices are right angles. But then the middle of the three triangles has two right angles, contradiction. Hence the non-strict vertex does not have type 3 : 1. If it has type 2 : 2, then similarly the angles at the interior vertices are right angles, so the union of the four copies of T has four vertices and cannot be a triangle ABC. Therefore the type of the non-strict vertex must be 2 : 1.

The non-strict vertex accounts for π of the 4π angles of the T_{i}, and since π is accounted for by the corners of ABC, that leaves either one strict interior vertex and no boundary vertices, or two boundary vertices and no strict interior vertices.

First assume there is one strict interior vertex Q and no boundary vertices. The maximal segment must run from a vertex (say B) to Q, since it cannot run to another vertex of ABC. The other edges emanating from Q must be AQ and BQ. There are 3 boundary segments and 9 doublecounted interior edges. The maximal segment, since it is of type (2 : 1), contains 3 of these edges, and the other 6 correspond to three additional interior edges. These are AQ, CQ, and the other edge emanating from P. The endpoint of that edge must be a vertex of ABC, which by relabeling we can assume is C. Then by the lemma, the angles at P are right angles, and γ = π/2. But only three angles meet at Q, so one of them must be greater than π/2. This contradiction disposes of the case of one strict interior vertex.

Therefore the second case must hold: there are two boundary vertices and no strict interior vertices. Then there are five boundary segments and seven double-counted interior segments, of which three lie on the maximal segment, so there are two additional interior segments, one of which has one end at the non-strict vertex P. That makes five ends of interior segments (two of the maximal segment, and three of the four ends of the two additional interior segments) that must terminate at two boundary vertices and/or the three corners of ABC. The maximal segment cannot connect two vertices of ABC, so it must have one end at a boundary vertex Q. The other end of the maximal segment is either at a vertex of ABC or at the other boundary vertex R.

Assume first that the maximal segment connects Q to a vertex of ABC. Relabeling, we can assume it is vertex B, and the Q lies on AC, and triangle CQB is congruent to T while triangle AQB is 3-tiled. By Theorem 2, it follows that AQB is a right triangle or is equilateral. First assume AQB is equilateral. Then angle CQB is π/6 and angle AQC = 5π/6, not a straight angle, contradiction. Hence AQB is not equilateral. Hence the other case holds, namely that AQB is a 30-60-90 triangle, tiled as in Fig. 1. Since QB is a single side of T in triangle QBC, the tiling of AQB must be oriented so that QB is the smallest side of QBA. The tiling then must be the third one shown in Fig. 3, or its reflection.

The only remaining case is that the maximal segment connects Q to the other boundary vertex R. Relabeling, we can assume that Q lies on AC and P lies on AB. Assume first that an interior segment emanates from P on the same side of P Q as A. Then its other endpoint must be A. Hence the angles it makes at P are right angles and QP = P R. The second interior segment cannot have endpoints at both Q and R, so at one of those points, only the maximal segment meets the boundary of ABC. By the lemma then, angle AQR or angle BQR is a right angle, contradiction, since that would make two right angles in one of those triangles. Hence no interior segment emanates from P on the same side of P Q as A. Then triangle T is congruent to AQR. Since not both angle CQR and angle BRQ can be right angles (else QC and RB would be parallel), an interior segment must emanate from one of them; relabeling, we can assume it is R. The other endpoint of this segment must be C. The other interior segment has one endpoint at P, and its other endpoint must also be at C, since there is no interior vertex. Then by the lemma, CP is perpendicular to QR. Also by the lemma, P Q must be perpendicular to AC. But then triangle CQP has right angles at both P and Q, contradiction. That completes the proof.

4 Strict 7-tilings

Theorem 4 There is no strict 7-tiling.

Proof. Consider a strict 7-tiling. Since it is composed of 7 triangles, the angles make a total of 7π. Of that 7π, there is π in the corners of the large triangle, and 2π for each interior vertex, and π for each boundary vertex (i.e. vertex lying on an edge of the large triangle but not in a corner). Therefore we have either: zero interior vertices and 6 boundary vertices, or one interior vertex and 4 boundary vertices, or two interior vertices and two boundary vertices. We consider these three cases one by one.

Case 1: Zero interior vertices and 6 boundary vertices. Since there are 6 boundary vertices, there are 9 sides of triangles on the boundary and (21 − 9)/2 = 6 interior edges. If at any boundary vertex, only one interior side terminates there, then γ must be a right angle. Assume that γ is not a right angle. Then consider a boundary vertex P on side AB, the next vertex to A. It must connect to vertex Q on side AC, the next vertex to A (else no side can escape from Q). The other edge from P must not connect to side AC, or else no edge can escape from Q. So P connects to a vertex R on side BC. The second edge from Q must connect either to R or to a vertex S on BC between C and R. Assume the latter. But then, a second edge from S must intersect either P R or P Q, so that case is impossible and Q connects to R. Consider another boundary vertex U. Relabeling, we can assume U lies on P B or RB. Assume first that U is on RB. The other end of both edges leaving U must be at P or at another vertex V on P B, since there are no interior vertices. Since two edges leave U, there is such a vertex V on P B and UV is an edge. Then either UP or RV must be an edge, to tile quadrilateral P RUV. Assume it is RV. Then the second edge leaving U must terminate at another boundary vertex W on V B, and there are no more edges to leave W, so the angles at W are right angles and γ = π/2. Similarly if UP is an edge instead of RV, the second edge leaving V must terminate at another boundary vertex W on UB, and there are no more edges to leave W. Hence the assumption that U is on RB has led to a contradiction (under the assumption that γ is not a right angle). The remaining alternative is that U is on P B. But then the argument proceeds similarly: the other end of both edges leaving U must be at R or at another vertex V on RB. Since two edges leave U, there is such a vertex V on RB and UV is an edge. From that point the argument is exactly the same as in the case U is on RB. Hence the assumption that γ is a right angle has led to a contradiction. Hence γ is a right angle.

Six boundary vertices means that 9 edges of triangles lie on the boundary of the large triangle. Since there are only seven triangles, that means that there exist two triangles with two edges on the boundary, i.e. there are two triangles located at vertices of triangle ABC which do not share that vertex with any other triangle. Hence the final triangle ABC is similar to triangle T. Since there are supposedly seven copies of T tiling ABC, the similarity factor is \sqrt{7}. Let us suppose that the final triangle ABC has angle α and A, angle β at B, and a right angle γ at C. Consider the triangle T_{1} of the tiling that has a vertex at A. It must have angle α at that vertex. Let P be its vertex on AC and Q its vertex on AB. Then P Q = a. Let T_{2} be the triangle that shares side P Q with T_{1}, and let R be its third vertex. Since there are no interior vertices, R must lie on AB, or on AC, or on BC. Case 1a, R lies on AC. Then both T_{1} and T_{2} have their right angle at P. R is not equal to C since it is 2b from A, while C is \sqrt{7}b from A. Consider the triangle T_{3} that shares side QR with T_{2}. Let S be its third vertex. What is angle QRS? It cannot be γ, since side c of triangle QRS is QR. It cannot be β, since that would make angle ARS equal to α + β, a right angle, and side RS equal to a, which would leave S in the interior of ABC. Hence angle QRS must be α, and hence angle ARS = 2α < π/2, so S must lie on AB. That makes the total angle π at Q equal to 3β, so β = π/3, α = π/6, so a = sin α = 1/2 and b = cos α = \sqrt{3}/2. Any linear combination of a, b, and c is thus of the form p + q\sqrt{3}; but AC = \sqrt{7} is the sum of several sides of the basic triangle, contradiction. This contradiction disposes of Case 1a. Case 1b, R lies on AB. Then the right angles of T_{1} and T_{2} are both at Q. Let T_{3} be the triangle sharing side P R with T_{2}, and let S be its third vertex. If S lies on AC, then angle RP S must be either β or α; if it is α then 2β + α = π, which is impossible since α + β = π/2 and β < π. If it is β then 3β = π, which is impossible as in the previous case. Hence S does not lie on AC. S cannot lie on AB as that would make angle P RS more than π/2. Now T_{1}, T_{2}, and T_{3} all share vertex P, but since S does not lie on AC, a fourth triangle T_{4} shares that vertex too, and has angle at least α there. If T_{3} has angle β at P then the total angle at P is at least 3β + α > 2β + 2α = π, contradiction. Hence T_{3} has angle α at P, and P QRS is a paralleogram. Let triangle T_{4} be the triangle sharing side P S = b with T_{3}. The angle of T_{4} at P must be either α or a right angle, but a right angle is too large, since the other angles at P total π/2 + β. The right angle of T_{4} must therefore be at S, making S an interior vertex, contradiction. That disposes of Case 1b, R lies on AB. But since the right angle of T_{1} must lie at either P or Q, Case 1a and Case 1b are exhaustive, so Case 1 has been shown to be impossible.

Case 2: one interior vertex, and four boundary vertices. Then there are (21 − 7)/2 = 7 interior edges. There can be at most one triangle T_{i} that has one vertex on each side of ABC. (Call such a triangle an “interior triangle”.) Hence six or seven triangles have one or more sides on the boundary of ABC. With four boundary vertices, there are seven boundary segments to be accounted for. If there is an interior triangle, then one of the remaining six must account for two boundary segments, so one of the vertices A, B, or C is not “split”, i.e. shared by two or more triangle T_{i}. If there is no interior triangle, then each of the seven triangles must account for exactly one boundary segment, which means that all of the vertices A, B, and C are “split”.

Case 2a: There is an interior triangle. Let T_{1} be that triangle, having vertex P on AB, vertex Q on BC, and vertex R on AC. That creates three triangles BP Q, AP R, and QRC. The single interior vertex must occur in the interior of one of these triangles (since this is a strict tiling, it cannot occur on the boundary of T_{1}). Relabeling the vertices if necessary we can assume that it occurs in triangle AP R. At least three edges leave that interior vertex, so triangle AP R is divided into at least three triangles congruent to T. In fact it must be divided into exactly three triangles, since at least three are needed for T_{1}, BP Q, and QRC, so the possibilities are three or four, but there is no 4-tiling with an interior vertex, by Theorem 3 Hence AP R is 3-tiled, and there is only one 3-tiling with an interior vertex, by Theorem 2. Therefore T is the triangle with α = β = π/6 and γ = 2π/3, and triangle AP R is equilateral. Consider the angles at R: angle ARP is π/3, and P RQ is π/6, so QR is perpendicular to AC and therefore angle QRC must be composed of three angles α. That means that triangle QRC contains three smaller triangles, which makes seven counting the three in AP R and T_{1}, leaving none to cover BP Q. This disposes of case 2a.

Case 2b: No interior triangle, and all vertices A, B, and C are split. Then there is an interior edge emanating from each of A, B, and C. Any pair of these must intersect in an interior vertex. But there is only one interior vertex, so they all intersect in a common point P, the interior vertex, forming three triangles ABP, ACP, and BCP. These triangles are tiled by T, without interior vertices, since there is only one interior vertex. Hence none of them is 3-tiled. A 4-tiling requires a boundary vertex on each side, which would mean another interior vertex, so none of them is 4-tiled either. They cannot all be 2-tiled, as that would use only six triangles. That leaves only the possibility that two of them are congruent to T and the other is 5-tiled; relabeling vertices if necessary, we can assume it is ABP that is 5-tiled. Since there are four boundary vertices they must all be on AB and the five triangles all share vertex P, with one side contained in AB. This is impossible for several reasons, for example, since just two triangles share each boundary vertex, all those angles must be right angles, contradicting the fact that all those edges meet at P. This disposes of Case 2b, and hence of Case 2.

Case 3: two interior vertices and two boundary vertices. Then there are (21−5)/2 = 8 interior edges. Suppose first that two boundary vertices P and Q occur on AB, with P adjacent to A and Q adjacent to B. Let U and V be the interior vertices. If γ is not a right angle, then two edges must leave P and two edges must leave Q. One of the edges from P, and one from Q, can go to an interior vertex, and one to vertex C. One more edge can go from P or from Q to an interior vertex, but after that we are blocked–there is no place to put the rest of the 8 edges. Hence the two boundary vertices do not occur on the same side of the large triangle. Say P occurs on AB and Q occurs on BC. Then each of P and Q can connect to both interior vertices, and one of them can connect to an opposite vertex of the large triangle, but that is not enough edges. Therefore P and Q do not connect to C. To use up 8 edges, we must have an edge connecting U and V, and one of U and V, say V, connects to both B and C, while U connects to A. But then, there are exactly four angles at V, totaling 2π. That means two of them must add to at least π, which means γ is a right angle.

If just three edges emanate from V then there are exactly three angles at V. If three angles add to 2π, they must all be γ, since 2γ +β < 2π. But then γ = 2π/3, contradicting our conclusion that γ is a right angle. Hence at least four edge emanate from U and four from V, for the required total of 8. Every one of the eight interior edges then has one end at U or one end at V. The four edges emanating from U go to the edge vertices P and Q and to two vertices of the large triangle, say A and B. V must lie in one of the four regions formed by the angles at U; three of those are triangles, which leave only room for three edges to emanate from V. Hence V must lie in the quadrilateral UP CQ, and must connect to all four corners of that quadrilateral. Now we have five edges emanating from V and four from U, and as above, all the angles at U are right angles. Now consider the angles at P. Angle AP U and angle UP V are either α or β, since those triangles have their right angle at U. Hence angle V P B is γ, a right angle. But then P C = c, since it is opposite the right angle at V, and on the other hand it is less than c, since it is opposite angle V BP, which is not a right angle. This contradiction eliminates Case 3.

That completes the proof of the theorem.

5 Non-strict 7-tilings

Lemma 2 Suppose that a 7-tiling contains a non-strict vertex of type 3 : 1. Then the tile is a right triangle, c = 3a, and the smallest angle α satisfies the equation sin α = 1/3.

Proof. Let the maximal segment be P Q, running from P in the “north” to Q in the “south.” Suppose three triangles T_{1}, T_{2}, and T_{3} occur on the west side of P Q, meeting P Q at vertices U and V, and T_{4} lies on the east side of P Q. Let c be the longest side of T; then c = 3a or c = 2a + b. It is impossible that the three congruent triangles have a common vertex S, so that SP U, SUV, and SV Q are congruent triangles. Hence there are two distinct points S and R such that T_{1} = SP U and T_{3} = RV Q are two of the triangles in the tiling. T_{2} may have a side SU in common with T_{1} or a side RV in common with T_{3}. In either case the common side is perpendicular to P Q and T is a right triangle, so γ = π/2 and α + β = π/2. We have P U = UV = a, and SV = c = 3a or 2a + b. If SV = 3a then sin α = UV/SV = 1/3 as claimed in the lemma. The case c = 2a + b is impossible, since no right triangle has sides a, b, and 2a + b.

If, on the other hand, T_{2} does not have a side in common with T_{1} or T_{3} then there will be altogether five triangles on the left of P Q sharing vertices U and V. Let W be the west vertex of T_{2}. Then W cannot lie on SP, since if it does, W U is longer than both SP and SU, but one of those sides must be the longest side of triangle T, since the third side of SP U is less than P Q, which is one side of the copy of T on the right of P Q. Similarly, W does not lie on RQ. Let T_{5} and T_{6} be the other copies of T in M sharing vertices U and V, respectively. Let M be this six-triangle configuration. We claim that the boundary of M contains at least five non-straight angles. At R there is either another non-strict vertex with a non-straight angle, or at least (if R is a vertex of T_{2}) the boundary of M is not straight. Similarly at S. There is an angle at the east vertex of T_{4} (the triangle on the right of P Q). We claim there are also non-straight angles at P and Q. For those to be straight angles, we would have to have the sum of two angles of T equal to π. But at (one of) P or Q, the angle of T_{4} is the small angle α, the angle from T_{1} or T_{3} is not α, since α is the angle at S or R; so the sum of the two angles is less than π (in fact it is π − β). At the other of P or Q we would need β to be a right angle to create a straight angle. In that case T_{2} would have a side in common with T_{1} or with T_{3}, contradiction. Hence there are vertices of M (at least) at P, Q, R, S, and the third vertex of T_{4}–five in total. Suppose it were possible to place one more copy of T next to M so as to form a triangle. If the triangle is placed to the right of P Q against one of the sides of T_{4}, that may eliminate a vertex at P or Q, but will leave a vertex at the other of P or Q, as well as creating one more new vertex and leaving three old ones–too many vertices for a triangle. If the triangle is placed to the left of P Q, against T_{1} or T_{3}, again it may eliminate a vertex at P or Q, and possibly at S or R, but it will create a new vertex and leave at least three old ones. Placing it anywhere else will leave vertices at all three vertices of T_{4}; but since part of M exists outside the convex hull of those vertices, those cannot be the vertices of a triangle containing M. Hence, a tiling with a vertex of type 3 : 1 in which T_{2} does not have a side in common with T_{1} or T_{3} cannot occur in a 7-tiling. That completes the proof of the lemma.

Lemma 3 A 7-tiling cannot contain a maximal segment of type 3 : 2.

Proof. Consider a non-strict tiling containing a maximal segment of type 3 : 2. Let T_{1}, T_{2}, and T_{3} occur on the left of the (vertical) maximal segment P Q, with vertices U and V on P Q. Already five triangles will have vertices on the maximal segment P Q. If they do not have sides in common then three more triangles will be required to fill in the gaps–more than seven altogether. Say then that T_{1} and T_{2} have a side in common.

Let U be the vertex that T_{1} and T_{2} share on P Q. By Lemma 1, γ = π/2 and T_{1} and T_{2} both have a right angle γ at R. Suppose, for proof by contradiction, that the sides of T_{1} and T_{2} on P Q are equal to a, and that T_{1} and T_{2} share a common side (but not necessarily a common vertex west of P Q). But then they do share a common vertex S west of P Q, since they have their angles at U equal (both right angles) and their west angles both equal to α, hence the angles SP U and SV U are both β, and the sides opposite are both b, so the common vertex S is b away from U. Then triangle SP U is congruent to SV U. Let T_{4} and T_{5} be the triangles on the east side of P Q, sharing vertex R on P Q, and let T_{6} be another triangle west of P Q sharing vertex V (there must be one since T_{2} does not have a right angle there.) Let E be the east vertex of T_{4}. If T_{4} and T_{5}, the two triangles on the east of P Q, do not share a side (and hence have right angles at R), then a seventh triangle must occur between them, and we have too many vertices: P, Q, W, and S at least. Hence T_{4} and T_{5} do share a side, and their angles at P and Q are acute, and their angles at R are right angles. Hence EP and EQ are equal to c and P Q is 2b, since it cannot be 2a as it is the sum of three sides of T_{1}, T_{2}, and T_{3}.

Then this six-triangle configuration M has vertices at P, E, Q, S (the shared west vertex of T_{1} and T_{2}), and the west vertex W of T_{3}. Triangle T_{6} must then fill the angle at V, or else the seventh triangle would need to touch V, leaving more than three exterior vertices, namely P, Q, E, and at least one vertex west of P Q.

Suppose, for proof by contradiction, that W lies on line SQ. Then angles V W Q and V W S are right angles, so V Q, the side opposite angle V W Q in triangle T_{3}, equals c, and triangle SV W is congruent to QV W. M then forms a quadrilateral, and along P Q we see 2a + c = 2b. Angle W SV = α, by the congruence of triangles W SV and W QV. The angles of triangle SP Q are β at P, α at Q, and hence γ = π/2 at S. This angle at S is also equal to 3α, since each of triangles T_{1}, T_{2}, and T_{6} has angle α there. Hence α = π/6, so a = \frac{1}{2} , b = \sqrt{3}/2, and c = 1. But then the equation 2a + c = 2b does not hold, contradiction. Hence W is does not lie on line SQ.

Hence there are two vertices of M west of P Q (either S and W or vertices of T_{6}). M thus has at least five vertices. To reduce this to three vertices by placing one more triangle east of P Q is impossible. Similarly it is impossible to reduce the number of vertices to three by placing a new triangle along P S or W Q. But then, no matter where else we place T_{7}, P, E, and Q will remain vertices, and there must be a fourth vertex west of P Q, so the result cannot be a triangle. This contradiction shows that we cannot have V U = P U = a as we assumed above.

Now we drop the contradictory assumption V U = P U = a and begin anew. Again we have right angles in T_{1} and T_{2} at U, but this time one of them has side b along P Q (not c, since that must be opposite the right angle at U). The other one has side a along P Q, since 2b ≥ a + b > c. Hence they do not share a common west vertex. Let S be the west vertex of T_{1} and X the west vertex of T_{2}. Again T_{6} will have to be placed between T_{2} and T_{3} with a vertex at V. That will give us a six-triangle configuration M with vertices P, E, Q, S, X, and W (the west vertex of T_{3}). We only have to show that placing one more triangle T_{7} cannot possibly produce a triangular configuration. (That is not prima facie impossible just because there are six vertices–it could happen if M had two collinear sides separated by two sides forming a “notch” into which T_{7} would just fit–so some further argument is required.) If the two triangles T_{4} and T_{5} east of P Q do not share a side, then T_{7} would have to be placed east of P Q between the two, leaving a vertex at P (since T_{1} has an acute angle at P), as well as vertices at X and S, which are distinct, and of course at least one vertex east of P Q, totaling more than three. Hence T_{4} and T_{5} do share a side; hence their right angles are both on P Q at vertex R. Hence their angles at P and Q are acute, and triangle P ER is congruent to triangle QER. Hence the two sides P R and RQ are equal, and either equal to a or to b. The length of P Q is thus either 2a or 2b (measured from the right side), and also either 2a + b or 2b + a (measured from the left side). Three of the four possible equations here are immediately impossible, leaving only the possibility 2b = 2a+b; hence b = 2a. Now if the angle of T_{1} at P is α, then the southwest vertex S of T_{1} lies on the north side of T_{2}, and S is thus not on the convex hull of M, and triangle T_{7} cannot fill the obtuse angle π − β exterior to M at S. Hence the angle of T_{1} at P is not α; so it must be β. Then the northwest vertex X of T_{2} lies on the south side of T_{1}, and thus not on the convex hull of M, but there is an obtuse exterior angle π − β at X that cannot be filled by T_{7}. We conclude that it is not possible to place T_{7} to create a triangle. That completes the proof of the lemma.

Lemma 4 A 7-tiling cannot contain a maximal segment of type 4 : 1.

Proof. Let P Q be a (north-south) maximal segment with four triangles on the left and one on the right. Of the four triangles on the left, we cannot have three sharing a vertex not on P Q, so the minimum “tile” (consisting of all the triangles touching P Q) contains at least six triangles, and contains seven without making a triangle, unless the four triangles on the left occur in two pairs, each pair having a common side. Let the northernmost of these pairs be T_{2} and T_{3}, and let T_{4} and T_{5} be the southern pair. Let V be the vertex on P Q shared by T_{3} and T_{4}. Let T_{1} be the triangle on the right of P Q. Let E be the eastern vertex of T_{1}. let R be the vertex between P and V (shared by T_{2} and T_{3}) and S the vertex between V and Q (shared by T_{4} and T_{5}). Let M be the figure formed by these five triangles. Note that we have not proved that triangles T_{2} and T_{3} have a common “west” vertex, i.e. their shared sides may be of different length, and the same goes for T_{4} and T_{5}.

By Lemma 1, γ = π/2 and T_{2} and T_{3} have right angles at R, and T_{4} and T_{5} have right angles at S. Then P and Q are vertices of M, as are E and the western vertices of T_{3} and T_{5}. Thus we cannot afford to insert two triangles between T_{3} and T_{4} (that is, anywhere inside the angle at V between T_{3} and T_{4}), as that will make seven altogether and the result will not be a triangle, as it must contain a westernmost vertex in addition to P, Q, and E. Hence there is just one triangle T_{6} between T_{3} and T_{4}. The longest side c of T_{3} (the hypotenuse) is shared with T_{6}, and since no other triangle can be inserted between T_{3} and T_{4}, the side of T_{6} shared with T_{3} must also have length c. Similarly, T_{6} must share side c with T_{4}; but then T_{6} has two two sides equal to the hypotenuse, which is a contradiction. This contradiction completes the proof of the lemma.

Lemma 5 No 7-tiling contains a non-strict vertex of type other than 3 : 1 or 2 : 1 or 2 : 2.

Proof. Let V be a non-strict vertex in a 7-tiling. Then for some integers m and n, there is a maximal segment S containing V of type m : n. We have m + n ≤ 7 since there are only 7 triangles in the tiling. Visualize S as oriented in the north-south direction, with n triangles west of S and m triangles east of S. Let M be the configuration of triangles in the final tiling that touch S. No more than two triangles on the same side of S can share a common vertex that is not on S. Hence if n (or m) is three, then at least four triangles must occur on the west (or east) of S; and if n (or m) is four, then at least five triangles must occur on the west (or east); and neither n nor m can be as much as five, since then at least seven triangles would be required on one side of S.

We may change “east” and “west” if necessary to ensure m ≤ n. Suppose n = 4. Since we have proved above that S cannot be of type 4 : 1, there are at least two triangles east of S, and as remarked above, at least five west of S. For this to be the case, the triangles T_{1} and T_{2} on the northwest must share a side and both have right angles where they meet S, and the same for triangles T_{3} and T_{4} on the southwest, and for triangles T_{5} and T_{6} on the east, and then T_{7} must share vertex V on S with T_{2} and T_{3}. That means that the largest angle γ is a right angle, and the seven-triangle configuration has vertices at the endpoints of S and at least one vertex east of S and at least one vertex west of S, and hence is not a triangle. Hence n = 4 is not possible.

Suppose n = 3. We have proved above that type 3 : 2 is impossible. We now consider type 3 : 3. But as remarked above, this would require four triangles east of S and four triangles west of S, making more than seven, so 3 : 3 is impossible. That completes the proof of the lemma.

The following figures show some possible configurations in which a maximal segment of type 2 : 1 could occur. In these figures, α has to be as shown, either π/6 or arctan \frac{1}{2} , so that 2a = c or 2a = b. In the first two figures, γ has to be a right angle. In the third figure, γ and β have one degree of freedom; the figure illustrates the case γ = 80 degrees.

Figure 7: Two three-triangle configurations

P Q V W  E  P Q V W  E

Figure 8: Two five-triangle configurations

P V Q  E R S W  R W S  P  E Q V

Figure 9: Two four-triangle configurations

Q P  E V S R  Q P  E V S R

Lemma 6 Suppose that a 7-tiling contains a maximal segment of type 2 : 1. Then the tiling contains one of the six configurations shown in the preceding figures. To state the conclusion without reference to a figure: the smallest angle α of the tile is π/6 or arcsin \frac{1}{2} , so 2a = c or 2a = b; the maximal segment has length 2a, with the non-strict vertex V at its midpoint, and one of the following holds.

(i) 3 triangles meet at the non-strict vertex, two of them having a right angle there. See Fig. 8.

(ii) 5 triangles meet at the non-strict vertex. Denoting the maximal segment by P Q, with midpoint V, the triangles “west” of P Q with vertices at P and Q have right angles at P and Q, and angle β at V, and the other two triangles west of P Q have angle α at V. See Fig. 9.

(iii) 3 triangles on one side of the maximal segment share the vertex V, with the middle one (the one that does not share a side with the maximal segment) having angle α at V, and sharing another vertex with each of the other two triangles with which it shares vertex V. Moreover, each two adjacent triangles of the three on one side of the maximal segment form a parallelogram. See Fig. 10.

Note that in cases (i) and (ii), the tile is a right triangle, while in case (iii) that is not asserted.

Proof. The previous lemmas have ruled out all possible types of maximal segments except 3 : 1, 2 : 2, and 2 : 1. We consider the possible configurations in which a maximal segment has type 2 : 1. For convenience of description, let us orient triangle ABC so the maximal segment P Q is north-south, with two triangles on the west and one on the east. Because there is just one triangle on the east, the length of P Q is either b or c. The two triangles on the left must divide the maximal segment equally, because if they did not, then the two segments would have lengths a and b and their sum (the side of the one triangle on the right) would necessarily be c, but of course a + b < c. It follows that the two segments have length a, and 2a = b or 2a = c, since if the segments had length b instead of a, we would have c = 2b ≥ a + b, so a triangle with sides a, b, and c would be impossible.

We consider the “minimal configuration” M containing all the T_{i} with a vertex at V. How many triangles will M contain? It contains at least three. We will analyze the possibilities.

Let the direction of P Q be “north-south” with P at the north. Let T_{1} and T_{2} be the west triangles and T_{3} the east triangle with sides on P Q, and V the midpoint of P Q, the shared vertex of T_{1} and T_{2}. If T_{1} and T_{2} share a side, then the largest angle γ is a right angle, so T_{1} and T_{2} must also share their west vertex W (at distance b from P Q). Then we have 2a = b or 2a = c, leading to the possibilities listed in part (i) of the conclusion and illustrated in Fig. 8.

We therefore may assume that T_{1} and T_{2} do not share a side, and at least one additional triangle T_{4} shares their common vertex V at the midpoint of P Q. Let S be the west vertex of T_{1} and R the west vertex of T_{2}, and E the east vertex of T_{3}. Then angles P SV and V RQ are both α, since they are opposite side a.

Assume first that there is exactly one triangle T_{4} between T_{1} and T_{2} sharing vertex V. One of the sides of T_{4} lying along SV or RV must be b or a and since the a sides of T_{1} and T_{2} lie on P Q, the sides SV or RV must each be b or c. Assume, for proof by contradiction, that triangle T_{4} does not share west vertices with T_{1} and T_{2}, i.e. it is not triangle SV R. Then one of its vertices lies on SV or on RV, since the b side of T_{4} cannot be longer than the b or the c sides of T_{1} and T_{3}. Interchanging “north” and “south” if necessary, we can assume that the north vertex X of T_{4} lies on SV, and hence is a non-strict vertex. Since SV must be larger than XV, we have either SV = c and XV = b, or SV = c and XV = a, or SV = b and XV = a. The maximal segment of this non-strict vertex has one end at V and extends westward along SV. By Lemma 5, its type must be 2 : 1 or 3 : 1 or 2 : 2.

Case 1, angles SV P and RQV are equal. Since these are corresponding angles made by the transversal P Q to SV and RQ, lines SV and RQ are parallel. Then the alternate interior angles SV R and V RQ are equal. Angle V RQ = α, since it is opposite V Q = a. Therefore angle SV R = α. Therefore XV = b or XV = c; but since X lies on SV and X ≠ S we have SV = c and XV = b. If the type of the maximal segment of X is 2 : 1, we have a + b = c, which is impossible since a + b > c, or 2b = c, which is impossible since 2b > a + b > c. If it is 3 : 1 we have 3b = c, or a + 2b = c, or 2a + b = c, all of which are impossible since a + b > c. By Lemma 5, the only remaining possibility for the type of the maximal segment of X is 2 : 2. If γ is not a right triangle, then there must be three triangles on each side of the maximal segment, making six triangles, which together with T_{2} and T_{3} is more than seven. Hence γ is a right angle. Since SV = c, angle SP V is a right angle and angle SV P = β.

Hence angle RQV = β and angle RV Q is a right angle. Since XV = b, the right angle of T_{4} is at X. Therefore the south side of T_{4} is c, and extends west of R on RV. We now have a third non-strict vertex at R. The exterior angle at R is more than π/2, so we will need at least two more triangles to sharing vertex R to be placed south of RV. But we must also place at least two more triangles with sides on line SV, since the maximal segment of X has type 2 : 2. That makes eight triangles altogether, which is more than seven. That disposes of Case 1.

Case 2, angles SV P and RQV are not equal. Since triangles SP V and RQV are congruent and P V = V Q = a, we must have angles SV P and RV Q equal; and they must both be equal to β, since the angle of T_{4} at V is at least α, and 2γ + α > α + β + γ = π. Then the angle of T_{4} at V cannot be γ, since 2β + γ > π. If it is α then we have 2β + α = π, which would make β = γ so angles SV P and RQV would be equal and Case 1 would apply. Therefore it is β and we have 3β = π. Then side XV of triangle T_{4} must be a, since it cannot be c, because it is less than SV. Consider the type of the maximal segment of X. Assume, for proof by contradiction, that it is 2 : 1. Then 2a = c and X is the midpoint of SV. Since β = π/3, the equation 2a = c implies γ is a right angle and α = π/6. Then the side of T_{4} on line RV must be the c side, since if the vertex of T_{4} on RV lies between R and V, the distance from X to that vertex is less than c. Hence T_{4} has R for its southwest vertex. Now we have a fourtriangle configuration with parallel north and south boundaries SP and RQ, a concave exterior vertex at X, and a right angle at the east vertex E. If three additional triangles are added sharing vertex X, the resulting configuration of seven triangles will not be a triangle, since it has three vertices P, Q, and E and more to the west of P Q. If two triangles are added sharing vertex X, they must not be placed so as to create new nonstrict vertices, as that would require placing a seventh triangle west of P Q, leaving at least four vertices. Therefore if two more triangles are added sharing vertex X, they share a west vertex W and are triangles SXW and RXW. The resulting six-triangle configuration is convex and has six vertices. Placing one more triangle can decrease the number of vertices of a convex configuration by at most one, so this configuration cannot be completed to a 7-tiling. Hence the concave exterior vertex at X must be filled by just one triangle. If this triangle T_{5} is not SXR then its west vertex W lies on SV extended, and it has side b along SV. Its south vertex U lies on RX. New concave exterior angles are created at S and U, each of which is more than π/2, and hence each will require placing at least two more triangles with vertices S and U respectively. But that will require a total of 9 triangles. Hence triangle T_{5} must be triangle SXR. We now have a five-triangle configuration including rectangle SP QR and triangle P QE. Either QE = b or P E = b. Suppose, for proof by contradiction, that QE = b. Then it cannot be that both RS and QE lie on sides of the final triangle, since the area to be filled south of RQ would require more than two triangles. With only two more triangles available, we cannot create more non-strict vertices. Consider placing a triangle T_{6} south of QE. Then it must share vertices Q and E. If we do not place the right angle at E then there will be a concave exterior vertex at Q that will require two more triangles to fill, contradiction. Hence T_{6} must have its right angle at E. Now we have a six-triangle convex configuration with five vertices. This cannot be completed to a 7-tiling since adding one triangle to a convex configuration can reduce the number of vertices by at most one. Hence QE is one of the sides of the final triangle. Placing a triangle T_{6} north of P E without creating a new non-strict vertex would require that T_{6} have side a along P E; that would create a concave exterior vertex at P greater than π/2, which could not be filled with one more triangle. Hence P E is also one of the sides of the final triangle. Then there must be a triangle T_{6} north of SP whose c side lies on P E extended. But now two triangles have sides on line RS, and we have already seen that not both RS and QE can be sides of the final triangle. So at least two more triangles will be required west of RS, but we have only one more available. This contradiction shows that QE ≠ b.

Therefore P E = b. Then it cannot be that both RS and P E lie on sides of the final triangle, since the area to be filled north of SP would require more than two triangles. With only two more triangles available, we cannot create more non-strict vertices. Consider placing a triangle T_{6} north of P E. Then it must share vertices P and E. If we do not place the right angle at E then there will be a concave exterior vertex at P that will require two more triangles to fill, contradiction. Hence T_{6} must have its right angle at E. Now we have a six-triangle convex configuration with five vertices. This cannot be completed to a 7-tiling since adding one triangle to a convex configuration can reduce the number of vertices by at most one. Hence P E is one of the sides of the final triangle. Placing a triangle T_{6} south of QE without creating a new non-strict vertex would require that T_{6} have side a along QE; that would create a concave exterior vertex at Q greater than π/2, which could not be filled with one more triangle. Hence QE is also one of the sides of the final triangle. Then there must be a triangle T_{6} south of RQ whose c side lies on QE extended. But now two triangles have sides on line RS, and we have already seen that not both RS and P E can be sides of the final triangle. So at least two more triangles will be required west of RS, but we have only one more available. This contradiction disposes of Case 2. That in turn completes the proof by contradiction that triangle T_{4} is triangle SV R.

Now that we know T_{4} is triangle SV R, we again have cases to consider. Case 1, angles SV P and RQV are equal. Since these are corresponding angles made by the transversal P Q to SV and RQ, lines SV and RQ are parallel. Then the alternate interior angles SV R and V RQ are equal. Angle V RQ = α, since it is opposite V Q = a. Therefore angle SV R = α. Since angles RSV and V QR are opposite side RV in their respective triangles, they are equal. Since V Q and SR are opposite angle α, they are equal. Hence sides SV and RQ are also equal. That makes SV QR a parallelogram. This is the configuration described in part (iii) of the lemma, so we are finished with Case 1.

Case 2, angles SP V and RQV are equal. Then since angle P SV = α, angles SP V and RQV are equal either to β or to γ. Then angles SV P and RV Q are also equal (either to γ or to β). They cannot be equal to γ, because in that case it would not be possible to place even one triangle T_{4} between T_{1} and T_{2}, since α + 2γ ≥ α + β + γ = π. Hence SV P and RV Q are equal to β. Then 2β plus angle SV R equals π. Since angles SP V and RQV are both equal to γ and SP and RQ are both equal to b, SR is parallel to P Q, and triangle SV R is isosceles, with both sides SV and RV equal to c. Hence angles SRV and RSV are both equal to γ. Angles SRV and RV Q are alternate interior angles of the transversal RV of parallel lines SR and P Q, so they are equal; but angle RV Q = β, so SRV = β also. Then angle SRV is opposite side c and hence equals γ; hence β = γ. Hence the triangles T_{i} are isosceles with 2a = b = c, and triangle SV R is congruent to the T_{i}, and hence is a fourth triangle T_{4} belonging to the tiling. In this case conclusion (iii) of the theorem holds.

That completes the proof of the lemma in case there is only one triangle between T_{1} and T_{2}.

We still have to consider the case in which there are two or more triangles, T_{4} (with a side on SV) and T_{5} (with a side on RV), and possibly still more triangles, between T_{1} and T_{2}. Changing south and north if necessary, we can assume angle P QE = α. Angle QP E is either β or γ. If γ is not a right angle, or if γ is a right angle but SP E is not a straight angle, then this 5-triangle configuration has vertices (at least) at S, P, E, Q, and R, and at least one more vertex W of T_{4}. If W is a concave vertex it will have to be removed by placing T_{6} with a vertex at W. That will leave at least five other vertices, all convex; that cannot be reduced to three by placing one more triangle; hence W is not a concave vertex. Unless W occurs on the line SR, we then have six convex vertices, which cannot be reduced to three by placing two more triangles. The only way W can occur on SR is if there are two triangles between T_{1} and T_{2}, and they share vertex W and have a right angle there. In that case γ is a right angle. Hence γ must be a right angle, and one of two cases holds: either 2a = c and angle QP E = β, so α = π/6 and β = π/3, and there are two triangles sharing a vertex at the midpoint of SR and another vertex at V, or QP E = γ, and SP E is a straight angle. Then 2a = b, rather than 2a = c, and α = arctan \frac{1}{2} . Now consider the angles at V. The angle of T_{1} and V is β. If T_{2} has angle γ there, that leaves room for only one triangle between them, with angle α; hence T_{2} has angle β at V. We have β = π/2 − arctan(1/2), which is about 63.43 degrees. Hence any triangles between T_{1} and T_{2} have angle α at V, since β is too big to fit. The angle to be filled is π − 2β = 2α, so more than two triangles cannot fit, but two fit nicely. If these two triangles are placed “naturally” then both of their west sides will lie on SR. These are the two configurations described in part (ii) of the lemma. The last two triangles must be placed in this configuration, because any other placement would place a side of length b along a side of length c in at least two places, each of which would require placing two more triangles to make these vertices have type 2 : 2, contradiction.

That completes the proof of the lemma.

Lemma 7 Suppose that a 7-tiling contains a non-strict vertex. Then the type of that vertex is 2 : 1, the tile is a right triangle whose smallest angle α is π/6 or arcsin \frac{1}{2} , and 3 triangles meet at the non-strict vertex. (See Figure 8.)

Proof. After the previous lemma, it only remains to rule out cases (ii) and (iii) of that lemma’s conclusion. We first rule out case (iii). We may suppose the non-strict vertex is V, the midpoint of the maximal segment P Q, with P at the north, Q at the south; that triangles T_{1}, T_{2}, and T_{4} are on the west of P Q, sharing vertex V, and that P is a vertex of T_{1} and Q is a vertex of T_{2}, and S is the shared west vertex of T_{1} and T_{2} and R is the shared west vertex of T_{4} and T_{2}. Triangle T_{3} is east of P Q, and its east vertex is called E. P V = V Q = SR = a. We have RQ parallel to SV and SP parallel to RV. By changing “north” and “south” if necessary, we can assume that angle RQV = β and angle SP V = γ. There are four cases to consider. Namely:

Case 1: 2a = b, angle E = β, angle V QE = α

Case 2: 2a = b, angle E = β, and angle V QE = γ

Case 3: 2a = c, angle E = γ, angle V QE = α, and b ≠ c

Case 4: 2a = c, angle E = γ, and angle V QE = β, and b ≠ c We call this four-triangle configuration M. All we have to do is prove

that it is not possible to add three more tiles to M and thereby create a triangle. This could be done by computer, but it is within reach to do it by hand.

M has five exterior vertices, all of which have angles less than π (because they are composed of two angles of the tile triangle, not both γ) except possibly P in case 1, where two angles γ share vertex P. In cases where M is convex and five-sided, placing three new triangles must leave at least two of the original five edges as part of the boundary of the final triangle. Hence two of the sides of the final triangle contain sides of M. There are 10 pairs of sides of M to consider; in each case we can ask whether it is possible to draw a third side and fill in the remaining area with copies of the tile.

Case 1 divides into Case 1a (when γ = π/2), Case 1b (when γ > π/2), and Case 1c (when γ < π/2). Before subdividing into these cases, we first argue that SR cannot be a side of the final triangle. Assume, for contradiction, that both EQ and SR are sides of the final triangle. Extend EQ and SR to their intersection point L. The final triangle must include triangle SLE, hence must contain triangle RQL. But the area of triangle RQL is five tiles, not three, which we see as follows: Angle RQL = γ, since the other two angles at Q are α and β. Angle SRV = γ (because it is opposite SV which is oppositive angle SP V), and angle V RQ = α, since it is opposite V Q = a. S Angle L = α. Since the angles at R must add up to π, angle QRL = β. Then angle QLR = α, in order that the angles of triangle QLR add to π. So triangle QLR is similar to the tile, and since RQ = c, the similarity factor is c/a. The area of RQL is therefore c^{2}/a^{2} times the area of a single tile. To complete a 7-tiling this way would thus require 3a^{2} = c^{2}. But c^{2} = a^{2} + b^{2} = a^{2} + (2a)^{2} = 5a^{2}, not 3a^{2}. (Nevertheless, it does not seem that we can complete this configuration to a 9-tiling, but that is irrelevant.) This contradiction shows that not both EQ and SR are sides of the final triangle. If SR is a side, then another triangle T_{5} must be placed east of EQ. If T_{5} does not share vertices E and Q, then at least one concave exterior vertex is created, which will require placing T_{6} south of EQ or on QE extended north of E. We will then have vertices at S, R, at P unless γ = π/2, and at one of Q or E, at most one of which can be removed by placing T_{7}, and at least two more on line QE, which is too many. Hence T_{5} must share vertices E and Q. The resulting five-triangle configuration is convex. Since in Case 1, angle E = β, the vertex at E remains a vertex. Since EQ = c, angle γ does not occur in T_{5} at Q, so the vertex at Q remains a vertex too. Then the five-triangle convex configuration has six vertices. T_{6} must share the existing vertices, as if we create a new non-strict vertex we do not have enough triangles to fill the exterior angles thus created and tile a triangle. But if T_{6} shares existing vertices, the resulting six-triangle configuration will be convex, and will have at least five vertices, so cannot be completed to a triangle by adding one more triangle T_{7}. This contradiction proves that SR is not a side of the final triangle.

Now we subdivide Case 1. First consider case 1a, when γ = π/2, and hence SP E is a straight line. This is the only case in which M is a quadrilateral, rather than having five sides. As proved above, we will have to add one triangle T_{5} west of SR, with westernmost vertex W, with one of its sides containing segment SR (which has length a). If we place this triangle so that W lies on SE extended and on RQ extended, then we will create a 5-tiling. Placing T_{5} with vertices at S and R but with angle γ at R instead of β will create a concave vertex at R with an exterior angle greater than π/2, so two tiles would be required at vertex R, leaving vertices at Q, E, S, and W at least. Placing triangle T_{5} with one side along line SR but extending north of S would require placing at least two more triangles north of SE, leaving vertices R, Q, E, and W. Placing triangle T_{5} with one side along SR but with northernmost vertex on SR south of S would require us to place T_{6} on SR north of T_{5}, and since SR = a, T_{6} would extend south of R, requiring T_{7} to share vertex R. Then seven triangles would be used and more than three vertices would still exist, e.g. Q, E, W, and the south vertex of T_{6}. Hence the triangle west of SR must be placed with W on SE extended, forming a 5-tiling. We are then asked to add two more triangles to produce a 7-tiling.

Side W E has length 2b + a = 5a > 2c = 2\sqrt{5}a and hence cannot be entirely covered by placing two triangles north of W E. Hence no triangle can be placed north of W E, which is thus a side of the final triangle. Side W Q has length 2c and hence if one triangle is placed south of W Q, a second one must be placed there as well; if these are placed so as to cover all of W Q then the result is not a triangle. Hence no triangles can be placed south of W Q, which must be a second side of the final triangle. We are thus asked to place two triangles east of EQ and complete a 7tiling. If these two triangles can be placed so that Q remains a vertex of the final triangle then the vertices of the final triangle will be W, Q, and a third vertex U east of E on W E extended. Triangle QEU must be composed of two copies of the tile T. These two triangles share vertices E and another vertex X on QU. The angle at X is a right angle by Lemma 1. Angle P EQ = β. Angle EQX = angle EUX since both are opposite side EX. Hence angle QEX = angle UEX, and these angles are not γ since the right angles occur at X. If they are β then, adding the three angles at E, we have 3β = π; but since α = arctan(1/2), we do not have β = π/3. If angles QEX and UEX are both α then we have 2α + β = π, but that is impossible since α + β + γ = π and γ > α. Hence it is not the case that Q is a vertex of the final triangle.

Therefore we will have to extend side RQ past Q by adding another triangle south of QE. Extending side RQ past Q will require creating a right angle at Q; if that is done with one triangle, then the side it shares with QE will have length b, creating a non-strict vertex at distance b along QE, which has length c. That will create a concave vertex with exterior angle more than π/2, which cannot be filled with our one remaining triangle. Hence both remaining triangles will have to share vertex Q, using angles α and β. Hence the shared side of those two triangles cannot have length a or b in both triangles, and it cannot have length c either since the first one has its side of length c along QE. Hence these two new triangles do not even share a vertex along their shared side, and cannot form a 7-tiling. That disposes of case 1a.

Now we take up case 1b, when γ > π/2. Then the boundary of M is concave at P, so in addition to adding a triangle T_{5} west of SR, with westernmost vertex W, we must add T_{6} north of P, with a vertex at P. Suppose, for proof by contradiction, that γ ≠ 2π/3, in which case T_{6} does not fill the vertex at P, or that γ = 2π/3 but T_{6} is not placed with angle γ at P. Then we must also add T_{7} with a vertex at P. That would mean that no triangles can be added south of P E or west of W, so that W and Q must both be vertices of the final triangle, and R must not be a vertex, and hence lies on W Q. Moreover, nothing can be added touching QE, so QE must lie on one side of the final triangle. Therefore the third vertex is either E or lies northeast of E on QE extended. It follows that triangle T_{7} must have side a along P E, since it has a vertex at P and cannot extend beyond E along line P E, and P E = a is the shortest side of the tile. E cannot be a vertex of the final triangle, since W and Q are vertices and triangle T_{7} does not lie inside triangle W QE, since it extends north of E along line QE. Therefore only two tiles meet at E. Hence by Lemma 1, γ = π/2. But in Case 1b, γ > π/2, so this is a contradiction. This contradiction proves that γ = 2π/3 and T_{6} is placed with angle γ at P.

Triangle T_{5} has to be placed west of SR, since we proved above that SR cannot be a side of the final triangle. Assume, for proof by contradiction, that it is placed with its a side along SR. Consider the three interior angles at R. They are angle QRV = α, angle V RS = γ, and angle W RS which might be γ or β, but not α since angle W = α, because it is opposite SR. If angle W RS = γ then a concave exterior angle exists at R that would have to be filled by one more triangle T_{7}, leaving four vertices W, S, E, and Q still present. S would be a vertex since the angles there would be α from T_{6}, α from T_{1}, β from T_{4}, and β from T_{5}, and their sum is 2α + 2β = 2π − 2γ = 2π/3 ≠ π. Four vertices remaining means a triangle is not created. Hence angle W RS ≠ γ. Hence angle W RS = β. Then W RQ is a straight line. Consider the angles at S. They are α from T_{6}, α from T_{1}, β from T_{4}, and this time γ from T_{5}. Their sum is 2α + β + γ = α + π > π. Hence a concave exterior vertex exists at S after the placement of T_{5} and T_{6}. The exterior angle is β + γ, too large to be filled by the placement of one more triangle T_{7}. This contradiction shows that T_{5} cannot be placed with its a side along SR.

Hence triangle T_{5} must be placed west of SR in such a way that not both S and R are vertices. Suppose, for proof by contradiction, that T_{5} is placed so that it does not have a vertex at R. It must have two vertices on line SR, of which say U is the southernmost. We then have a concave exterior vertex either at R (if U is south of R) or at U (if U is north of R). T_{7} will have to be placed to fill this exterior concavity. Since Q and E will remain vertices after the placement of T_{7}, the third vertex must be W; hence S lies on line W E. That implies that T_{7} has S for a vertex and the sum of the angles at S must be π. Those angles are α from T_{6}, α from T_{1}, β from T_{4}, and an unknown angle from T_{5}. The unknown angle must be π − 2α − β = γ − α. This cannot be γ, nor can it be α, since 2α = γ = 2π/3 implies α = π/3 which in turn implies β = 0, a contradiction. Hence the angle of T_{5} at S is β = γ − α. Hence α + β = γ; but also α + β = π − γ, which contradicts γ = 2π/3. (So we do not even have to analyze the impossible situation near R.) This contradiction shows that R must be a vertex of T_{5} and S is not a vertex. Since SR = a is the shortest side of the tile, triangle T_{5} extends north of S along SR. Then T_{7} must be placed with a vertex at S north of SE. There will then be vertices E, Q, W, and a vertex north of S on line SE. That is four vertices at least, so no triangle is created. That disposes of Case 1b.

Now we take up case 1c, when γ < π/2. Triangle T_{5} must be placed west of SR. Suppose, for proof by contradiction, that it is placed with angle β at R and angle γ at S. Then the first five triangles form a quadrilateral with vertices at W, Q, E, and P, and straight angles at R and S. Consider the possibility that W SP is a side of the final triangle. Then triangle T_{6} must be placed with a side along P E and a vertex at P. But the angle at P will then be at least 2γ + α, which is more than π. Hence W SP cannot be a side of the final triangle. But the length of W P is 2b, and we have 2b > c since 2b = 4a > 3a = a + b > c. Hence at least two triangles will have to be placed north of W SP. But that will leave vertices W, Q, E, and another vertex north of W SP –more than three. This contradiction shows that T_{5} cannot be placed west of SR with angle β at R and angle γ at S. If it is instead placed with angle γ at R and angle β at S, there will be a concave exterior angle at R, and convex vertices at W, S, P, E, and Q. Even if it is possible to fill the exterior angle at R with T_{6}, that still leaves five vertices in a convex configuration, or more than five vertices with some concave exterior angles. In either case, a triangle cannot be created by adding T_{7}. Hence T_{5} cannot be placed in either orientation with its a side along SR. Therefore T_{5} must be placed with a side extending past SR, either north or S or south of R along line SR (or both). Suppose, for proof by contradiction, that T_{5} extends north of S to a vertex N on RS extended, but has R for a vertex. Then T_{6} must be placed north of P S to fill the concave exterior angle at S. That leaves convex vertices at W, N, E, and Q at least. There will also be a vertex at R unless T_{5} has angle β there; that must be the case since five convex vertices is too many to create a triangle by placing T_{7}. Then T_{5} has angle α at N and γ at W. Hence NS has length c − a. Triangle T_{6} must also eliminate vertex P, or there will again be five vertices, which is too many. To do that, T_{6} must fill the entire angle NSP = γ, and must supply an angle equal to π − 2γ at P. Since SP = b, T_{6} has angle β at N, so it must have α at P. Therefore α = π − 2γ. Adding β + γ to both sides of this equation we have π = π + β − γ, which implies β = γ. We now have a 6-tiling of quadrilateral NEQW. But this quadrilateral is a parallelogram: NP E is parallel to W RQ because transversal P Q makes equal alternate interior angles RQP = β and QP E = γ, and QE is parallel to W N because transversal NE makes complementary corresponding interior angles W NP = α + β and QEP = β = γ. By placing one more triangle T_{7}, one cannot turn a parallelogram into a triangle. This contradiction shows that T_{5} cannot be placed with R for a vertex.

Now suppose, for proof by contradiction, that T_{5} is placed with one vertex at S and a second vertex U south of R on SR extended. Then T_{6} must be placed south of RQ to fill the concave exterior angle at R. That leaves convex vertices at W, U, E, and P at least. There will also be a vertex at S unless T_{5} has angle γ there; that must be the case since five convex vertices is too many to create a triangle by placing T_{7}. Then T_{5} has angle α at U and β at W. Hence UR has length b−a = a, so the angle of T_{6} opposite UR must be α. Triangle T_{6} must also eliminate vertex Q, or there will again be five vertices, which is too many. To do that, T_{6} must have a vertex at Q, and must fill the entire angle URQ = β, and must supply an angle equal to π − α − γβ at R. Since RQ = c, T_{6} has angle γ at U, so it must have α at Q. Therefore to eliminate vertex Q we must have 2α + β = π. This implies α = γ, so the tile is an equilateral triangle. Now we have a 6-tile convex configuration with vertices at W, U, E, P, and S. Placing T_{7} can decrease the number of vertices by at most one, since the configuration is convex. Hence no final triangle can be created. This contradiction shows that triangle T_{5} cannot be placed with a vertex at R or at S.

Hence triangle T_{5} is placed west of SR, with two vertices on line SR somewhere, but not at R or at S. Then at least two concave exterior vertices will be created somewhere on line SR, which must be filled by placing triangles T_{6} and T_{7} with one vertex each on line SR and a side contained in line SR. That will create two new vertices on line SR, say N to the north and U to the south. We then have vertices W, E, N, and U, even if straight angles at P and Q are created by placing T_{6} and T_{7}. Four vertices do not make a triangle, so this placement of T_{5} is also contradictory. That disposes of case 1c, and with it, of case 1.

Now we take up Case 2. In that case SR is parallel to P Q since the alternate interior angles SV P and RSV are both equal to β. We ask which pairs of sides of the pentagon SP EQR could be (contained in) sides of the final triangle.

Case 2a: SR and P E are sides. Let N be the intersection point of SR extended and EP extended. Then triangle NSP has angle β at S and γ at P, because the angles at S and P must add to π. Then angle N is α, and triangle NSP is similar to the tile T. But it has side b = 2a opposite angle α, so its area is four times that of the tile. It therefore requires four triangles congruent to T_{1} to cover triangle NSP, which is eight total, more than seven.

Case 2b: SR and QE are sides of the final triangle. Then we have to add triangle T_{5} southwest of RQ, and the third side will have to be east of P E and require at most two triangles to complete the figure, one of which, say T_{6}, will have to be north of SP, say SP N and one, say T_{7}, will have to be east of P E, say P EX. But it will not be possible for T_{7}, adding just one angle at E, to have a side extending QE, since then by Lemma 1, the tile would contain a right angle, so γ = π/2, and triangle P EQ would have two right angles, one at Q and one at E. This contradiction disposes of case 2b.

Case 2c: SR and SP are sides. Since P E and QE are not sides (by Case 2a and Case 2b), there must be two new triangles sharing vertex E, one on each side of the existing triangle P QE. These triangles must share (respectively) side P E and side QE, since otherwise additional triangles will have to be placed sharing vertex P or Q, and a triangle will not be created. Thus we have triangle T_{5} = P EF and triangle T_{6} = QEG, with F EG and RQG and SP F straight lines. Then angle P F E = γ (being opposite P E which has length c) and angle EP F = γ (so that SP F is straight, since angle V P S = β and angle QP E = α), so angle P EF = α and the tile is isosceles with β = γ, since triangle P EF has two angles γ. Angle QEG = γ since the sum of angles at E is π. Angle G = α since it is opposite QE = a. Hence angle EQG = β = γ. Now there are three angles γ at Q and since RQG is a straight angle, γ = π/3. Hence the isosceles tile is actually equilateral. But that contradicts the equation b = 2a. That disposes of Case 2c.

Now suppose, for proof by contradiction, that SR is (contained in) a side of the final triangle. Then since SP is not an edge, we must place a triangle, say T_{5}, north of SP, and since P E is not an edge, we must place T_{6} east of P E, and since QE is not an edge, we must place T_{7} south of QE. Then we count vertices: We have the north vertex N of T_{5} north of SP, and the south vertex of T_{7}, south of QE, and R. In addition, S will be a vertex unless N lies on RS extended; and N must also be the third vertex triangle T_{6}, which has P E for one side. Consider the angles at P. Angle QP E = α because angle P QE = γ and angle E is opposite P Q = b = 2a. Angle SP Q = γ. Angle SP N must be γ, since if it is α or β, the remaining angle NP E will be π or more, but it is a single angle of triangle T_{6}. Since there are four angles at P and one of them is α, we have γ > π/2. Since in triangle NSP, γ is used at P, angle NSP is either α or β. Then the angles at S are β, α, and angle NSP; the total is at most α + 2β which is less than π, since β < π/2 < γ. Hence N does not lie on RS extended, after all. This contradiction shows that SR is not contained in a side of the final triangle.

We therefore must place T_{5} west of SR. Let W be its western vertex. Unless T_{5} is placed so that at least one of vertices S and R are eliminated, i.e. W SP is a straight angle or W RQ is a straight angle, then there will be six vertices, too many to allow the creation of a triangle by placing two more copies of T_{1}. Suppose, for proof by contradiction, that W SP is a straight angle. Angle V SP = α and angle V SR = β. Hence angle W SR = γ. Since SR = a because it is opposite to angle SV R = α, angle SW R = α. Hence angle W RS = β and side W S = b. We will show that no triangle can be placed north of W SP. If T_{6} is placed north of W SP with north vertex N then there will be vertices Q, E, and N. Assume, for proof by contradiction, that T_{6} has W P for a side. The length of side W SP is 2b, so we must have c = 2b. But b = 2a, so c = 4a. This is impossible, since then 4a = c < a + b = 3a. This contradiction shows that T_{6} cannot have W P for a side. Hence two triangles, T_{6} and T_{7}, must be placed north of W SP. If any side of length a is placed along W SP then we need at least three triangles (which is too many). Suppose triangles T_{6} and T_{7} are placed north of W SP with their b sides along W S and W P. Then there will be vertices Q and E, and in order to create a final triangle, there must be straight angles at P and at W, and triangles T_{6} and T_{7} must share a side. By Lemma 1, triangles T_{6} and T_{7} have right angles at the shared vertex on W SP. Hence γ = π/2. Then T_{6} and T_{7} have acute angles at P and W. Hence there is a vertex at W, as well as at Q, E, and the north vertex of T_{6}, so no triangle is created. This contradiction shows that W SP is not a straight angle after the placement of T_{5}.

It follows that W RQ is a straight angle after the placement of T_{5}. Since angle RQV = β and angle QRV = α, we have angle QV R = γ and hence RQ = c. We have angle V RS = γ and since angle W RQ = π we have angle W RS = β. If angle W = α then T_{5} has SR = a for a side and W SP is a striaght angle, which have already disproved. Hence angle W = γ and triangle T_{5} has side c along RS extended, creating a concave exterior vertex at S, which must be filled by T_{6}. Let the north vertex of T_{5} be N. Then NS = a, so it is possible to place triangle T_{6} = NSP. If this is done, we have a six-triangle convex configuration with five vertices W QEP N. (There is a vertex at P since the angles at P are 2α + γ < π; there is a vertex at N since the angles there are α + β < π.) Such a configuration cannot be completed to a triangle by placing T_{7}. Hence triangle T_{6} is not NSP. But any other way of placing T_{6} with a vertex at S will create another convex exterior vertex north of or on W SP and east or or on SR extended, so T_{7} will have to be placed north of W SP and east of SR extended. That will leave vertices at Q, E, and W, as well as at least one vertex north of SP, so a triangle will not be created. This contradiction finally disposes of Case 2.

We now take up Case 3. Since the angles at P are γ and β, the possibility that SP E is straight does not arise; M is a convex pentagon. Adding three triangles will still leave two of the five sides on the boundary; hence at least two sides of M are (contained in) sides of the final triangle. We will consider each of the ten pairs of two sides and show that those two sides cannot be sides of the final triangle.

Case 3a: SR and QE are sides of the final triangle. Then let X be their intersection point. The transversal RQ makes alternate interior angles RQX = γ and QRS = γ + α with QE and SR, so X lies to the south of RF. The triangle RF X can be covered exactly by four copies of T_{1} (since it is similar to the tile but has side c = 2a opposite angle α), but it must be contained in the final triangle, which is contradictory, since only three more triangles are available. That disposes of Case 3a.

Case 3b: SR and P E are sides of the final triangle. Then let X be their intersection point; triangle T_{5} will be required to cover SXP. Then XP E must be a side of the final triangle, or else we will need to place T_{6} and T_{7} north of XP E, leaving vertices R, Q, and at least two north of or on XP E, contradiction. Therefore XP E is a side. Since EQ is not a side, we must place triangle T_{6} with a side on line QE. Suppose, for proof by contradiction, that T_{6} does not have E for a vertex. Then a convex exterior vertex is created on line QE, which must be filled by T_{7}. If T_{6} extends north of E along QE then T_{7} must be placed with a side on XE; but that would create another exterior vertex on XE, because XE has length greater than c, and so no triangle would be created. If, on the other hand, T_{6} has its north vertex on QE south of E, then T_{7} must be placed east of QE, leaving parallel sides RQ and XE, so no triangle is created. Hence T_{6} does have E for a vertex. Now suppose, for proof by contradiction, that T_{6} does not have Q for a vertex. If the south vertex of T_{6} is north of Q, then T_{7} must be placed east of QE, leaving parallel sides RQ and XE, so no triangle is created. If the southwest vertex U of T_{6} lies on QE south of Q, then since QE = b and T_{6} has E for a vertex, T_{6} must have side c along QE, and T_{7} will have to be placed south of RQ and west of QE, in order to fill the concave exterior angle at Q. That will leave vertices at X, R, U, and the southeast vertex of T_{6}, so no triangle is created. This contradiction proves that T_{6} has Q for a vertex, as well as E. In other words, T_{6} has QE for a side.

Let Y be the third vertex of T_{6}. Since QE = b, angle EY Q = β. If angle Y QE = γ, then EY is parallel to XR, and we have two pairs of parallel sides in the six-triangle configuration, a problem that cannot be fixed by placing T_{7}. Hence angle Y QE = α. Assume, for proof by contradiction, that γ is not a right angle. Then we have five vertices X, E, Y, Q, and R. If γ < π/2 then this is a convex pentagon, and cannot be completed to a triangle by placing T_{7}. If γ > π/2, then there is a concave exterior angle at E, which possibly could be filled by T_{7} if γ = 2π/3, but then another concave exterior angle would be created on XP E somewhere, since the length of XP E is more than c. This contradiction proves that γ is a right angle. Therefore P EY is a straight angle. Now, however, we have parallel sides XP EY and RQ; since XP EY has length more than c, we must place T_{7} south of RQ with its c side along RQ. Suppose, for proof by contradiction, that T_{7} has angle α at R. Then there is a vertex at R, since the sum of the angles at R is 2α + γ < π, since we know γ = π/2 and hence, in view of c = 2a, α = π/6. We then have vertices at X, R, Y, and the third vertex of T_{7}, so no triangle is created. This contradiction proves that T_{7} does not have angle α at R. Hence it has angle β at R, making a straight angle there. Then T_{7} has angle α at Q. The sum of the angles at Q is then β + 3α = π/2 + 2α < π. Hence there is a vertex at Q. Since there are also vertices at X, Y, and the third vertex of T_{7}, making at least four vertices, no triangle is formed. That disposes of Case 3b.

Case 3c: SR and RQ are both sides of the final triangle. Since the intersection of lines P S and QR lies west of SR, P S is not a side of the final triangle. That requires the placement of a triangle T_{5} north of P S. Since P E is parallel to RQ, we must place a triangle T_{6} north of P E. This triangle T_{6} cannot reach all the way to SR extended, requiring the placement of T_{7} north of T_{5}. This cannot leave a triangle, as we have vertices at Q, R, and a north vertex X on RS extended, and in addition a fourth vertex either at E, or if T_{6} created a straight angle at E, then at another vertex of T_{6} on QE extended. That vertex lies east of QP extended and hence cannot coincide with X. This disposes of Case 3c.

Case 3d: SR and SP are both sides of the final triangle. Then by Case 3b we would have to place T_{5} north of P E. It would have to have angle α at P in order not to extend north of P S. Then E would become a non-strict vertex, with T_{5} extending past E. Since by Case 3c, we must place one triangle T_{6} south of RQ, we must fill the concave exterior angle at E with a single triangle T_{7}. Hence by Lemma 1, γ = π/2. T_{7} and T_{5} must share a common east vertex Y, or else further concave exterior angles are formed. At Y the sum of angles is 2β = 2π/3 < π, so there are vertices at S, Y, and the northeast vertex Z of T_{5}. Since Q does not lie inside triangle SY Z, there is a fourth vertex and no triangle is formed. This disposes of Case 3d.

Case 3e: RQ and P E are both sides of the final triangle. This is impossible since they are parallel, because transversal P Q makes equal alternate interior angles β with RQ and P E. That disposes of Case 3e.

Case 3f: RQ and SP are both sides of the final triangle. The intersection point of lines RQ and SP, say W, lies to the west of SP and triangle SP W is congruent to the tile. Since by Cases 3a to 3d, SR is not a side of the final triangle, we must place T_{5} as triangle SP W. Since P E is parallel to RQ, we must place T_{6} north of P E and south of SP extended. This is only possible if T_{6} has P for a vertex and has angle α there, so the north side of T_{6} extends segment SP, i.e. there is a straight angle at P. That creates a concave exterior angle at E, since P E = a < b (we cannot have a = b since c = 2a). Let N be the vertex of T_{6} on SP extended and let X be the south vertex of T_{6}. We must therefore place T_{7} with a vertex at E and a side along QE and a side along P X. Unless T_{7} is triangle QEX, more exterior concave angles (and hence no triangle) will be formed; hence T_{7} is triangle QEX. Then considering the triangles meeting at E, by Lemma 1, γ = π/2. Triangle QEX has angle α at Q and β at X. The sum of the angles at Q is β + 2α < π, so there is a vertex at Q. The sum of the angles at X is 2β < π, so there is a vertex at X. There are also vertices at N and W, so no triangle is formed. That disposes of Case 3f.

Case 3g: RQ and EQ are both sides of the final triangle. By Cases 3a to 3d, SR is not a side of the final triangle, so we must place T_{5} west of SR but touching segment SR. Assume, for proof by contradiction, that R is not a vertex of T_{5}. Then the southern vertex of T_{5} on SR lies north of R and south of S. But SR = a, so the portion of SR south of T_{5} has length less than a and cannot be covered by a triangle lying north of RQ extended; that will leave a concave exterior angle that cannot be filled. This contradiction proves that R is a vertex of T_{5}. There are two possible orientations of T_{5}: either it has its a side along SR, so that W lies on SP extended, or it has its c side along SR, so that S is the midpoint of the east side of T_{5}. Assume, for proof by contradiction, that S is the midpoint of the east side of T_{5}. Let N be the north vertex of T_{5}, and W its west vertex. Then W N is parallel to QE, since the corresponding interior angles made by transversal W Q are angle NW Q = γ and angle EQR = α+β, which are supplementary. Since EQ is (in Case 3g) a side of the final triangle, NW cannot be, so triangle T_{6} must be placed northwest of NW. Then T_{7} has to have a vertex at S to fill the concave exterior angle there. But by Case 3e, P E is not a side, so some triangle has to be placed north of P E; but we have no more triangles, so a contradiction has been reached. That disposes of Case 3g.

Case 3h: SP and QE are both sides of the final triangle. We note that the intersection point X of those two sides lies to the northeast of P E, because of the alternate interior angles made by the transversal P Q. Hence triangle T_{6} will have to be placed with its a side along P E. But then its α angle is not at P and it cannot lie south of SP extended, contradiction. That disposes of Case 3h.

Case 3i: SP and P E are both sides of the final triangle. By Case 3e, RQ is not a side of the final triangle, and by Case 3h, QE is not a side. By Cases 3a to 3d, SR is not a side. Therefore we will have to place T_{5} west of SR, T_{6} south of RQ, and T_{7} southeast of QE. T_{7} will have to be placed with a vertex at E in order to avoid creating a concave exterior vertex on QE. Let W be the west vertex of T_{5}; this must be the west vertex of the final triangle. It must therefore lie on SP extended. Hence T_{5} has W S for its north side. If T_{5} has R for a vertex, then there is a concave exterior angle at the western vertex of T_{6} on RQ or RQ extended. Hence T_{5} has its southern vertex Y south of R on SR extended. T_{5} has angle γ at S, since the sum of angles there must be π, and angle β at W, since the side opposite W is greater than SR = a. Hence T_{5} has angle α at Y. But W R is parallel to P E, so the intersection point of lines W Y and P E will lie north of SP. Hence SP, P E, and W Y cannot be sides of a triangle including any points (such as V) south of SP. But in Case 3i, by hypothesis SP and P E are sides of the final triangle, and W Y must be a side since it is on the boundary of the seven-triangle configuration. This contradiction disposes of Case 3i.

Case 3j: P E and QE are both sides of the final triangle, then as we have already shown, none of SP, SR, and RQ can be sides. Hence we will have to place the remaining three triangles on those sides, say T_{5} on SR, T_{6} on RQ, and T_{7} on SP. T_{6} must be placed with its c side on RQ (or else a concave exterior angle will be created), so its angle at Q will not be γ, and the seven-triangle configuration has a vertex at Q, since the angle sum there is less than π. Assume, for proof by contradiction, that T_{5} and T_{7} do not have the same third vertex. Then we have four vertices–those two plus E and Q. That contradiction shows that T_{5} and T_{7} must share their third vertex, say N. Let X be the southern vertex of T_{6}. For a triangle to be formed, the sides must be NP E, XQE, and NRX. Now RS = a so in T_{5}, NS must be b or c. In T_{7}, SP = b so NS must be a or c. Therefore NS = c, and angles NRS and NP S are both γ. The angle sum at P is then 2γ + β = π. But since α + γ + β = π, this yields γ = α, so the tile is equilateral, contradicting 2a = c. That contradiction disposes of Case 3j.

We have now shown that no pair of sides of M can be sides of the final triangle. This completes Case 3.

Now we take up Case 4. As in Case 3, M is convex.

We first prove that the three sides of the final triangle cannot be SP, P E, and RQ. Suppose, for proof by contradiction, that those are the three sides. The intersection point W of lines SP and RQ lies to the west of SR, because the transversal SR makes corresponding interior angles RSP = β + α and SRQ = γ + α, and in Case 4 we have c > b, which implies γ > β. Triangle W SP has angle β at R (since the angle sum at R is π) and angle γ at S (since the angle sum at S is π); hence it has angle α at W, opposite side SR = a. Hence triangle W SP is congruent to the tile and we can call it T_{5}. Let X be the intersection point of lines P E and RQ, which exists because we have assumed the third side is RQ. Then triangle QEX is 2-tiled by triangles T_{6} and T_{7}. Hence the tile is a right triangle, γ = π/2. Since there is a straight angle at Q, we have β = π/3 and hence α = π/6. Consider the angle sum of triangle W P X: the angle at W is α, that at P is γ + α, so that at X is π − γ − 2α = π/6 = α. Hence triangle QEX is congruent to the tile; it cannot be 2-tiled. This contradiction proves that the three sides of the final triangle cannot be SP, P E, and RQ.

We shall now argue that none of the ten pairs of sides of M can be (contained in) sides of the final triangle.

Case 4a: SP and P E are both sides of the final triangle. Then QE does not lie on the third side, because of the alternate interior angles made by transversal P Q to SP and P E. Hence triangle T_{5} is required south of QE. T_{5} must be placed along QE with a vertex at E, since P E is a side of the final triangle. Let X be the east vertex of T_{5}. SR cannot lie on a side of the final triangle, since the intersection point of SR and P E lies north of SP, because the transversal SP of RS and P E makes corresponding interior angles RSP = β + α and SP E = γ + α, whose sum is π + α > π. Hence triangle T_{6} must be placed west of SR. Let W be the west vertex of T_{6}. Since RQ cannot be a side of the final triangle when SP and P E are sides (as shown above), T_{7} must be placed south of RQ. Now, what can be the vertices of the final triangle? P is one of them; the others must be X (the east vertex of T_{5}), and W (the west vertex of T_{6}). Then S is not a vertex; hence triangle T_{6} has angle γ at S. Since RQ is not a side, W lies to the west of the intersection point Y of lines SR and RQ. But Y S = b since angle Y RS = β. Hence W S = c. Then the γ angle of T_{6} must be opposite W S; but we already proved T_{6} has angle γ at S. Since in Case 4, we have b ≠ c, we have β < γ so T_{6} has only one angle equal to γ. This contradiction disposes of Case 4a.

Case 4b: SP and QE are sides of the final triangle. The intersection point X of these two sides lies to the east, because of the alternate interior angles made by the transversal P Q. Suppose, for proof by contradiction, that γ is a right angle. Since 2a = c we have α = π/6 and β = π/3. Considering the angle sum of triangle P QX, we have α + β at P and β at Q, so we have π − α − 2β = α at X. Hence triangle P XE is similar to the tile, but with α opposite its b side P E. Hence the similarity factor is b/a = sin(π/3)/ sin(π/6) = \sqrt{3}. Hence the area of triangle P XE is three times that of the tile, and after it is tiled there will be no more triangles, but there will still be vertices at X, S, R, and Q. This contradiction shows that γ is not a right angle.

Let W be the intersection point of lines RQ and SP. Suppose, for proof by contradiction, that RQ is the third side. Then triangle W SR is congruent to the tile, having angle β at R, angle γ at S, and angle α at R. Give triangle W SR the name T_{5}. Then triangle XP E is 2-tiled by T_{6} and T_{7}; hence the tile is a right triangle, which we have shown is not the case. This contradiction shows that RQ is not the third side. Since triangle W SR is congruent to the tile, but RQ is not the third side, at least two triangles will be required west of line SR, and at least one south of RQ. That will use seven triangles, and leave none to tile triangle P EX. That disposes of case 4b.

Case 4c: SP and RQ are sides of the final triangle. These lines intersect in the west vertex W. Triangle W SR is congruent to the tile, and we call it T_{5}. Then, by cases 4a and 4b, triangles T_{6}, and T_{7} will have to have sides P E and QE, respectively. (If these triangles do not have vertices matching already existing vertices, no triangle will be formed.) Let X be the third vertex of T_{7}, and N the third vertex of T_{6}. Then the final triangle is W XN. A straight angle is formed at P, and since angle SP V = γ and angle QP E = α, we must have angle EP N = β. But P E = b so the angle at N must also be β. But in Case 4, we have b ≠ c so β ≠ γ. Hence β = α, since T_{6} has two angles equal to β. Then angle P EN = γ. Now suppose, for proof by contradiction, that angle QEX = α = β. Then considering the angle sum at E we have 2γ + α = π. But

2γ + α = 2(π − α − β) + α
= 2π − 3α

which implies α = π/3. But if α and β are both π/3 then so is γ, contradiction, since b ≠ c. This contradiction proves that angle QEX is not β. Therefore it is γ, and the angle sum at E tells us 3γ = π, or γ = π/3. But again that implies α = β = γ, contradiction. That disposes of Case 4c.

Case 4d: SP and SR are sides of the final triangle. Then by Case 4a, triangle T_{5} will be required east of P E; by Case 4b, triangle T_{6} will be required south of QE; and by Case 4c, triangle T_{7} will be required south of RQ. The three vertices of the final triangle must be S, a point X east of P on SP extended, which must be the east vertex of T_{5}, and a point Y south of R on SR extended, which must be the south vertex of T_{7}. Since triangle SXY must include T_{6}, located south of QE, triangles T_{7}, T_{6}, and T_{5} must all share a vertex on line Y X, with an angle sum of π there. But then the side of T_{7} along RQ must be longer than RQ = c, contradiction. That disposes of Case 4d.

Case 4e: SR and P E are sides of the final triangle. Let N be the intersection of P E and SR, which lies to the north of SP. Then triangle NSP is similar to triangle T_{1}, since it has angle β at P and angle γ at S (and hence α at N), but the side opposite angle α is b, not a. So triangle NSP could be tiled by an integral number K of copies of T_{1} only if K = (b/a)^{2} is an integer. If K = 3 then a triangle is not formed, since we have vertices at N, E, Q, and R. If K = 2 then NSP is 2-tiled, so γ = π/2. Since c = 2a in Case 4, we have α = π/6 and β = π/3. Then b/a = sin β/ sin α = \sqrt{3}, so K = (b/a)^{2} = 3, not 2. Hence K ≠ 2. But these are all the possibilities for K. That disposes of Case 4e.

Case 4f: P E and QE are sides of the final triangle. Then we require T_{5} south of QR by Case 4c, T_{6} north of SP by Case 4a, and T_{7} west of SR by Case 4e. In order that a triangle be formed, the new triangles must share existing vertices along the sides mentioned, and we must have straight angles at Q and P. The vertices of the final triangle are E, the north vertex N of T_{6}, on P E extended, and the south vertex X of T_{5}, on EQ extended. Then the three new triangles must share a vertex W west of SR and the angle sum there must be π. But then the north side of T_{5} will be W Q, which is longer than c, since RQ = c. That contradiction disposes of Case 4f.

Case 4g: P E and RQ are sides of the final triangle. Then we must place T_{5} south of QE with its a side along QE. In case γ = π/2 this can be done, with the third vertex X at the intersection point of P E and RQ. Otherwise more triangles will be required south of QE. Since neither SP nor SR is a side when P E is, by Case 4a and Case 4d, triangles T_{6} and T_{7} will be required north of SP and west of SR, so there can be no second triangle south of QE. Hence γ = π/2. Since c = 2a, we then have α = π/6 and β = π/3. The three vertices of the final triangle are X, and the north vertex N of T_{6}, which lies on P E extended, and the west vertex W of T_{7}, which lies on QR extended and is also a vertex of T_{6}. If T_{6} does not have vertices W and P or T_{7} does not have vertices S and R, no final triangle is formed. We have SP = b; angle RSV = β; angle W RS = β because the angle sum at R is π; hence W S = b. Hence W P, the south side of T_{6}, is 2b. Hence c = 2b. But c = 2a. Hence a = b and α = β. This contradicts α = π/6 and β = π/3. This contradiction disposes of Case 4g.

Case 4h: SR and QE are sides of the final triangle. Because P Q is parallel to SR, the intersection point X of QE and SR lies to the south of R. At least one triangle T_{5} will have to be placed inside triangle RQX. Since SR is a side, P E and SP are not sides, by Case 4e and Case 4d, so triangles T_{6} and T_{7} are required north of SP and northeast of P E, areas which cannot intersect triangle RQX. Hence triangle T_{5} must be exactly triangle RQX. Angle RXQ = β because angle P QE = β and P Q is parallel to SRX. But angle RXQ is opposite side RQ = c, so angle RQX = γ. Hence β = γ. Hence b = c. But in Case 4 we have b ≠ c. This contradiction disposes of Case 4h.

Case 4i: Suppose QE and RQ are sides of the final triangle. Since QE is a side, none of P E, SP, and SR are sides, by Case 4f, Case 4b, and Case 4h, so triangles T_{5}, T_{6}, and T_{7} must be placed on those three sides, respectively. Let X be the east vertex of T_{5}, and let Y be the northwest vertex of T_{6}. Then these must be the vertices of the final triangle, so Y is also the west vertex of T_{7}, and T_{7} is triangle SRY, T_{6} is triangle P SY, and T_{5} is triangle P EX. Since only two triangles meet at E, by Lemma 1 γ = π/2. Then because c = 2a we have α = π/6 and β = π/3. But at S, there are four angles, two of which are α and β, so the angles of T_{6} and T_{7} at S must add to 3π/2, which is impossible. That disposes of Case 4i.

Case 4j: RQ and SR are sides of the final triangle. By Case 4c, Case 4g, and Case 4i, neither SP, P E, nor QE can be sides along with RQ. Therefore we must place triangles T_{5}, T_{6}, and T_{7} along those sides, respectively. Let X be the southeast vertex of T_{7}, and N the north vertex of T_{5}. Then X, N, and R must be the vertices of the final triangle. Hence X lies on RQ extended and N lies on RS extended. The vertex Y of T_{6} that does not lie on P E extended must lie on NX. Then Y must lie on SP extended, so that SP Y is the south side of T_{5}; and Y must lie on QE extended, so that QEY is the north side of T_{7}. That is, lines SP and QE meet at Y. Since only two triangles lie to the north of E and QEY is a straight line, by Lemma 1 we have γ = π/2. Then because c = 2a we have α = π/6 and β = π/3. Because the angle sum at P of the angles SP V, QP E, and EP Y is π, we must have angle EP Y = β. Because the sum of angles QEP and Y EP must be π, we have angle Y EP = γ. Hence angle EY P = α. But side P E = b since it is opposite angle P QE. Hence angle EY P = α = β, which is a contradiction since α = π/6 and β = π/3. That disposes of Case 4j.

That completes all ten sub-cases of Case 4, and with them, the proof that case (iii) of the previous lemma’s conclusion is impossible.

We now take up showing that case (ii) of the previous lemma’s conclusion is impossible. In that case we start with a five-triangle configuration M, and we must show it is not possible to make it into a triangle by adding two more copies of T. Since SP is parallel to RQ, those two sides cannot both be sides of the final triangle. Suppose, for proof by contradiction, that triangle T_{6} is placed south of RQ. Let X be the south vertex of T_{6}. Then we have vertices at S, E, and X. Suppose, for proof by contradiction, that RQ is not a side of T_{6}. Then we will have to place T_{7} with a vertex on RQ. Then we still have vertices at X, S, and E, so the final triangle must be SXE. Then T_{6} and T_{7} must 2-tile triangle RQX. Then angle QRX is less than a right angle, so angle W RX is not a straight angle, since angle W RQ is a right angle. This contradiction shows that RQ is a side of T_{6}. Since RQ = b, T_{6} has a right angle at R and α at Q, or the other way around. Unless the right angle of T_{6} is at R and SP E is straight, the resulting six-triangle configuration will have five or more convex vertices, and cannot become a triangle by placing one more copy of T. Therefore T_{6} must be placed with its right angle at R. Let X be its south vertex. Then SX is longer than c since SR = 2a = c. Hence we cannot create a triangle by placing T_{7} west of SR. Also SP E, even if it is a straight line, has length b + a > c, so we cannot create a triangle by placing T_{7} north of SP E. The only remaining possibilities are south of T_{6} or east of QE. In either case the triangle T_{7} will share vertex Q. The angle already at Q is π/2 + 2α, so to make a triangle we must add π/2 − 2α. If SP E is not straight, we cannot possibly create a triangle; hence SP E is straight. Then α = arctan \frac{1}{2} , not π/6, so π/2−2α is neither α nor β, and it is not possible to eliminate the vertex at Q by placing T_{7}. This contradiction shows that T_{6} cannot be placed south of RQ and completed to a triangle.

Therefore RQ will be one of the sides of the final triangle. Then SP cannot be a side of the final triangle, since it is parallel to RQ. Hence we must place T_{6} north of SP. We must not create a concave vertex anywhere on SP by placing T_{6}, so the vertices of T_{6} must include S and P, unless SP E is straight and SE is one side of T_{6}; but SE = b + a > c, so that is not possible. Hence SP is a side of T_{6}. If SP E is straight, i.e. α = arctan \frac{1}{2} , then we have created a concave vertex at P, so we must have α = π/6. In that case the total angle at P after placing the α angle of T_{6} there is π/2 + α + β = π, so if N is the north vertex of T_{6} we have NP E straight. We cannot make a triangle by placing T_{7} west of NR, or north of NP E, since these segments are longer than c, so that leaves east of QE as the only possibility, since we know RQ must be a side of the final triangle. Since QE = b, we must place the b side of T_{7} along QE, with the right angle at E, because the sides of the final triangle must be NSR, NP E (extended), and RQ (extended). The angle of T_{7} at Q must then be α. The total angle at Q is then π/2 + 2α = 5π/6, not enough to eliminate the vertex at Q. That completes the proof that case (ii) of the previous lemma’s conclusion is impossible, and that completes the proof of the lemma.

Lemma 8 A 7-tiling cannot contain more than one non-strict vertex.

Proof. The previous lemmas have shown that each non-strict vertex is of type 2 : 1 and occurs in a certain 3-triangle configuration (shown in Figure 8.) Suppose a 7-tiling contains two (or more) non-strict vertices. We have to consider the following cases: Case 1, the two 3-triangle configurations overlap (share a triangle), thus requiring five or fewer triangles; Case 2, the two 3-triangle configurations do not share a triangle, but share a side; Case 3, the two 3-triangle configurations do not share a triangle or a side.

W  P Q V  Y X E

Figure 10: Two non-strict vertices in five triangles, first configuration

We take up Case 1. Each of the two 3-triangle configurations exposes only one side a on its boundary. If the (or a) shared triangle is one of the two triangles whose a sides are on the maximal segment in one configuration, then that triangle must be the T_{3} of the other configuration. There are just two such configurations possible, when α = π/6. When α = arctan \frac{1}{2} , we will show no such configuration is possible: let P Q be the maximal segment, with triangles T_{1} and T_{2} west of P Q and T_{3} east of P Q with its b side equal to P Q and its right angle at P. Then the north side of T_{3} is the only exposed side of length a, and it cannot occur as part of a pair in another configuration, since its b side is already used.

Therefore α = π/6. We will show that the two configurations with α = π/6 cannot be completed to a 7-tiling. Let P Q be a (north-south) maximal segment of length c, and T_{1} and T_{2} west of P Q with their a sides together matching P Q and shared west vertex W; then T_{3} is east of P Q, with angle α at Q and east vertex E, and T_{4} shares side QE and has angle α at Q and east vertex X, and T_{5} shares side P X, and can be placed in either of two orientations. Let Y be the north vertex of T_{5}. Suppose, for proof by contradiction, that the β angle of T_{5} is placed at X, and the α angle at P. (See Figure 11) Then W P XQ is a parallelogram: since angle XP Q = angle P QW, P X is parallel to W Q, and since angle XQP = angle QP W, QX is parallel to W P. Since only two more triangles can be placed, the three sides of the final triangle are among the five sides of the pentagon W P Y XQ. The remaining two triangles must therefore each share a side with one of the existing triangles. Since W P is parallel to XQ, we must place a triangle on one of those sides. It is not possible to place T_{6} on any existing side in such a way as to create a concave exterior angle of π/3 or less. Hence, T_{6} must be placed so as to create a quadrilateral by creating straight angles where two vertices were before. There is only one position in which that is possible: triangle T_{6} must be placed along P Y with its right angle at Y. (Technically, we should compute the angle sums for all the other possibly positions of T_{6}, rather than rely on inspection of Figure 11 for a “proof”.) Then straight angles are created at Y and P. Then since W P and QX are parallel, we must place T_{7} along QX. To make a straight angle at X we need to place the β angle of T_{7} at X; but to make a straight angle at Q, we need to place the β angle there. Indeed this six-triangle configuration can be completed to an 8-tiling, but not to a 7-tiling. This contradiction shows that T_{5} cannot be successfully placed with its β angle at X.

W  P Q V  Y  X E

Figure 11: Two non-strict vertices in five triangles, second configuration

Therefore the β angle of T_{5} is placed at P. (See Figure 12). Then the total angle there is 3β = π, and W P Y is straight. The quadrilateral W Y XQ is convex. If we place a triangle to share any of its sides, and we do not match the side exactly, then another triangle will be required on that side, and a large triangle cannot be created. Hence when we place a triangle, it must match one of the sides of the quadrilateral exactly. That means that two of the sides of the quadrilateral must be sides of the final triangle. If any triangle is placed north of W P Y then we must place two triangles there, since the length of W P Y is c + a. That will leave at least four vertices; hence W P Y is one of the sides of the final triangle. Since QX is parallel to W P Y, it cannot be a side of the final triangle. Hence T_{6} must be placed south of QX, sharing side QX. There are two possible orientations. If the β angle of T_{6} is at X, then the resulting figure has vertices at W, Y, X, Q, and the south vertex Z of T_{6}. It is convex, so it is not possible to complete it to a triangle by placing T_{7}. Therefore, the α angle of T_{6} is at X. That creates a straight angle is created at Q, so the resulting figure is a quadrilateral. Now we ask if it is possible to place T_{7} along any side of this quadrilateral so as to create a triangle. Sides W P Y and W QZ are longer than c, so it is not possible to place T_{7} there. If T_{7} is placed along Y X, then the β angle would go opposite Y X, the right angle would have to go at Y to avoid creating a fourth vertex there, leaving the α angle to go at X, where the angle sum would then be 3α + β = 5α < π, so a triangle is not created. Hence T_{7} must be placed south of T_{6}. That means the sides of the final triangle are W Y, Y X extended, and W Q extended. Let Z be the intersection point of lines W Q and Y X. This point lies to the southeast, because of the angles made by the transversal W Y. Triangle ZXQ is similar to the tile, because it has angle α at Z (because the angle sum of triangle W Y Z must be π), and angle β at Q (because W QZ is a straight angle). But it has side QX = c = 2a opposite angle α. Hence its area is four times that of the tile, and thus it cannot be tiled by T_{6} and T_{7}. That contradiction completes the proof of Case 1.

Now we take up Case 2, in which the two configurations share a side but not a triangle. When we join two convex quadrilaterals along a side, we get a figure with at least six convex vertices, and possibly with one or two more vertices, which might be concave. Placing one triangle can reduce the number of convex vertices by at most 2 (that can happen if a concave vertex occurs in just the right position). But even if that happens, there will still be four vertices left after placing T_{7}. That completes the proof in Case 2.

Finally we consider Case 3, in which the two configurations do not even share a side. Since the quadrilaterals are convex, they share at most one point. At that shared point there may be a vertex with a concave exterior angle, so placing one triangle could possibly eliminate two of the six remaining vertices, but that would still leave four, too many for a triangle. That completes the proof of the lemma.

Theorem 5 (Main Theorem) There is no 7-tiling.

Proof. Suppose triangle ABC is 7-tiled by seven copies of triangle T. Then according to our previous theorems, it is not a strict tiling, and there is exactly one non-strict vertex V, and triangle T has a right angle, and its small angle α is either π/6 or arctan(1/2), and the non-strict vertex occurs in one of two specific configurations of three triangles (one for each α). (Those configurations are illustrated in Figure 8.) To finish the proof, we have to show that it is impossible, starting from either of those configurations, to add four more copies of T to create a triangle. We need only consider placements of new copies of T that share sides with existing copies, since no additional non-strict vertices can occur in a 7-tiling.

There are 7π angles total in the seven copies of T. Of these, π are used at the vertices A, B, and C, and π at the non-strict vertex V, leaving 5π to be used at boundary and interior vertices (other than V). An interior vertex uses 2π, and a boundary vertex uses π. The possibilities are thus: one boundary and two interior vertices, or three boundary and one interior vertex, or five boundary vertices and no interior vertex. In particular there are at most two interior vertices.

First we take up the case α = π/6. The starting configuration is the first one shown in Figure 8. The non-strict vertex V is at the midpoint of north-south line P Q. Triangles T_{1} and T_{2} are west of P Q, with a shared west vertex W, a right angle at their shared vertex V, and angle α at W. Triangle T_{3} is east of P Q, with angle P QE = α, and angle QP E = β.

Consider adding T_{4} north of P E with its third vertex N on QE extended. If we then add T_{5} north of P N, two additional triangles T_{6} and T_{7} will be required to fill the angle 2π at P. Vertices Q and N remain, so if this were to create a triangle, the west vertex Y would have to lie on QW extended. That would require at least two more triangles to share vertex W, one of which might be T_{6}, but there is no second one available. Hence the indicated placement of T_{5} fails. Similarly, if we add T_{5} southeast of EN, with southeast vertex X, then the exterior angle at vertex E will be concave, so we will have to add T_{6} sharing vertex E. That can fill vertex E to 2π only if both T_{5} and T_{6} have a right angle at E, which will make a six-triangle convex pentagon; such a configuration cannot be completed to a triangle. If instead T_{5}, T_{6}, and T_{7} are all placed with a vertex at E, the result cannot be a triangle since there are vertices at W, N, Q, and at least one more southeast of QN. Hence it fails to place T_{5} southeast of EN. Since T_{5} cannot be placed east of EN or north of P N, with this placement of T_{4}, two sides of ABC must be W P N and NEQ. We must then add the remaining three triangles to the southwest of W Q. Since P and E will now be boundary vertices, we are allowed only one interior vertex in the process. That vertex S will be created when we add triangle T_{5} immediately south of W Q, with southwest vertex S. There are two ways to place T_{5}; first consider placing its α angle at Q. Then W S = a and we must place T_{6} west of W S with its right angle at S (since otherwise W S must be the third side of the final triangle, and there is not enough area south of SQ and north of NQ extended to hold two more copies of the tile). Let X be the west vertex of T_{6} (placed west of W S). We now have a non-strict 6-tiling of triangle XNQ (Figure 13). This cannot be made into a triangle by adding one more triangle south of XSQ. Hence the indicated placement of T_{5} fails.

W  PQ V X  S  N E

Figure 12: A six-tiling that cannot be completed to a 7-tiling

Therefore T_{5} has to be placed with its β angle at Q, and its α angle at W. Let S be the third vertex of T_{5}. Then we cannot place T_{6} with its α angle at Q to have its side extend NEQ, since that creates a non-strict vertex at S. Hence QS must be the third side of the final triangle. But that is not possible, either, since QS is parallel to W P N (since transversal P Q makes equal alternate interior angles SQP and QP N, both equal to 2β). this placement of T_{5} also fails. Hence the indicated placement of T_{4} (north of P E with its third vertex N on QE extended) fails.

Now consider adding T_{4} north of P E with is third vertex N not on EQ extended, i.e. its right angle is at P instead of E. Then the exterior angle at vertex P is concave, with total interior angle 7π/6. At least two more triangles T_{5} and T_{6} must share vertex P. If we use T_{7} also at P, then we will still have vertices four vertices W, Q, E, and N, so we must use only T_{5} and T_{6} at P. We must therefore place T_{5} along NP = b with its right angle at P. Let Y be its third vertex. Then T_{6} must be placed with its c side along W P and its β angle at P, so Y P = a is matched. The angle of T_{6} at Y is π/2, so the total angle at Y is 5π/6, not π, and our 6-triangle configuration has vertices at W, Y, N, E, and Q. (Figure 14.) Since this configuration is convex, the best we could hope to do by placing T_{7} is to reduce the number of vertices by one to four. Hence this placement of T_{4} also fails.

W  P Q V  N Y  E

Figure 13: Another configuration that cannot be completed to a 7-tiling

Thus T_{4} cannot be placed north of P E in either orientation. Hence P E lies on the boundary of triangle ABC. Since W Q is parallel to P E, it follows that W Q is not a side of triangle ABC. We must therefore place T_{4} south of W Q; call its third vertex S. There are two orientations to consider: either the angle of T_{4} at Q is β or it is α. First suppose T_{4} has angle α at Q. Then SW is a north-south line. Consider placing T_{5} along W P with α at P. Let X be the third vertex of T_{5}. Then XP is parallel to SQ, so T_{6} must be placed either north of XP or south of SQ. In either case we are committed to making W S extended a side of ABC, since placing another triangle west of W S extended will create concave vertices. We must therefore definitely add T_{6} north of XP to reach the intersection point Y of P E and W S extended. Now we have vertices Y, E, Q, and S, and we must remove the vertex at Q by placing T_{7}. The angle at Q is presently 2α + β = 2π/3. To remove it we would have to put the β angle of T_{7} at Q, but the two sides SQ and QE are both b, so the β angle of T_{7} cannot be placed at Q. Hence the placement of T_{5} along W P with α at P fails. Now consider placing T_{5} along W P with β at P. Then the third vertex X of T_{5} lies on P E extended, and XW is perpendicular to P E. If we place T_{6} along XW, we again reach the intersection point Y of P E and W S, and we have the same convex quadrilateral 6-tiled as with the previous placement of T_{5}, and again T_{7} cannot be placed to make a triangle. Hence we cannot place T_{6} along XW. But then XW must be a side of the final triangle ABC. Since P E is a side, the vertex at W will have to be eliminated, but this is not possible, since we would have to place angle α at W, but the side W S is a, which cannot be adjacent to angle α. Hence the placement of T_{5} along W P with β at P fails. Now both possible placements of T_{5} along W P have failed. Hence W P is contained in one of the sides of triangle ABC. We therefore must add T_{5} west of W S, matching its a side to W S. Let X be the third (westernmost) vertex of T_{5}. If XSQ is not a side of ABC, we would have to add two more triangles south of XSQ, but that would not make a triangle. Hence XSQ is the third side of ABC. Let Y be the intersection point of EP and SQ. Then the remaining two triangles would have to tile triangle EQY. Triangle EQY is similar to T, since it has angle β at Q and a right angle at E, but the side opposite angle α is EQ = b = \sqrt{3}/2. So the area of triangle EQY is 3 times the area of T, not twice the area of T. Hence the placement of T_{4} with angle α at Q fails.

Now consider the other possible placement of T_{4}, namely south of W Q with angle β at Q. Let S be the third vertex of T_{4}. If W S is a side of ABC, then we will need to use three more triangles north of W P to reach the intersection point Y of W S and EP, and that will leave four vertices Y, S, E, and Q. Hence W S is not a side of ABC, so we must place T_{5} along W S = b. There are two possible orientations, with the angle of T_{5} at W either α or a right angle. Consider first placing T_{5} on W S with a right angle at W. Let X be the third vertex of T_{5}. Then X lies on W P extended and XS is parallel to W Q and P E. Then XS cannot be a side of ABC (since P E is a side), so we must place T_{6} south of XS. Let Y be the third vertex of T_{6}. Then we have vertices Y, X, P, E, Q, and possibly S. Since this figure is convex, we cannot possibly reduce the number of vertices to three by placing T_{7}. So the placement of T_{5} on W S with a right angle at W fails. Now consider the other possible placement of T_{5}, on W S with angle α at W. Let X be the third vertex of T_{5}. Now W X cannot be a side of ABC, since in that case three more triangles would be needed to reach the intersection point Y of P E and W X. Therefore we must add T_{6} west of W X. Let Z be the westernmost vertex of T_{6}. Then we have vertices Z, X, Q, E, P at least, and the figure is convex, so it cannot be made into a triangle by placing T_{7}. Hence both orientations of T_{5} on W S fail. Hence the second possible orientation of T_{4} (south of W Q with angle β at Q) fails. That exhausts the possibilities, and completes the proof in case α = π/6.

Now we consider the case α = arctan \frac{1}{2} , which is about 26.565 degrees. Then β = arctan 2 is about 63.435 degrees. The starting configuration is shown in the second part of Figure 8. The non-strict vertex V is at the midpoint of north-south line P Q. Triangles T_{1} and T_{2} are west of P Q, with a shared west vertex W, a right angle at their shared vertex V, and angle α at W. Triangle T_{3} is east of P Q, with angle P QE = α, and angle QP E is a right angle.

We first consider placing T_{4} north of P E with a right angle at P. Let N be its northern vertex. That creates a concave vertex at P with exterior angle π − β. That will require T_{5} and T_{6} to be used north of W P and west of NP,respectively. Even if we managed to solve the problem of the concave vertex at P, we would then have only one more triangle T_{7} to place, and we cannot reduce the number of vertices by placing it on NE, QE, or W Q, but it must be placed on one of those sides since not all three can be sides of the final triangle ABC. Hence the placement of T_{4} north of P E with a right angle at P fails. Next consider placing T_{4} north of P E with a right angle at E, and let N be its northern vertex. This also creates a concave vertex at P, with exterior angle 3π/2 − 2β. We cannot use all three remaining triangles at P, as that will leave vertices W, Q, E, and N. Since we can place only two new triangles with vertices at P, they must have their c sides along W P and P N, so they cannot have right angles at P. Therefore their maximum contribution to the angle sum at P is 2β, which is not enough to fill the angle at P, since 4β + π/2 is about 343.74 degrees, not 360. Hence the placement of T_{4} north of P E with a right angle at E fails. Hence T_{4} cannot be placed north of P E at all. Hence P E is (contained in) a side of the final triangle ABC. Suppose, for proof by contradiction, that W Q is a side of the final triangle. Then we must place T_{4} north of W P, and T_{5} west of T_{4}. Let X be the west vertex of T_{5}. Then XQE is 5-tiled. We must place T_{6} along QE, since XW Q and XP E are sides of the final triangle. There are two possible orientations of T_{6}, with angle α at E or angle β at E. First consider placing T_{6} on QE with angle α at E. Let Y be the third vertex of T_{6}. Since QE = c, the angle of T_{6} at Y is a right angle and QY is parallel to side P E of the final triangle. Hence T_{7} must be placed south of QY = a, and its right angle must go at Y or a vertex will be created there. Hence the angle of T_{7} at Q will be β, making the total angle at Q equal to 3β + α = π/2 + 2β > π, contradicting our assumption that W Q is a side of the final triangle. Hence the placement of T_{6} on QE with angle α at E fails. Next consider the other possible placement of T_{6}, on QE with angle β at E. Let Y be the eastern vertex of T_{6}. We cannot place T_{7} north of EY, since 3β > π and T_{7} would then extend north of P E. Hence the third side of ABC must be EY, and T_{7} must be placed south of QY with its right angle at Y. Then the angle of T_{7} at Q is α and the total angle at Q is 3α + β = π/2 + 2α < π, so a triangle has not been created. Hence the placement of T_{6} on QE with angle β at E fails. Now both possible placements of T_{6} have failed. This contradicts our assumption that W Q is a side of the final triangle. Hence W Q is not a side of the final triangle.

Therefore we must place T_{4} south of W Q. First consider placing T_{4} along W Q with angle α at Q, as shown in the first part of Figure 15. Let R be the third vertex of T_{4}. Suppose, for proof by contradiction, that RW is a side of the final triangle. Then we must place T_{5} north of W P. Call its north vertex N. Since RQ is parallel to P E, RQ is not a side of the final triangle, and we must place T_{6} south of RQ. Call its south vertex X. Since P E is (contained in) a side of the final triangle, the east vertex of the final triangle must lie on P E (extended); but since the exterior angle between EQ and P E extended is more than π/2, triangle T_{7} cannot extend to the east of E on P E extended. Hence E is a vertex of the final triangle. It is not possible to create a final triangle by placing triangle T_{7} along the east side of EQ, since then we will have four distinct vertices

R W  P  E Q V ^{R W  P  E Q V}

Figure 14: Two possible placements of T_{4}

N, E, X, and the east vertex of T_{7}. The latter two cannot coincide since X is south of RQ and west of P Q extended, while the east vertex of T_{7} would be (in either possible orientation of T_{7}) east of P Q extended. Since T_{7} cannot be placed east of EQ, the third side of the triangle must be EQ extended. Hence triangle T_{6} does not have its right angle at Q (or it would extend east of EQ extended). Since RQ = b, triangle T_{6} has angle β at X. Hence it has either angle α at Q. Then the angle sum at Q is 3α + β = 2α + π/2 < π. Now we have a six-triangle configuration with four vertices N, E, Q, and X, and NR, NE, and QE are sides of the final triangle. But T_{7} cannot be placed south of XQ so as to create a triangle, since to do so it would need an obtuse angle at X. This contradicts our assumption that RW is a side of the final triangle. Therefore RW is not a side of the final triangle, and we are back to where only T_{4} has been placed (along W Q with angle α at Q), as shown in the first part of Figure 15.

Since RQ is parallel to P E, RQ is not a side of the final triangle, and we must place T_{5} south of RQ. If its right angle is placed at R, then since RW is not a side of ABC, we will require both T_{6} and T_{7} west of SW extended, since we are not allowed to create another non-strict vertex. That will leave vertices at P, E, Q, and points west. Hence T_{5} cannot be placed with its right angle at R. But then the right angle of T_{5} is at Q. This creates a concave vertex at Q with exterior angle greater than π/2, so two more triangles are required at Q, leaving none to place west of RW, where we need one since RW is not a side of ABC. This contradiction shows that placing T_{4} along W Q with angle α at Q fails.

Hence we must place T_{4} south of W Q with angle β at Q, as shown in the second part of Figure 15. Let R be the third vertex of T_{4}. Suppose, for proof by contradiction, that W P is not a side of triangle ABC. Then we must add T_{5} with vertex α at P, side c along W P, third vertex N on P E extended, with angle W NP = π/2. The total angle at W is now 3α+β = π/2+2α. To eliminate the vertex at W (making a straight angle at W) we would need an angle of π − (π/2 + 2α) = π/2 − 2α, which is about 90 − 2 · 26.5 = 37 degrees, more than α, but less than both 2α and β, and hence impossible to supply. Possibly W might be eliminated as a vertex of ABC by becoming an internal vertex. Then the total angle at W would have to be made equal to 2π. After placing T_{5} it is π/2 + 2α, so we would need 2π−(π/2+2α) = 3π/4−2α more, which is about 217 degrees. Even if we used the right angles of T_{6} and T_{7} we could not make it. Hence W is definitely a vertex of triangle ABC (under the assumption that W P is not a side.) This is, however, impossible, since two of the vertices must be on line P E extended, and W cannot be the southernmost vertex, since for example point Q lies farther to the south than W. This contradiction shows that W P is, in fact, one of the sides of the final triangle ABC, along with P E.

Thus P is one of the vertices of ABC, and the other two lie on P E (extended) and P W (extended). W cannot be a vertex of ABC, since then W would be the southernmost vertex of ABC, but R lies farther south than W. Hence the vertex at W must be eliminated by placing more triangles with a vertex at W. The angle sum at W (from the three triangles already there) is 3α, about 79.5 degrees. To reach an angle sum of π at W, we could place three triangles with angle β at W, but that would use all seven triangles and still leave vertices at P, E, and Q, as well as somewhere southwest of W on P W extended, so no triangle would be formed. At least two must be used since 3α + π/2 < π. The possible angle sums resulting from placing two more angles at W are among

5α < π
4α + β = 3α + π/2 < π
4α + π/2 > π
3α + 2β = π + α > π
3α + β + π/2 = 2α + π > π

None of these possibilities would succeed to eliminate vertex W. This final contradiction completes the proof of the theorem.

References

[1] Beeson, M., Tiling a Triangle with Congruent Triangles, to apppear.

[2] 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. QA167 .B6513

[3] 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). QA447 .B573

[4] Goldberg, M., and Stewart, B. M., A dissection problem for sets of polygons, Amer. Math. Monthly 71 (1964) 1077–1095.

[5] Golomb, Solomon W., Replicating figures in the plane, The Mathematical Gazette48 No. 366. (Dec., 1964), pp. 403-412.