/^creative ^commor
ARS MATHEMATICA CONTEMPORANEA
Also available at http://amc-journal.eu ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)
ARS MATHEMATICA CONTEMPORANEA 12 (2017) 415-424
Spherical quadrangles with three equal sides and
rational angles
Kris Coolsaet
Department of Applied Mathematics Computer Science and Statistics, Ghent University, Krijgslaan 281-S9, B-9000 Gent, Belgium
Received 14 June 2016, accepted 25 June 2016, published online 19 February 2017
Abstract
When the condition of having three equal sides is imposed upon a (convex) spherical quadrangle, the four angles of that quadrangle cannot longer be freely chosen but must satisfy an identity. We derive two simple identities of this kind, one involving ratios of sines, and one involving ratios of tangents, and improve upon an earlier identity by Ueno and Agaoka.
The simple form of these identities enable us to further investigate the case in which all of the angles are rational multiples of n and produce a full classification, consisting of 7 infinite classes and 29 sporadic examples. Apart from being interesting in its own right, these quadrangles play an important role in the study of spherical tilings by congruent quadrangles.
Keywords: Spherical quadrangle, rational angle, spherical tiling. Math. Subj. Class.: 51M09, 52C20, 11Y50
1 Introduction
In general there will be an infinite number of non-congruent spherical quadrangles with given (ordered) quadruple of angles a, 7, S, provided that 2n < a + / + 7 + S < 6n. By imposing restrictions on the sides of a quadrangle this is reduced to a finite number. In this paper we shall investigate the case of a convex quadrangle ABCD with (at least) three equal sides, say |AB| = |BC| = |CD| = a and derive two simple identities which must be satisfied by the angles of that quadrangle as a consequence of this restriction (cf. Theorem 2.1).
A similar, but more complicated identity for this case was already published by Ueno and Agaoka in [3]. We shall show that our identities are stronger (cf. Section 2) in a sense to be made clear below.
E-mail address: kris.coolsaet@ugent.be (Kris Coolsaet)
©® This work is licensed under http://creativecommons.org/licenses/by/3.0/
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Ars Math. Contemp. 12 (2017) 383-413
The identity of Ueno and Agaoka arose in the search for tilings of the sphere by congruent quadrangles. In that context it is particularly relevant to consider quadrangles all of whose angles are rational multiples of n (henceforth simply called rational angles).
Indeed, consider a vertex P of such a tiling. P belongs to a certain number (say NA) of quadrangles in the tiling for which P corresponds to vertex A of the quadrangle. Likewise there will be NB quadrangles for which P corresponds to B, and similar numbers NC, ND for C and D. Because the sum of all angles in P must be 2n, we find
NAa + Nb 3 + NC Y + ND S = 2n,	(1.1)
where a, 3, Y,S are the corresponding angles of the quadrangle (see Figure 1 for naming conventions). A different vertex P' of the tiling will lead to a similar identity, but generally with different values of NA, NB, NC, ND.
We may treat the set of identities (1.1) that arise from all vertices of a given tiling as a system of equations with unknowns a, 3, Y, S. Note that all coefficients in these equations are integers, while every right hand side is equal to 2n. In particular, if this system has rank 4, there will be exactly one solution and it will consist entirely of rational angles.
In Theorem 3.2 we give a full classification of all convex spherical quadrangles with three equal sides whose angles are rational. There turn out to be 7 infinite families of such quadrangles and 29 sporadic examples. The proof of Theorem 3.2 hinges on the fact that our identity (3.2) can be rewritten as an equality between two products of two sines and that instances of such identities with rational angles were already classified by Myerson [2] (cf. Theorem 3.1).
Figure 1: Naming conventions for a spherical quadrangle ABCD with three equal sides.
2 Relations between the angles
In what follows we shall consider a spherical quadrangle ABCD with corresponding angles a, P, y, S, sides a, a, a, b and diagonals x, y as indicated in Figure 1. The spherical quadrangle will be called convex if it satisfies 0 < a,b, x, y, a, P,Y,S < n. In particular this means that all constituent spherical triangles ABC, ABD, ACD and BCD are 'proper' and satisfy the classical laws of spherical trigonometry.
We take the following inequalities for such quadrangles from [1, Lemma 2.1]:
a + S	<	n + P,
a + S	<	n + Y,
a + P	<	n + S,
Y + S	<	n + a.
K. Coolsaet: Spherical quadrangles with three equal sides and rational angles
417
As mentioned in the introduction, we also have
E = a + 3 + y + 5 — 2n> 0,
(2.2)
where E denotes the spherical excess of the quadrangle, which is equal to the area of the quadrangle on a unit sphere.
Finally, we note that a = S if and only if 3 = 7, cf. [1, Lemma 2.3].
Theorem 2.1. In a convex spherical quadrangle ABCD with three equal sides, the following identities hold:
sin(a — 2 ) sin(5 — 2 )
sin Y
sin 2
or equivalently,
tan ( 2 — 2 ) _ tan ( f — 2 )
tan 2
tan a
(2.3)
(2.4)
Proof. Consider the equilateral spherical triangle ABC. The (polar) cosine rule for side BC yields
cos $ + cos $ cos 3 1 + cos 3	3
cos a = -----= cot $ •--— = cot $ cot —,	(2.5)
sin $ sin ¡3	sin ¡3	2
where $ = ZBAC = ZACB as indicated in Figure 2. The sine rules for side AC in both
A
b
D
B
C
Figure 2: The spherical triangles ABC and ACD. ABC and ACD, yield
sin 3 sin 0 sin 5 sin(a — 0)
sin x sin a sin x
sin a
and hence
sin 5 sin(a — 0) sin 3 sin 0
sin a cot 0 — cos a.
Multiplying by cot 2 and using sin 3 = 2 cos | sin | and (2.5), yields
sin S	3
-tt—t = sin a cos a — cos a cot —,
2 sin2 2	2 ,
whence
cos a =
sin 5 + cos a sin 3
2 B .
2 sin a sin 2
(2.6)
a
418
Ars Math. Contemp. 12 (2017) 383-413
By repeating the argument above for triangles ABD and BCD, or equivalently, by interchanging a ^ S and fi ^ y, we find
sin a + cos S sin 7
Y 2
We shall now use (2.6-2.7) to compute
cos a = —	2 Y—.	(2.7)
2 sin S sin i
A = 2 cos a sin a sin S — sin2 a + cos2 S = 2 cos a sin a sin S + cos2 a — sin2 S
in two ways. From (2.6) we obtain
R	R
O R O	O R	o	o
A = csc2 — sin2 S + csc2 — cos a sin R sin S + cos2 a — sin2 S
2+2 R +
csc2 R — sin2 S + 2 cot R cos a sin S + cos2 a R	R	R
= cot2 — sin2 S + 2 cot — cos a sin S + cos2 a = cot — sin S + cos a I .
2+2	+	V 2 + 1
By symmetry, from (2.7) we obtain
A = ^cot Y sin a + cos sj .
(Note that the original definition of A remains unchanged when interchanging a and S.) Combining both values of A we end up with two possibilities:
R	( Y	\
cot — sin S + cos a = ± ^cot — sin a + cos SJ .	(2.8)
We consider both cases separately. The second case will turn out to be impossible.
Case 1. Assume there is a plus sign in the right hand side of (2.8). Then (2.8) rewrites
to
R	Y
cot — sin S — cos S = cot — sin a — cos a,	(2.9)
2 2 , V ^
which is equivalent to (2.3).
In general, if xx/yx = ^2/^2, then also (xi — yi)/(xi + yi) = (x2 — y2)/(x2 + ^2). Applying this to (2.3) yields
sin(S — 2) — sin 2 sin(a — Y) — sin Y sin(S — 2) + sin f sin(a — Y) + sin Y ,
which transforms to
cos 2 sin( 2 — 2) cos ia sin( a — 2) sin § cos( ^ — 2) = sin f cos( f — ^),
and hence
tan (2 — 2) _ tan (f — 2)
tan 2	tan f
K. Coolsaet: Spherical quadrangles with three equal sides and rational angles
419
Case 2. Assume there is a minus sign in the right hand side of (2.8), i.e., cot f sin 5 + cos a = — cot f sin a — cos 5. This formula can be obtained from the formula for the first case by replacing a by —a and 5 by n — 5. As a consequence, we now have the following identities:
sin(n - J - f ) sin(-a - f) tan (f - f - f) tan ( - f - f)
sin f
tan (f - f)
- tan a
equivalent to
sin( J + f ) sin f

sin(a + f ) sin f '
tan f
^ (f + f)
tan (a + f)
tan a
Because the tangent function is monotonous in the interval [0, n/2[, the latter is only possible if fi = 7 = 0 or if one of 1 (J + fi), f (a + 7) lies outside that interval. And because of the signs, this implies that both values must belong to the interval ] f n, n[. Hence
n < a + 7, fi + J. Now a + fi < n + J by (2.1). Hence a + fi < fi + 2J and hence a < 2J. By symmetry, also J < 2a, a contradiction.	□
In [3] Ueno and Agaoka derived the following identity for spherical quadrangles with three equal sides:
(1 - cos fi) cosf a - (1 - cos fi)(1 - cos 7) cos a cos J + (1 - cos 7) cosf J + cos fi cos 7 + sin a sin fi sin 7 sin J - 1 = 0.
(2.10)
We have
Lemma 2.2. Formula (2.10) is equivalent to
fi7
cot — sin J - cot — sin a = ±(cos a - cos J).
(2.11)
Proof. We express cos ^ and sin ^ in terms of cot f as follows:
sin fi =
2 cotf f cotf f + 1,
cos fi =
cotf f - 1 cotf f + 1,
1 - cos fi =
cotf f + 1
(2.12)
and similar for cos 7 and sin 7. Also note that
cos fi cos 7 - 1 =
(cotf f - 1)(cotf f - 1) (cotf f + 1)(cotf f + 1)
1
-2 cotf f - 2 cotf f (cotf f + 1 )(cotf 2 + 1) .
(2.13)
Applying (2.12-2.13) to the left hand side of (2.10) transforms it into
2 cos2 a
4 cos a cos J

+
2 cos2 J
cotf f + 1 (cot2 f + 1)(cot2 2 + 1) ' cot2 f + 1

2 cot2 f +2 cot2 2
+
4 sin a cot f cot f sin J
(cot2 2 + 1)(cot2 f + 1) (cot2 f + 1)(cot2 2 + 1)
sin 2
2
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Ars Math. Contemp. 12 (2017) 383-413
which after multiplying by the common denominator and dividing by 2, reduces to
cos2 a (cot2 Y + l) - 2 cos a cos S + cos2 S ^cot2 P +
2 P 2 Y	P Y
— cot2--cot2 —+2 sin a cot — cot — sin S
2 2 2 2
= cos2 a — 2 cos a cos S + cos2 S
2 ,2 Y . 2 c , 2 P ,2 P ,2 Y
+ cos a cot —+ cos S cot--cot--cot —
2 2 2 2
P Y + 2 sin a cot — cot — sin S 2 2
( P	Y ^ 2
= (cos a — cos S)2 — ( cot — sin S — cot — sin a
□
Remark that choosing the minus sign in (2.11) yields our formula (2.9) from the proof of Theorem 2.1. This shows that Theorem 2.1 is stronger than the result of Ueno and Agaka, as they also allow solutions with a plus sign in the right hand side of (2.11).
3 Rational angles
In what follows we shall investigate spherical quadrangles with three equal sides with the additional property that the four angles a, P, y, S are rational. Our main tool is the following theorem from [2].
Theorem 3.1 (Myerson). For all 0 we have
■ n . „ . 0 . (n 0 \
sin — sin 0 = sin — sin---.	(3.1)
6 2^22/
All other solutions of
sin nxi sin nx2 = sin nx3 sin nx4	(3.2)
with rational numbers x1, x2, x3, x4 such that 0 < x1 < x3 < x4 < x2 < 1/2, are given in Table 1.
Table 1: Nongeneric solutions to (3.2).
x1	X2	x3	x4	x1	x2	X3	x4
1/21	8/21	1/14	3/14	4/15	7/15	3/10	11/30
1/14	5/14	2/21	5/21	1/30	11/30	1/10	1/10
4/21	10/21	3/14	5/14	7/30	13/30	3/10	3/10
1/20	9/20	1/15	4/15	1/15	4/15	1/10	1/6
2/15	7/15	3/20	7/20	2/15	8/15	1/6	3/10
1/30	3/10	1/15	2/15	1/12	5/12	1/10	3/10
1/15	7/15	1/10	7/30	1/10	3/10	1/6	1/6
1/10	13/30	2/15	4/15				
K. Coolsaet: Spherical quadrangles with three equal sides and rational angles
421
Althought our condition (2.3) is almost a direct match with equations (3.1) and (3.2) of the theorem, there are some additional complications that must be taken into account. Most importantly, Theorem 3.1 imposes extra conditions on the angles nxi,..., nx4 which are too stringent for the angles a - §, §, §,6 - § of (2.3).
First, all angles in Theorem 3.1 must lie in the interval ]0, n/2[ while of the four angles in (2.3) only § and § satisfy this restriction, while the other two (a - §, 6 - §) are only known to lie in the interval ] - n/2, n[. This means that we have to 'renormalize' these angles by using the identities
sin(n - nxj) = sin nxj, sin(-nxj) = - sin nxj.
As a consequence, we shall always need to consider the following five cases:
{nxi,nx§,nx3 ,nx4} = <
(a _ y â Y 6- âl {a §,§,§,6 §},
{n - a + §, §, §, 6 - §}, {a - §, §, §, n - 6 + §}, {n - a + §, §, § ,n - 6 + § }
SX _ a â 1 â _ 6j { § a, § , § , § 6}.
(3.3)
(Note that a - 2 < 0 automatically implies 5 - 2 < 0 because the signs of both sides of (2.3) must be the same.)
Furthermore Theorem 3.1 assumes a specific ordering of the variables xi; x2, x3, x4, which again we cannot guarantee. In principle we must therefore consider eight different ways to assign the angles on the right hand side of (3.3) to the angles of the left hand side, yielding 40 possibilities in total. We can reduce this amount by half by taking into account the symmetry a ^ 5, ft ^ 7.
Finally, Theorem 3.1 does not consider the 'trivial' cases where one (and then at least two) of the angles is zero, or where {sin nxi, sin nx2} = {sin nx3, sin nx4}.
Taking all of this into consideration leads to
Theorem 3.2. Consider a convex spherical quadrangle with three equal sides, with angles and sides as indicated in Figure 1.
If the angles a, ft, 7,5 are rational multiples of n, then they must satisfy one of the following properties, or a property derived from these by interchanging a ^ 5 and ft ^ 7:
a = 7 and ft = 5 (and all four sides are equal),
a = 5 and ft = 7,
a = § and 6 = §, with a + 6 < n,
a = , ft = § and 5 = 2n - 2, with 2 < 7 < 2n. a = 6 + 2, ft = 27 and 5 = 2 + 2, with n < 7 < 2, a = 6 + 2, ft = 27 and 5 = 2 + fl (= 3a), with 4f < 7 < f, a = 6 + 2, ft = 2n - 27 and 5 = ff - f-, with f < 7 < 52, (sporadic cases) a/n, ft/n, 7/n, 5/n are as listed in Table 2.
8
422	Ars Math. Contemp. 12 (2017) 383-413
Table 2: Sporadic cases of Theorem 3.2.
a/n	P/n	Y/n	S/n	a/n	P/n	Y/n	S/n
29/42	8/21	3/7	23/42	5/6	8/15	3/5	19/30
31/42	8/21	3/7	23/42	23/30	8/15	3/5	19/30
5/6	8/21	5/7	17/42	5/6	8/15	11/15	17/30
37/42	8/21	5/7	17/42	9/10	8/15	11/15	17/30
5/6	3/7	20/21	17/42	23/60	8/15	9/10	13/60
11/42	5/7	20/21	1/6	31/60	8/15	9/10	19/60
29/42	5/7	20/21	23/42	17/30	8/15	13/15	11/30
49/60	4/15	7/10	17/60	31/60	3/5	5/6	23/60
53/60	4/15	7/10	17/60	11/15	3/5	13/15	8/15
7/10	4/15	13/15	7/30	19/30	3/5	14/15	13/30
49/60	3/10	14/15	17/60	5/6	3/5	14/15	17/30
23/30	1/3	14/15	3/10	19/60	7/10	14/15	13/60
11/15	7/15	3/5	8/15	37/60	7/10	14/15	29/60
13/15	7/15	3/5	8/15	23/30	11/15	14/15	19/30
17/30	7/15	14/15	3/10				
Proof. (The proofs of the inequalities in the statement of the Theorem are left to the reader. They are immediate consequences of (2.1-2.2).) We split the proof into three parts.
Part 1. We first consider the 'trivial' cases. The only angles in (2.3) which are allowed to be zero, are a - 2 and S -f. This corresponds to case 3 in the statement of this theorem. Next, {sinnxi, sinnx2} = {sinnx3, sinnx4} corresponds to either
a - 2) = sin 2 and	sin ^ - 0 = sin P, (3.4) or
( Y) ( P\	Pi
sin (a--J = sin S----and sin — = sin —. (3.5)
V 2 / V2/	2 2
Equation (3.4) further splits into 4 different cases:
a - 2 = 2 and S - f = f,
a - 2 = ±n - 2 and S - 2 = f,
a - 2 = 2 and S - f = ±n - f,
a - 2 = ±n - 2 and S - f = ±n - f.
The first case corresponds to case 1 in the statement of this theorem, the other three are not allowed because then a = ±n or P = ±n.
Similarly, equation (3.5) splits into the following cases:
a - 2 = S - 2	and	f = 2,
a - 2 = ±n - S + f	and	f = 2,
a - 2 = S - 2	and	f = n - 2,
a - 2 = ±n - S + f	and	f = n - 2.
K. Coolsaet: Spherical quadrangles with three equal sides and rational angles	423
The first of these reduces to a = S and 0 = 7, i.e., case 2 in the statement of this theorem. The second implies 0 = 7, a + S = n + 7 which is disallowed by (2.1). The third leads to a + n = 7 + S, again forbidden by (2.1). The last yields either a = 2n - S or a = -S, which again is not allowed.
Part 2. Let us now consider formula (3.1). As mentioned above, this formula must be applied to our problem in 20 different ways, 4 permutations of the angles in (3.1) each time matched to the 5 cases listed in (3.3).
In Table 3 we list each of these 20 possibilities (columns 1-4), and the corresponding values of a, 0,7, S (columns 5-8). In each of the four tables, columns 1-4 contain the same values but correspond to different angles of (3.1), as indicated in the column headers.
Table 3: The generic case of Theorem 3.2.
	e		n 6	6 2	n 2	6 2		a	ß	Y	8		
a	_ Y 2		ß 2	Y 2	8	ß 2	ß 2	36 2	n 3	e	2n 3	6 2	a +-----+ 8 < 2n
a	Y 2		ß 2	Y 2	n —	8 +		36 2	n 3	e	2n 1 3 +	6 2	a + 8 < ß + n ^ e < n but a +-----+ 8 > 2n ^ e> |
n —	a +	Y 2	ß 2	Y 2	8	ß 2	ß 2	n— 2	n 3	e	2n 3	6 2	a +-----+ 8 = 2n
n —	a +	Y 2	ß 2	Y 2	n —	8 +		n— §	n 3	e	2n 1 3 +	6 2	a + 8 > n + ß
Y 2	a		ß 2	Y 2	ß 2	8		6 2	n 3	e	-n + 3 +	6 2	a < 0
Y
a - Y a 2
a - Y a 2
n — a + : n — a + -
Y - a 2 a
Y 2
Y 2
Y 2
Y 2
Y 2
8 - ß
8 2 n — 8 + ß
8-ß ' 8 2
n — 8 + ß
ß-8' 2 8
n 1 6
2	+ 2
n I 6
2	+ 2
3n
2
3n 2
36
" 2 36 " 2 36
2
n — e n — e n — e n — e n - e
n 1 6
6 + 2
7n
6
6 ' 2 n I 6
6 + 2 7n 6 6 2 n _ 6 6 2
a +-----+ 8 = 2n
a + 8 = n + ß
a + 8 > n + ß a +-----+ 8 < 2n
e
ß
8
a
6
2
3
3
3
3
2
3
	n 6		e	6 2	n 2	6 2		a	ß	Y	8	
a	_ Y 2		ß 2	Y 2	8	ß 2		n 1 6 6 + 2	2e	e	n 1 6 2 + 2	
a	Y 2		ß 2	Y 2	n —	8 +	ß 2	n 1 6 6 + 2	2e	e	n 1 36 2 + 2	
n —	a +	Y 2	ß 2	Y 2	8	ß 2		5n + 6 6 + 2	2e	e	n 1 6 2 + 2	a + 8 > n + y
n —	a +	Y 2	ß 2	Y 2	n —	8 +	ß 2	5n , 6 6 + 2	2e	e	n 1 36 2 + 2	a + 8 > n + ß
Y 2	a		ß 2	Y 2	ß 2	8		6 + 2	2e	e	+ 36 2 + 2	a +-----+ 8 < 2n
n 6		e	n 6 2 2		6 2		a		ß	Y		8		
a - Y a 2		ß 2	Y 2	8	ß 2	ß 2	2n 3	6 2	2e	n —	e	36 2		
a - Y a 2		ß 2	Y 2	n —	8 +		2n 3	6 2	2e	n —	e	n +	6 2	Y + 8 > n + a
n — a +	Y 2	ß 2	Y 2	8	ß 2		4n 3	6 2	2e	n —	e	36 2		a + ß > n + 8
n — a +	Y 2	ß 2	Y 2	n —	8 +	ß 2	4n 3	6 2	2e	n —	e	n +	6 2	a + 8 > n + y
Y - a 2 a		ß 2	Y 2	ß 2	8		n 3	6 2	2e	n —	e	6 2		a +-----+ 8 < 2n
It turns out that 16 of these options are disallowed by the inequalities (2.1-2.2). We list the corresponding details in the right hand column of each table. The four possibilities that remain are listed as cases 4-7 in the statement of the theorem.
424
Ars Math. Contemp. 12 (2017) 383-413
Part 3. For the sporadic examples from Table 1 we could proceed in the same manner as in part 2 of this proof. Although there are some shortcuts which could be taken to avoid to have to consider each of the 20 x 15 = 300 cases separately, we thought it less error prone to enlist the help of a computer.
Recall that i sin belongs to the cyclotomic field Q(Zn) where Zn is a primitive n-th root of unity. Modern computer algebra systems can do exact arithmetic over such cyclotomic fields, hence we may use such a system to directly check all instances of equation (2.3) in which a, ft/2, y/2, S are integral multiples of and are in the required range. From Table 1 we may derive all values of n which we are required to try: for each row let n denote twice the least common multiple of the four denominators. In fact, many rows will yield the same value of n and it turns out to be sufficient to do the computations only for n = 84 and n = 120. We used this method to obtain the values in Table 2. (The source code for these computations is available from http://caagt.ugent.be/ratquad/.)
The same method, with n = 12p, p a prime > 7, was used to verify the results of part 2 of this proof.	□
Note that in case 3 of Theorem 3.2 the quadrangle is a union of three disjoint congruent triangles with angles a, S and n/3 — cf. Figure 3.
Figure 3: The special case a = 2, S = 2.
References
[1]	Y. Akama and N. Van Cleemput, Spherical tilings by congruent quadrangles: Forbidden cases and substructures, Ars Math. Contemp. 8 (2015), 297-318, http://amc-journal.eu/ index.php/amc/article/view/581/7 67.
[2]	G. Myerson, Rational products of sines of rational angles, Aequationes Math. 45 (1993), 70-82, doi:10.1007/bf01844426.
[3]	Y. Ueno and Y. Agaoka, Examples of spherical tilings by congruent quadrangles, Mem. Fac. Integrated Arts and Sci., Hiroshima Univ., Ser IV 27 (2001), 135-144, doi:10.15027/781.