Download ON ANGLES WHOSE SQUARED

ON ANGLES WHOSE SQUARED TRIGONOMETRIC
FUNCTIONS ARE RATIONAL
by
John H. Conway1 * Charles Radin2 ** and Lorenzo Sadun3 ***
1 Department of Mathematics, Princeton University, Princeton, NJ 08544
2;3 Department of Mathematics, University of Texas, Austin, TX 78712
Abstract
We consider the rational linear relations between real numbers whose squared
trigonometric values are rational, angles we call \geodetic". We construct a convenient basis for the vector space over Q generated by these angles. Geodetic angles,
and rational linear combinations of geodetic angles, appear naturally in Euclidean
geometry; for illustration we apply our results to equidecomposability of polyhedra.
* Research supported in part by NSF Grant No. DMS-9701444
** Research supported in part by NSF Grant No. DMS-9531584 and
Texas ARP Grant 003658-152
*** Research supported in part by NSF Grant No. DMS-9626698 and
Texas ARP Grant 003658-152
0. Introduction
Many well known geometric objects involve angles that are irrational when
measured in degrees or are irrational multiples of in radian measure. For instance
we might mention the dihedral angle ( 70310 4400) of the regular tetrahedron,
whose supplement ( 1092801600 ) is known to chemists as the carbon valence bond
angle. A goodly number of these angles have the property that their six trigonometric functions have rational squares. For instance,
sin2 = 98 ; cos2 = 91 ; tan2 = 8; cot2 = 81 ; sec2 = 9; csc2 = 89 1)
and, for the dihedral angle ( 144440 800 ) of the cuboctahedron,
sin2 = 32 ; cos2 = 13 ; tan2 = 2; cot2 = 21 ; sec2 = 3; csc2 = 23 : 2)
There are many additive relations between angles of this kind; for instance and
satisfy + =2 = . In this paper we nd all relations between these angles.
To be precise, we shall say that is a \pure geodetic angle" if any one (and
therefore all) of its six squared trigonometric functions is rational (or innite), and
use the term \mixed geodetic angle" to mean a linear combination of pure geodetic
angles with rational coecients. The mixed geodetic angles form a vector space
over the rationals and we shall nd an explicit basis for this space.
Another aim of this paper is to introduce an elegant notation for these angles
which we hope will nd general acceptance. Namely, if 0 r 1 is rational we
dene
6 r = sin?1 pr:
3)
(We feel free to write these angles in either degrees or radians.) The well-known
particular cases are:
6 0 = 0 = 0; 6 1 = 30 = ; 6 1 = 45 = ; 6 3 = 60 = ; 6 1 = 90 = : 4)
4
6 2
4 4
3
We extend this notation by writing, for all integers n,
6 n = 90n = n ; 6 (n + r) = 6 n + 6 r:
2
2
5)
Our basis contains certain angles hpid for prime p and square-free d. (Some
exercise in the notation is provided in Tables 1 and 2, which show the rst few
elements in the basis.) hpid is dened just when ?d is congruent to a square modulo
p and is found as follows. Express 4ps as a2 + db2 for the smallest possible positive
s. Then
1
s
s
r 2
2
1
a
1
?1
?1 db :
=
cos
=
tan
6)
s
s
s 4p s
4p s
4p s
a2
The expression is unique except when d = 1 or 3, when we make it so by demanding
that b be even (if d = 3) or divisible by four (if d = 1). Our main result is then
Theorem 1. Every pure geodetic angle is uniquely expressible as a rational multiple
of plus an integral linear combination of the angles hpid . So the angles hpid ,
supplemented by (or 6 1 or 1), form a basis for the space of mixed geodetic
angles.
It is easy to nd the representation of any pure geodetic angle in terms of
the basis.
p
Theorem 2. If tan = ab d for integers a; b; d, with d square-free and with a and
b relatively prime, and if the prime factorization of a2 + db2 is p1 p2 pn (including
multiplicity), then we have
= t hp1id hp2 id hpn id
7)
for some rational
p t. (In fact the denominator of t will be a divisor of the class
number of Q ( ?d).)
p
For example, for tan = 45 3 we nd a2 + db2 = 16 + 75 = 91 = 7 13 and
indeed = n ? h7i3 ? h13i3.
Our results have many applications but we shall only mention one outstanding
one here. In 1901 Dehn [Deh] solved Hilbert's 3rd problem by giving a necessary
condition for the mutual equidecomposability of polyhedra in terms of their dihedral
angles. In 1965, Sydler proved [Syd] that Dehn's criterion was also sucient. For
polyhedra with geodetic dihedral angles our Theorem 1 makes the Dehn-Sydler
criterion eective. At the end of this paper we shall apply our theory to the nonsnub Archimedean polyhedra (whose dihedral angles are all geodetic.)
2
hpid = 1 6 dbs = 1 sin?1
db2
1. Angles with polyquadratic tangents and the Splitting Theorem.
The addition formula for tangents enables us to show that the tangent of any
sum ofp purepgeodetic
p of the
p angles is a \polyquadratic number", that is a number
form ap+ b + c + , with a; b; c rational. For instance, if tan = 2=2 and
tan = 3=3, then
p
p
p
p
2
=
2
+
p p 3=3 = 54 2 + 35 3;
tan( + ) =
1 ? 2 3=6
8)
p 3p
4
tan( ? ) = 5 2 ? 5 3:
We now suppose the sum of a number of pure geodetic angles is an integral
multiple of ; let us say 1 + 2 + + 1 + 2 + = m, where we have
2
p
p
chosen the notation so thatpthe tangents
p ofp 1; 2; p are in Qp( d1; p ; dn ) and
those of 1 ; 2; are in Q ( d1; ; dn ) d, where d 2= Q ( d1; ; dn ). Then
tan(1 + 2 + p
+ 1 + 2 +p ) and tan(1 +p2 + p? 1 ? 2 ? ) will be of
d and a ? b d where a; b 2 Q ( d1; ; dn ). But by assumption
the form
a
+
b
p
a + b d = 0, so a = b = 0, from which it follows that 1 + 2 + ? 1 ? 2 ? is also an integral multiple of . Adding and subtracting we deduce that the two
subsums 1 + 2 + and 1 + 2 + are integral multiples of =2. Repeating
this argument we obtain
Theorem 3 (The Splitting Theorem). If the value of a rational linear combination of pure geodetic angles is a rational multiple of then so is the value of its
restriction to those angles whose tangents are rational multiples of any given square
root.
Sketch of proof. We exhibit the argument in the instance that
X
X
X
X
Aii + Bj j + Ck k + D` ` = E;
9)
i
j
k
`
p p p
where the tangents of the 's, 's, 's and 's are rational multiples of 1; 2; 3; 6
respectively and the A's, B 's, C 's, D's and E are rational numbers. By clearing
denominators, we may suppose that the A's, B 's, C 's, DP
's and E are
P all integers.
Applying
our P
argument with d = 3 we see rst that 2( i Ai i + j Bj j ) and
P
2( k Ck k +P ` D` ` ) are
multiples of . P
Applying it again with d = 2
P integral P
shows that 4 i Ai i, 4 j Bj j , 4 k Ck k and 4 ` D` ` are integral multiples
of as well.
We remark that the same method can be used to show that any angle whose
tangent is polyquadratic is a mixed geodetic angle. If is an angle whose tangent
is polyquadratic, we express 2 as the sum of two angles each of whose tangents are
simpler polyquadratic numbers; repeating the process we obtain pas a p
niteplinear
6+
3+ 2+1.
combination of pure geodetic angles. For
example,
suppose
tan
=
p p p
0
0
Let be an angle
p with
p tan = ? 6 + 3 ?p 2 + 1.p tan() and tan(00) are
conjugates in Q ( 2; 3) under the operation 2 ! ? 2. Let = + and
= ? 0 . By the addition formula for tangents,
p
) + tan(0) = 2 + 2p3 = ?2 + 6 p3;
tan( ) = 1tan(
? tan() tan(0 ) 5 +p2 3 p13 13
10)
0)
p p
2(1
+
3)
?
2
tan(
)
?
tan(
2
p
tan() = 1 + tan() tan(0 ) =
= 3 6 ? 2 2:
3+2 3
Thus 2 = + is the sum of two angles, each of whose tangents is the sum of
two square roots. By a similar reduction 2 and 2 are each the sum of two angles
whose tangents are a single square root, i.e. pure geodetic angles, so 4 is the sum
432
96
2592
of four pure geodetic angles. Specically, 4 = 6 (1 + 441
457 ) + 6 457 + 6 457 + 6 4113 .
3
2. Strmer theory and its generalization.
The Splitting Theorem reducespthe study of the rational linear relationships
between angles of the form tan?1 ( ab d) topthose with a xed d. These angles are
the arguments of algebraic integers a + b ?d and their theory is essentially
p the
factorization theory of numbers in Od , the ring of algebraic integers of Q ( ?d).
The method was rst used by C. Strmer [St] (in the case d = 1) who found all
the linear relations between the arctangents of rational numbers using the unique
factorization of Gaussian integers. (See also [Con].) We recall Strmer's analysis of
the case d = 1 and then generalize it to arbitrary d, which will prove Theorems 1
and 2.
It is known that the Gaussian integers have unique factorization up to multiplication by the 4 units: 1; ?1; i; ?i. It is also known how each rational prime p
factorizes in the Gaussian integers. Namely: 1) if p ?1 (mod 4) then p remains
prime; 2) if p +1 (mod 4), p = a2 + b2 is the product of the distinct Gaussian
primes a + ib and a ? ib (for uniqueness we choose a odd, b even, both positive);
and 3) 2 = ?i(1 + i)2 \ramies", that is to say it is (a unit times) the square of a
Gaussian prime.
Now let = 1 23 be the prime factorization of a Gaussian integer .
Then plainly arg() arg(1)+arg(2)+arg(3)+ (mod 2). So the arguments
of Gaussian primes (together with ) span the subspace of mixed geodetic angles
generated by the pure geodetic angles with rational tangent. However,
1) If p = 4k ? 1; arg(p) = 0 and can be ignored
2) If p = 4k + 1 = a2 + b2, then arg(a + bi) + arg(a ? bi) = 0. We dene
hpi1 = arg(a + bi) = ? arg(a ? bi).
3) arg(1 + i) = =4.
4) The arguments of the units are multiples of =2, and so are multiples of
arg(1 + i).
Thus the argument of any Gaussian integer is an integral linear combination of =4
and the angles hpi1, with p 1 (mod 4).
It is also easy to see that the numbers and the hpi1 are rationally independent.
Otherwise some integral linear combination of them would be an integral multiple
of . But suppose for instance that 2hp1i1 ? 3hp2 i1 + 5hp3i1 = 0. The left hand
side is the argument of 12 2335 which must therefore be a real number and hence
equal to its conjugate 122335. But this contradicts unique factorization of Gaussian
integers.
The analogue of the Strmer theory for the general case is complicated by the
fact that elements of Od may not have unique factorization. However the ideals do.
Instead of assigning arguments to numbers, we simply assign an angle to each ideal
4
I by
the rule
) if I = () is principal;
11)
arg(I) = arg(
1 arg(Is) if I is not principal,
s
where s is the smallest exponent for which Is is a principal ideal, and () denotes
the principal ideal generated by . Recall that for every d the ideal class group
is nite, so such an s exists for every ideal, and s divides the class number of Od .
Since the generator of a principal ideal is dened up to multiplication by a unit,
the argument of an ideal must be understood to be modulo the argument of a unit
divided by the class number of Od .
It is not hard to check that, for general ideals I and J,
arg(IJ) = arg(I) + arg(J);
12)
modulo the ambiguity, by a rational multiple of , in the denition of arg. Thus
the argument of any ideal (and in particular the principal ideal generated by any
algebraic integer) is an integral linear combination of the arguments of the prime
ideals. What remains is to determine the nontrivial arguments of prime ideals. As
in the case d = 1, there will be one such angle for each rational prime p for which
(p) splits as the product of distinct ideals.
We illustrate the procedure by working in O5 , for which the ideal factorizations
of the rst few rational primes are:
p
(2) =(2; 1 + ?5)2
p
p
(3) =(3; 1 + ?5)(3; 1 ? ?5)
p
(5) =( ?5)2
p
p
(7) =(7; 3 + ?5)(7; 3 ? ?5)
(11) =(11)
13)
(13) =(13)
(17) =(17)
(19) =(19)
p
p
(23) =(23; 22 + 3 ?5)(23; 22 ? 3 ?5)
p
p
(29) =(3 + 2 ?5)(3 ? 2 ?5):
Here (x; y) denotes the ideal generated by x and y. The reader will see that (2)
ramies as the square of a non-principal ideal and (5) as the square of a principal
ideal, (3), (7) and (23) split into products of non-principal ideals, (29) splits as the
product of distinct principal ideals, while (11), (13), (17) and (19) remain prime.
As in the Strmer case the principal ideals generated by rational primes that
remain prime have argument zero and can be ignored. We also ignore those that
ramify, since their angles will be rational multiples of . Otherwise we dene hpid to
be the argument of one of the two ideal factors of (p), making it unique by requiring
0 < hpid < =2 if the factors of (p) are principal and 0 < hpid < =4 if not.
5
To illustrate this we determine h3i5. The ideal factors of (3) are non-principal
so we square them:
p
p
p
(3; 1 + ?5)2 = (32; 3(1 + ?5); (1 + ?5)2 )
p
p
= (9; 3 + 3 ?5; ?4 + 2 ?5);
p
p
14)
p
which reduces
to (2 ? ?5). Similarly, (3; 1 ? ?5)2 = (2 + ?5). So h3i5 =
1 arg(2 + p?5) = 1 tan?1 ( 1 p5) = 1 6 5 .
2
2
2
2 9
In the general case (d an arbitrary square-free positive integer) we assign an
angle hpid to every rational prime p for which (p) splits as the product of two distinct
prime ideals I and J. Let s be the smallest integer for which Is (and
p therefore Js) is
b
a
principal. Recall that the elements of Od are of the form 2 + 2 ?d, where a and
b are rational integers. If d 6= 3 (mod 4), then a and b must be even; if d = 3 (mod
4) then a and b are either both even or both odd. We can therefore write
a p a bp b ?d ;
s
s
?
d
;
J =
15)
+
?
I =
2 2
2 2
where we can distinguish p
between I and
J by supposing that a and b are positive.
2d
1
1
b
b
?
1
6
We take hpid = s tan ( a d) = s a2 +b2 d .
The above denes hpid uniquely for all d other than 1 and 3, because then the
only units
are 1, so that the only generators of Is and Js are the four numbers
p
a2 2b d. When d = 1 we have the additional unit i which eectively allows us to
interchange a and b: we then achieve uniqueness by demanding that the generators
of I be a + bi with a and b positive integers with b even. In the case d = 3 the
eld has six units and
p the corresponding condition is that the generators of I should
have the form a + b ?3 where a and b are positive integers.
Theorem 4. For xed square-free d, consider the subspace
of mixed geodetic angles
p
generated by arctangents of rational multiples of d. This subspace is spanned by
the nonzero angles hpid and , where p ranges over the rational primes for which
(p) splits as a product of distinct ideals in Od .
Proof. The proof is essentially that of the Strmer decomposition, only substituting
the arguments of ideals for the arguments of algebraic integers.
Combining Theorem 4 with the Splitting Theorem, we obtain Theorem 1.
Proof of Theorem 2. Recall that, if I is any ideal (principal or not) in Od , then
II is a principal ideal with a positive integer generator that we call the norm of I,
and that norms are multiplicative. From this it follows that the prime ideals are
precisely the factors of (p), with p ranging over the rational primes, and that every
prime ideal that is not generated by
p a rational prime has rational prime norm.
b
Now suppose that tan() = a d with d square-free and with a and
p b relatively
prime. Consider the factorization of the principal ideal I = (a + b ?d). If I =
12 n , where each ideal i is prime, then none of the i 's are generated by
6
rational primes, insofar as a and b are relatively prime. Thus each i satises
i i = (pi ) for some rational prime pi . We then have
(p1 )(p2) (pn ) = (a2 + b2d) = II = 11 n n :
16)
So, up to permutation of the indices 1; : : :; n, wepmust have i i = (pi ), and so
the argument of i is hpi id . But = arg(a + b ?d) = arg(I) is the sum of the
arguments of the i 's, up to a rational multiple of that comes from the ambiguity
in the denition of the argument of an ideal.
All that remains is to identify the pairs (p; q) for which hpid is nonzero. The
following theorems give the criteria. These may be easily checked, by hand for small
d and p, and by computer for larger values. The theorems themselves are standard
results, and we abbreviate the proofs.
Theorem 5. Let p be an odd rational prime. The ideal (p) of Od splits as a product
of distinct ideals if and only if we can write
4ps = a2 + b2d
17)
for integers a and b (neither a multiple of p) and for an exponent s that divides the
class number of Od . If d 6= 3 (mod 4), or if d = 3, then the factor of 4 is unnecessary,
and the criterion for splitting reduces to
ps = a2 + b2d;
18)
for a and b nonzero (mod p). The ideal (p) is prime in Od if and only if ?d is not
equal to a square modulo p. If (p) is not prime and does not split, then (p) ramies.
Proof. Assume that d is not 1 or 3, so that the only units in Od are 1. If (p) = IJ,
let s be the order of I in the ideal class group. Then (ps ) = Isp
Js, where Is is a
principal ideal. Since the elements of Od are all of the form a2 + 2b ?d, with a and
b rational integers, we can write
p a b p a2 b2 a
b
s
s
s
s
(p ) = (p) = I J = 2 + 2 ?d 2 ? 2 ?d = 4 + 4 d ; 19)
which implies that 4ps = a2 + b2 d. Now, if d 6= 3 (mod 4), then the only way
to make a2 + b2 d divisible by 4 is to have both a and b even, in which case we
cancel the factors of 2pand write pps in the form a2 + b2 d. Conversely, suppose
4ps = a2 + b2d = [a + b ?d][a ? b ?d]. The left hand side is divisible by p, while
each factor in pthe right hand side
p is not, so p cannot be prime. Since the only units
are 1, (a + b ?d) and (a ? b ?d) are distinct ideals, so the factors I and J of (p)
must be distinct, so (p) splits.
2 (mod p), where x is a rational integer, then x2 + d = [x + p?d][x ?
If
?
d
=
x
p
p
?d] is divisible by p. However, neither x ?d is divisible by p, so y(pp) cannot
be prime. Conversely, if (p) is not prime, then there are elements x2 + 2 ?d and
7
p
b
by p, whose product is divisible by p. But then
2 + 2 ?d of2Od , 2neither divisible
the norms [x + y d]=4 and [a2 + b2 d]=4 multiply to a multiple of p2, so at least
one of these norms is divisible by p. Suppose [x2 + y2d]=4 is divisible by p, so
x2 = ?y2 d p(mod p). If y = 0 (mod p), then x = 0 (mod p), contradicting the fact
that x2 + y2 ?d is not in (p). Since y 6= 0 (mod p), we can nd an integer z such
that yz = 1 (mod p), and then ?d = (xz)2 (mod p).
The cases d = 1 and d = 3 are complicated slightly by the existence of units
other than 1, but the analysis is essentially the same.
Theorem 5 gives criteria for all odd primes. The prime p = 2 is somewhat
dierent. Since both 0 and 1 are squares, every ?d is congruent to a square modulo
2. However, there are values of d for which (2) is prime.
Theorem 6. If d 6= 3 (mod 4), then (2) ramies in Od . If d = 3 (mod 8), then (2)
is prime. If d = 7 (mod 8), then (2) splits and we can write a power of 2 as a2 +pbd2 .
Proof. If d 6= 3 (mod 4), then the algebraicpintegers all have thepform a +pb ?d,
with a and b rational integers. Either ?d = ( ?d)2 or 1+ d = [1+ ?d][1 ? ?d] is
divisible by 2, but none of the factors are divisible by 2, so 2 is not prime. However,
(2) does not split, for if 4 2s = a2 + db2, with s minimal, then the left hand side
is divisible by 4, while the right hand side is only divisible by 4 if both a and b are
even, in which case s is not minimal. Thus (2) ramies. (This argument breaks
down if d = 1, but then (2) = ([1 + i]=2)2, so the conclusion still holds).
If d = 3 (mod 8), then the algebraic integers that are not divisible by 2 are
precisely those with odd norm. Since the product of any two such numbers is
another such number, (2) is prime.
p
p
If d = 7 (mod 8), then [d + 1]=4 = [ 21 + 21 ?d][ 12 ? 21 ?d] is divisible by 2,
while neither factor on the right hand side is, so (2) is not prime. The fact that (2)
splits, as opposed to ramifying, is more dicult, and is not proven here.
a
3. Applications to the Dehn-Sydler criterion of Archimedean polyhedra.
The Dehn invariant of a polyhedron
whose ith edge has length `i and dihedral
P
angle i is the formal expression i `iV [i] where the \vectors" V [i ] are subject
to the relations
V [r + s] = rV [] + sV []; V [r] = 0;
20)
for all rational numbers r and s. The V [] satisfy the same rational linear relations
satised by the angles in the rational vector space they generate, together with
the additional relation V [] = 0; however we allow their coecients to be arbitrary
real numbers.
If every dihedral angle of a polyhedron is geodetic we can write
= r + r1 hp1 id1 + + rj hpj idj
21)
for rational numbers r; r1; ; rj , so
V [] = r1 V [hp1id1 ] + + rj V [hpj idj ]:
22)
8
Its Dehn invariant will then be a rational linear combination of the V [hpid ].
Now each face of an Archimedean polyhedron (other than the snub cube and
snub dodecahedron) is orthogonal to a rotation axis of one of the Platonic solids,
and the rotation groups of all the Platonic solids are contained in the icosahedral
group I . It follows that the dihedral angles of all the polyhedra are found among
the angles between thep rotation axes of I .
Now let = (1+ 5)=2 and = ?1 = ? 1. The 12 vectors whose coordinates
are cyclic permutations of 0; 1; lie along the pentad axes. Similarly the 20
vectors obtained by cyclicly permutating 1; 1; 1 and 0; ; lie along the
triad axes, and the 30 cyclic permutations of 2; 0; 0 and 1; ; lie along the
dyad ones. The cosines of the angles between the axes have the form v w=jvj jwj
where v and w are chosen from these vectors. These cosines are enumerated in
Fig. 1. The angles that correspond to them are those shown in Fig. 2, together with
their supplements.
Table 3 gives the components of the Dehn invariants for the non-snub
Archimedean polyhedra of edge lengths 1. For instance the dihedral angles of the
truncated tetrahedron are ? 2h3i2 at six edges and 2h3i2 at the remaining 12, so
that its Dehn invariant is
6V [ ? 2h3i2] + 12V [2h3i2] = 12V [h3i2];
23)
since V [] = 0. In the values we abbreviate V [hpid ] to hpid .
We note that the Dehn invariants of the icosahedron, dodecahedron and icosidodecahedron with unit edge lengths, namely 60h3i5; ?30h5i1, and 30h5i1 ? 60h3i5,
respectively, have zero sum, so Sydler's theorem shows that it is possible to dissect
them into nitely many pieces that can be reassembled to form a large cube. This
might make an intriguing wooden puzzle if an explicit dissection could be found.
(We have no idea how to do this.)
4. Angles with algebraic trigonometric functions.
It is natural to consider a generalization of our theory that gives a basis for the
rational vector space generated by all the angles whose six trigonometric functions
are algebraic. What is missing here is the analogue of our Splitting Theorem. If
such an analogue were found, the ideal theory would probably go through quite
easily.
We ask a precise question: Does there exist an algorithm that nds all the
rational linear relations between a nite number of such angles? The nicest answer
would be one giving an explicit basis, analogous to our hpid .
9
References
[Con] J.H. Conway, R.K. Guy, The book of numbers, Copernicus, New York, 1996.
[Deh] M. Dehn, Uber den Rauminhalt, Math. Ann., 55 (1901), 465-478.
[Syd] J.P. Sydler, Conditions necessaires et susantes pour l'equivalence des polyedres
l'espace euclidien a trois dimensions, Comm. Math. Helv., 40 (1965), 43-80.
[St] C. Strmer: Sur l'application de la theorie des nombres entiers complexes a la
solution en nombres rationnels x1; x2 : : : xn c1 c2 : : :cn ; k de l'equation c1 arc tg x1 +
c2 arc tg x2 + cn arc tg xn = k 4 , Arch. Math. Naturvid., 19 No. 3 (1896).
10
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4
1
2
0
19
68
6
189
289
88
169
0
828
2197
0
6
6
1
2
1
4
6
1
2
336
361
6
6
6
6
1
2
352
361
6
0
6
1
2
0
0
0
0
0
27
31
12
37
0
11
92
0
0
240
529
0
187425
279841
0
19
23
0
525
529
0
1
3
6
6
1
2
6
792
841
4508 1
24389 3
6
6
6
6
0
5
41
6
1
2
0
18
43
0
27
43
0
405 1
1849 2
0
0
0
28
37
0
7
43
640
1369
40
41
0
99
148
0
0
0
0
0
0
0
0
0
15
31
0
0
0
0
0
0
0
171
172
6
1
2
6
6
0
1
2
32
41
6
0
0
907137
923521
6
16
41
6
47
6
31
637
961
6
6
1
4
36
37
6
43
0
99
124
6
41
0
0
13
29
6
14
23
6
28
29
6
1
2
6
216
841
6
37
4
29
360
529
22
23
31
20
29
6
0
1
2
6
6
15
19
6
6
45
529
6
1
2
24696
130321
6
6
1
2
325
361
6
29
0
0
13
17
6
4536
28561
6
6
0
3825 1
14641 2
6
6
0
1
2
19
28
6
8
17
23
12
13
1377 1
2401 4
6
4
13
6
19
3
7
0
1
2
17
0
40
49
6
13
2
11
6
1
2
17
81
6
11
6
7
6
504
625
6
1
4
6
11
20
6
0
1
4
1
2
0
11
12
6
6
24
25
6
0
0
6
1
2
0
0
1
3
7
0
0
5
9
6
19
23
0
15
16
6
4
5
6
2
3
6
7
8
6
5
336
961
22
31
207
29791
6
6
6
0
21
37
1
2
0
0
6
6
6
6
525
1681
0
1
3
6
23552
68921
2205
2209
6
0
0
1
2
360
2209
6
11
47
6
1
2
1053
2209
6
0
1
2
2160
2209
6
0
19
188
6
0
1
2
6
0
1408
1849
0
6
1
3
6
22
47
103247
103823
Table 1. Basis elements hpid for some p and d. 0 indicates that (p) is prime in Od ,
while * indicates that (p) ramies.
11
h5i1 = 6 45 63260600
h13i1 = 6 134 334102400
0 00
h17i1 = 6 16
17 75 57 50
h29i1 = 6 294 21480500
0 00
h37i1 = 6 36
37 80 32 16
16
h41i1 = 6 41 383903500
h3i2 = 6 23 54440800
h11i2 = 6 112 251402200
h17i2 = 6 178 431805000
0 00
h19i2 = 6 18
19 76 44 14
30 4200
62
h41i2 = 6 32
41
18
6
h43i2 = 43 401805500
h7i3 = 6 37 405303600
0 00
h13i3 = 6 12
13 73 53 52
3 23 240 4800
6
h19i3 = 19
0 00
h31i3 = 6 27
31 68 56 54
420 5400
h37i3 = 6 12
34
37
27
6
h43i3 = 43 522403900
h3i5 = 12 6 59 2450 4100
0 00
h7i5 = 12 6 45
49 36 41 57
1 6 45
h23i5 = 2 529 8 2804300
0 00
h29i5 = 6 20
29 56 8 44
h41i5 = 6 415 202602200
405
135701000
h43i5 = 12 6 1849
0 00
h47i5 = 12 6 2205
2209 43 46 50
0 00
h5i6 = 21 6 24
25 39 13 53
6
6
h7i6 = 7 67 4703200
96 31 280 5600
h11i6 = 12 6 121
1
216
h29i6 = 2 6 841 15 1303100
h31i6 = 6 316 2660000
h2i7 = 6 78 691704300
h11i7 = 6 117 525404800
h23i7 = 6 237 332805600
0 00
h29i7 = 6 28
29 79 17 54
28
h37i7 = 6 37 602605700
h43i7 = 6 437 234704400
0 00
h7i10 = 12 6 40
49 32 18 42
270 600
h11i10 = 6 10
72
11
0 00
h13i10 = 21 6 160
169 38 19 44
300 3100
h19i10 = 6 10
46
19
0 00
h23i10 = 21 6 360
529 27 47 29
1 6 640
h37i10 = 2 1369 21 340500
0 00
h41i10 = 6 40
41 81 0 54
360
h47i10 = 21 6 2209
11 5401700
0 00
h3i11 = 6 11
12 73 13 17
520 1100
h5i11 = 6 11
47
20
130 4600
h23i11 = 6 11
20
92
99
6
h31i11 = 124 631901100
99
h37i11 = 6 148
545202100
h47i11 = 6 11
28 550 5700
47
h
h
h
h
h
h
h
i
7 13 = 12 6 13
15 300 500
49
11 13 = 21 6 117
39 450 4400
121
580 5800
17 13 = 6 13
60
17
19 13 = 21 6 325
35 470 4500
361
10 5200
29 13 = 6 13
42
29
31 13 = 21 6 637
27 150 700
961
1053
1
47 13 = 2 6 2209 21 490 5500
i
i
i
i
i
i
0 00
h3i14 = 14 6 56
81 14 3 46
580 2700
h5i14 = 14 6 504
15
625
4536
1
6
h13i14 = 4 28561 55201700
24696
h19i14 = 14 6 130321
6 270500
160 4100
h23i14 = 6 14
51
23
0 00
h2i15 = 12 6 15
16 37 45 40
500 3200
h17i15 = 12 6 240
32
289
410 1800
h19i15 = 6 15
62
19
0 00
h23i15 = 21 6 240
529 21 10 17
40 3300
44
h31i15 = 6 15
31
0 00
h47i15 = 21 6 2160
2209 40 43 2
0 00
h3i17 = 41 6 17
81 6 48 59
h7i17 = 41 6 1377
12 180 2400
2401
1 6 3825
h11i17 = 4 14641 7410500
0 00
h13i17 = 12 6 153
169 36 2 24
h23i17 = 14 6 187425
13 430 5100
279841
1 6 907137
h31i17 = 4 923521 203501100
0 00
h5i19 = 6 19
20 77 4 45
270 4500
h7i19 = 6 19
55
28
19
6
h11i19 = 44 4140 5300
0 00
h17i19 = 6 19
68 31 54 38
210 1000
h23i19 = 6 19
65
23
171
6
h43i19 = 172 853703700
19
h47i19 = 6 188
183201100
0 00
h5i21 = 21 6 21
25 33 12 39
1 6 21
h11i21 = 2 121 121803600
0 00
h17i21 = 12 6 189
289 26 59 3
220 1600
h19i21 = 12 6 336
37
361
525
1
6
h23i21 = 2 529 423002100
0 00
h31i21 = 12 6 336
961 18 7 29
530 000
h37i21 = 6 21
48
37
525
h41i21 = 12 6 1681
16 5901700
88
h13i22 = 12 6 169
23503600
0 00
h19i22 = 12 6 352
361 40 27 28
22
6
h23i22 = 23 77 5705300
0 00
h29i22 = 12 6 792
841 38 0 58
230 4900
h31i22 = 6 22
57
31
0 00
h43i22 = 12 6 1408
1849 30 22 59
22
6
h47i22 = 47 43 1001100
0 00
h2i23 = 31 6 23
32 19 19 27
1
23
6
h3i23 = 3 27 222701500
828
h13i23 = 13 6 2197
12 3702800
4508 8 290 1600
h29i23 = 13 6 24389
207 1 350 3800
h31i23 = 13 6 29791
0 00
h41i23 = 13 6 23552
68921 11 55 27
340 3400
h47i23 = 13 6 103247
28
103823
Table 2. The values of some basis elements hpid .
12
Tetrahedron
Truncated tetrahedron
6 ? 12h3i2
6 + 12h3i2
Cube
Truncated cube
6
30 ? 24h3i2
Octahedron
Truncated octahedron
Rhombicuboctahedron
24h3i2
24
9 ? 24h3i2
Cuboctahedron
Truncated cuboctahedron
24 ? 24h3i2
54
Icosahedron
Truncated icosahedron
15 + 60h3i5
60 + 30h5i1
Dodecahedron
Truncated dodecahedron
Rhombicosidodecahedron
?30h5i1
75 ? 60h3i5
105 + 60h3i5 ? 30h5i1
Icosidodecahedron
Truncated icosidodecahedron
45 ? 60h3i5 + 30h5i1
150
Table 3. The Dehn invariant for the non-snub unit edge Archimedean polyhedra.
13
1; 2 ; 12 ; 2
p
0; 35 ; 13
11
1 1 1
11
11
dyad 1 1 1 1
11
11
q q
11
p3 ; 1; p3 ; 0 p5 ; p5 ; 0
11
11
11
11
11
11
11
11
11
triad
pentad 1 1 1 1
1
q 3 q 3
3p5 ; 3p5
p
1; 55
Figure 1. Cosines of angles between axes of xed rotational symmetry (shown at
corners), and between axes of dierent rotational symmetry (shown at edges).
14
0; 5 ; 3 ; 25 ; 2
11
1 1 1
1
dyad 1 1 1
11
11
11
11
; + h3i ; h3i ; ? h3i
1 1 21 h5i1 ; 2 ? 12 h5i1 ; 2
5
2
5
2 4
4
11
11
11
11
11
11
11
11
11
pentad 1 1 1
0; 2 ? 2h3i5; triad
11
? 2h3i2
3 ? h3i5 ? 1 h5i1 ; 1 + h3i5 ? 1 h5i1
4
2
4
2
h5i1; 0
Figure 2. Angles between axes of xed rotational symmetry (shown at corners),
and between axes of dierent rotational symmetry (shown at edges).
15