toán tiếng anh cho trẻ em Maths gazette

The
Mathematical
Gazette

Sir Wifred H. Cockcroft
1923-1999

Volume 84:

Number 499

THE MATHEMATICAL ASSOCIATION

March 2000
£16.00


CONTENTS (continued)

THE MATHEMATICAL ASSOCIATION

Notes 84.01 to 84.28 (continued)
AN ASSOCIATION OF TEACHERS AND STUDENTS
OF ELEMENTARY MATHEMATICS

Unexpected symmetry in a derived
Fibonacci sequence

Alexander J. Gray

87

A recurrence relation among Fibonacci sums Alexander J. Gray

89

Some unusual iterations

Mark Thornber

90

When the sum equals the product

Leo Kurlandchik and
Andrzej Nowicki

91

Never say never: some mistaken identities

Mark J. Cooker

94

A curious property of the integer 24

M. H. Eggar

96

What do cycles of a given length generate?

Mowaffaq Hajja

97

A game with positive and negative numbers

M. H. Eggar

98

An inductive proof of the arithmetic mean
− geometric mean inequality

Zbigniew Urmanin

101

Weighted mean in a trapezium

Larry Hoehn

102

(Chair)

A formula for integrating inverse functions

S. Schnell and C. Mendoza 103

(Chair)

Mathematician versus machine

P. Glaister

105

On a conjecture of Paul Thompson

Tim Jameson

107

(Chair)

Maximal volume of curved folding boxes

Kenzi Odani

110

(Chair)

More on a sine product formula

Walther Janous and
Jeremy King

113

On a limit for prime numbers

J. A. Scott

115

SHM and projections

P. Glaister

116

Another cautionary chi-square calculation

Nick Lord

119

More on dual Van Aubel generalisations

Michael de Villiers

121

Obituary
Sir Wilfred Cockcroft 1923-1999

Peter Reynolds

123

'I hold every man a debtor to his profession, from the which as
men of course do seek to receive countenance and profit, so ought
they of duty to endeavour themselves by way of amends to be a
help and an ornament thereunto.'
BACON

THE COUNCIL
PRESIDENT

Professor John Berry
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CHAIR OF COUNCIL
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OFFICE MANAGER

Professor Chris Robson
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(Chair)
(Chair)

Ms Trish Morgan
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EDITORIAL COMMITTEE OF THE MATHEMATICAL GAZETTE
Editor Mr Steve Abbott
Production Editor Mr Bill Richardson
Reviews Editors Mr Bud Winteridge
Mrs Rosalie McCrossan
Problems Editors Mr Graham Hoare Mr Tim Cross
Assistant Editor Mr Gerry Leversha

Correspondence

125

Notices

127

Problem corner

G. T. Q. Hoare

128

Student problems

Tim Cross

135

Other Journals

Anne C. Baker

139

Book Reviews

140
© The Mathematical Association 2000


CONTENTS
Editorial

The
Mathematical
Gazette

1

One hundred years on

Graham T. Q. Hoare

2

Lewis Carroll − mathematician
and teacher of children

Canon D. B. Eperson

9

Snubbing with and without eta

H. Martyn Cundy

14

The Fermat-Torricelli points of n lines

Roy Barbara

24

Continued fractions

Robert Macmillan

30

A construction of magic cubes

Marián Trenkler

36

The factorial function: Stirling's formula

David Fowler

42

A simple energy-conserving model

Richard Bridges

51

The Hale-Bopp comet explored
with A level mathematics

H. R. Corbishley

58

THE MATHEMATICAL GAZETTE

Articles

Notes 84.01 to 84.28

Circumradius of a cyclic quadrilateral

Larry Hoehn

69

A neglected Pythagorean-like formula

Larry Hoehn

71

An unexpected reduced cubic equation

J. A. Scott

74

Touching hyperspheres

D. F. Lawden

75

Comments on note 82.53—a generalised
test for divisibility

Andrejs Dunkels

79

A matrix method for a system of
linear Diophantine equations

A. J. B. Ward

81

On the application of Whittaker's theorem

J. A. Scott

84

Digital roots and reciprocals of primes

Alexander J. Gray

86

(The contents are continued inside the back cover.)
Printed in Great Britain by J. W. Arrowsmith Ltd
ISSN 0025-5572

Vol. 84 No. 499

66

MARCH 2000

A portrayal of right-angled triangles which
I. Grattan-Guinness
generate rectangles with sides in integral ratio

Sir Wifred H. Cockcroft
1923-1999

Volume 84:

Number 499

THE MATHEMATICAL ASSOCIATION

March 2000
£16.00


1

The

Mathematical Gazette
A JOURNAL OF THE MATHEMATICAL ASSOCIATION
Vol. 84

March 2000

No. 499

Editorial: It's voting time again!
The time has come to vote for the Fifth Annual Mathematical Gazette
Writing Awards. Please use the address carrier from this issue of the Gazette
to identify the articles and notes of 1999 that impressed you most. The Index
for 1999 will remind you of the many good submissions.
There will again be a prize draw among those who respond. The
prize, worth about £30, will be a copy of the book Mathematics: frontiers

and perspectives, edited by Vladimir Arnold, Michael Atiyah, Peter Lax and
Barry Mazur (AMS, 2000).
Previous Annual Mathematical Gazette Writing Awards
Year

Best Article

Best Note

1996

Colin Fletcher
Two prime centenaries

David Fowler
A simple approach to
the factorial function

Ann Hirst and Keith Lloyd

Colin Dixon

Cassini, his ovals and a
space probe to Saturn

Geometry and the
cosine rule

Robert M. Young

Robert J. Clarke

Probability, pi, and
the primes

The quadratic equation
formula

1997

1998

Please indicate, in the spaces provided on the voting form, the titles of
your 3 favourite Articles and your 3 favourite Notes. Note that Matters for
Debate count as Articles. Alternatively, you can just write your choices in a
letter or on a postcard. Each vote will be given equal weighting. The results
will be announced in the July 2000 issue.
Return the form as soon as you can, and definitely by 31st May 2000 to:

Gazette Poll, 91 High Road West, Felixstowe IP11 9AB, UK
STEVE ABBOTT


2

THE MATHEMATICAL GAZETTE

One hundred years on
GRAHAM T. Q. HOARE
David Hilbert, one of the giants of mathematics, delivered a lecture at

the International Congress of Mathematics at Paris in 1900. The first part of
the lecture, a preamble to his announcement of the now-famous 23
problems, began with the words:
‘Who of us would not be glad to lift the veil behind which the future
lies hidden; to cast a glance at the next advances of our science and
at the secrets of its development during future centuries? What
particular goals will there be toward which the leading mathematical
spirits of coming generations will thrive? What new methods and
new facts in the wide and rich fields of mathematical thought will the
new centuries disclose?
History teaches the continuity of the development of science. We
know that every age has its own problems, which the following age
either solves or casts aside as profitless and replaces by new ones. If
we would obtain an idea of the probable development of
mathematical knowledge in the immediate future, we must let the
unsettled questions pass before our minds and look over the
problems which the science of today sets and whose solution we
expect from the future. To such a review of problems the present
day, lying at the meeting of the centuries, seems to me well adapted.
For the close of a great epoch not only invites us to look back into
the past but also directs our thoughts to the unknown future.
The deep significance of certain problems for the advance of
mathematical science in general and the important role which they
play in the work of the individual investigator are not to be denied.
As long as a branch of science offers an abundance of problems, so
long is it alive; a lack of problems foreshadows extinction or the
cessation of independent development. Just as every human
undertaking pursues certain objects, so also mathematical research
requires its problems. It is by the solution of problems that the
investigator tests the temper of his steel; he finds new methods and

new outlooks, and gains a wider and freer horizon.’
Later we find the oft-quoted passage:
‘This conviction of the solvability of every mathematical problem is a
powerful incentive to the worker. We hear within us the perpetual
call: There is the problem. Seek its solution. You can find it by
pure reason, for in mathematics there is no ignorabimus.’
Hilbert considered that the 23 problems he had chosen were those most
likely to stimulate important new advances in mathematics. It redounds to
his perspicacity that much fruitful mathematical activity resulted in
addressing these problems in the twentieth century. As we shall see from
the list below, which we give together with short commentaries and notes,


ONE HUNDRED YEARS ON

3

the so-called problems vary from specific mathematical questions to
programmes of research. Some have been reformulated or extended without
losing their identity. We note the importance Hilbert attached to algebraic
number theory, since the 8th problem, partly, and the 9th, 11th and 12th,
entirely, are devoted to it. Problems 1, 2 and 10 belong to mathematical
logic, whereas 6, 19, 20 and 23 fall within the provinces of applications.
Observe too that topology, then at an early stage of its development, features
strongly. Readers will appreciate that we cannot do justice to Hilbert's
vision in a short article such as this.
Hilbert's Problems
1. Cantor's continuum hypothesis (CH) and well-ordering
1(a). Is 2¼0 = 1? Undecidable. Assuming the consistency of the
Zermelo-Fraenkel axioms for set theory (ZF), the work of K. Gödel (1938)

and P. Cohen (1963) established that both the statement of the hypothesis
(that 2¼0 = 1) and its negation are consistent with ZF. Thus the
hypothesis is completely independent of the axioms of set theory.
1(b). Hilbert also asked whether the continuum of numbers can be wellordered. This problem is related to the Axiom of Choice (AC), but in 1963
P. Cohen proved the independence of AC from the other axioms of set
theory, so the problem remains unresolved.
Note: Gödel believed that the AC and the CH were either true or false
and that ZF did not encapsulate what was ‘obviously’ true about set theory.
The task was to think of some new axiom which would determine AC and
CH. He did not succeed in devising such an axiom (the existence of
measurable cardinals was proposed as such, but was not in any sense
‘obvious’) so this remains an unresolved consequence of the Hilbert
challenge.

¼

¼

2. To establish the consistency of the axioms of arithmetic
Gödel's two theorems shattered the Hilbert programme. The second of
these proves that the consistency of a theory at least as strong as arithmetic
cannot be proved within the theory.
3. To show, using only the congruence axioms, whether two tetrahedra
having the same altitude and base area have the same volume
Proved false by M. Dehn (1900).
4. To investigate geometries (metrics) in which the line segment between
any pair of points gives the shortest path between the pair (geodesic)
Considered too vague.
5. Can the assumption of differentiability for the functions defining a
continuous transformation group be avoided?

Reformulated to encompass a larger domain of topological groups, the
problem was solved in the form that a locally Euclidean topological group is


4

THE MATHEMATICAL GAZETTE

a Lie group by A. Gleason (1952) and by D. Montgomery and L. Zippin
(1955).
Note: If each point of a topological group G has a neighbourhood
homeomorphic to an open set of a given Euclidean space, then G is called a
locally Euclidean group. If the underlying topological space of a
topological group has the structure of a real analytic manifold, where the
group operations (x, y) → xy, x → x−1 are real analytic mappings, then G
is a Lie group.
S. Lie envisaged an approach to solving partial differential equations
analogous to Galois' group-theoretic resolution of algebraic equations.
6. The mathematical axiomatisation of physics
Hilbert considered that physics was too difficult to be left to physicists.
Progress has been minimal, not least because the meaning of 6 is unclear.
Again, Hilbert could not have foreseen the many developments in 20th
century physics. We can record, however, that the axiomatisation of
probability theory was accomplished by A. Kolmogorov and that of
quantum physics by A. Wightman.
7. To establish the transcendence of certain numbers
The following generalisation of Lindemann's theorem was conjectured
by A. O. Gelfond (1929) and proved by A. Baker (1966).
If α1, α2, … , αr, β1, β2, … βr are non-zero algebraic numbers such
that ln α1, ln α2, … , ln αr are linearly independent over the rationals then

β1 ln α1 + β2 ln α2 + … +βr ln αr ≠ 0.
A special case of this, found independently by Gelfond and T.
Schneider (1934), which answers Hilbert's enquiry about the nature of 2 2,
states that if α is an algebraic number ≠ 0, 1 and β is an irrational number,
then αβ is a transcendental number.
8. To investigate problems concerning the distribution of prime numbers; in
particular, to show the correctness of the Riemann hypothesis
Tantalisingly, the Riemann hypothesis evades resolution.
∞
Note: The Riemann zeta function is defined by ζ (s) = ∑1 n−s for
s = σ + iτ ∈ c and σ > 1. This converges when σ > 1, and can be
continued to all of c by a formula giving ζ (1 − s) in terms of ζ (s). The
Riemann hypothesis states that the non-trivial roots of the Riemann zeta
function all lie on the line σ = 12 . Riemann had already noted that, if
ζ (s) = 0, then 0 ≤ Re (s) ≤ 1. He believed, for example, that a proof of
the hypothesis might establish the existence of an infinity of twin primes.
9. To find the most general law of reciprocity in an algebraic number field
Hilbert contributed to this, but it was E. Artin (1927) who established it
for Abelian extensions of q; the non-Abelian case is still open.


ONE HUNDRED YEARS ON

5

Note: The quadratic reciprocity law state that if p, q are different odd
primes then
p q
.
= (−1)(p − 1)(q − 1)/4 ,

q p
a
where
, Legendre's symbol, is defined for any integer a and any odd
p
prime p as

()()

()

 1 if x2 ≡ a (mod p) is solvable for x

a
=  −1 if x2 ≡ a (mod p) is not solvable for x
p
 0 if a ≡ 0 (mod p) .

Gauss was the first to solve the quadratic and cubic reciprocity laws.

()

10. To find an algorithm for deciding whether any given Diophantine
equation has a solution
Following pioneering work by M. Davis, H. Putnam and J. Robinson,
the problem was finally solved, negatively, by Y.
(1970).

Matijaseviè

11. To investigate the theory of quadratic forms over an arbitrary algebraic
number field of finite degree
H. Hasse (1929) and C. L. Siegel (1936, 1951) obtained important
results. A. Weil and T. Ono (1964-1965) demonstrated a connexion
between the problem and algebraic groups. Generally, still incomplete.
12. Extension of Kronecker's theorem on Abelian fields to an arbitrary
algebraic field
Poorly posed by Hilbert, the problem was corrected and solved by T.
Takagi. In 1922 he proved the following fundamental theorem: every
Abelian extension of an algebraic number field F is a class field for the field
(corresponding to a congruence class group in F) and, conversely, every
class field E of F is an Abelian extension of F.
Note: Given a group G of automorphisms of a given field L, and K a
subfield of L, the group consisting of all automorphisms of L leaving every
element of K invariant is denoted by G (L / K). A Galois extension is called
an Abelian extension when G (L / K) is Abelian. Kronecker's theorem states
that cyclotomic fields are Abelian extensions of q and, conversely, every
Abelian extension of q is a subfield of a cyclotomic field.
The problem is related to finding functions which, for an arbitrary field,
play the same role as the exponential function for the rational field and
elliptic modular functions for imaginary quadratic fields.
13. To show the impossibility of the solution of the general algebraic
equation of the 7th degree by compositions of continuous functions of two
variables


6

THE MATHEMATICAL GAZETTE

Solved by V. I Arnol'd (1957) for continuous functions; still unsolved if
analyticity is required.
Note: We mention, in passing, the beautiful results of Kolmogorov and
Arnol'd that arbitrary real-valued continuous functions of any number of
variables can be represented exactly as compositions of a finite number of
such functions of only two variables.
14. To consider invariants which arise when only the transformations of a
subgroup of the totality of linear transformations, the projective linear
group, are permitted
By producing a counter-example, M. Nagata (1958) showed that the
invariants need not be finitely generated.
Note: An invariant is a mathematical object which remains unchanged
under certain kinds of transformation.
Recently there has been renewed activity in invariant theory; it has
widened its scope and has entered the realm of abstract algebra. Indeed
Problem 14, in algebraic language, can be rendered as:
Given fields k , k (x1, … , xn), and K , where k ⊆ K ⊆ k (x1, … , xn),
the problem is to determine whether the ring K ∩ k [ x1, … , xn] , is finitely
generated over K . Here k (x1, … , xn) is the field of rational functions in
(x1, … , xn) with coefficients in k, and k [ x1, … , xn] is the ring of
polynomials with coefficients in k .
15. To establish the foundations of algebraic geometry, in particular,
H. Schubert's enumerative calculus
Solved by B. L. van der Waerden (1938-1940), A. Weil (1950) and
others. In the late 1950s and 1960s, A. Grothendieck rewrote the
foundations of algebraic geometry after Weil.
Note: Algebraic geometry is the study of algebraic curves, algebraic
varieties and their generalisations to n dimensions. Suppose V is an ndimensional vector space with scalars in some field F. If W is a subset of V
composed of all points (x1, … , xn) which satisfy each of a set of polynomial
equations {pi (x1, … , xn) = 0}, i ∈ z+, with coefficients in F, then W is

an algebraic variety.
Originally, enumerative calculus was developed for counting the
number of curves touching a given set of curves, and enumerative geometry
refers to Schubert's application of the conservation of number principle [1].
16. To study the topology of real algebraic curves and surfaces
Sporadic results.
17. Suppose f (x1, … , xn) is a rational function with real coefficients that
takes a positive value for any n-tuple (x1, … , xn). The problem is to
determine whether the function f can be written as a sum of squares of
rational functions


ONE HUNDRED YEARS ON

7

Solved, affirmatively, by Artin (1926-1927) for real closed fields. In
1967 DuBois gave a negative solution to the general case. In the same year
Pfister gave the number of squares required.
18. To investigate the existence of non-regular space-filling polyhedra
K. Reinhardt (1928), a student of Hilbert, showed that such a ‘tiling’
exists. In 1910, L. Bieberbach proved that, up to equivalence, there are only
finitely many n-dimensional crystallographic groups.
19. To determine whether the solution of regular problems in the calculus
of variations are necessarily analytic
Solved by S. Bernstein, I. G. Petrovskii, and others.
20. To investigate the existence of solutions of partial differential equations
with prescribed boundary conditions
Hilbert contributed here by resurrecting Dirichlet's problem; a vast
amount of work has been done in this area pre- and post-Hilbert.

Note: This ‘problem’ is closely linked to the 19th.
∂ 2u
2
A typical boundary problem takes the form
− ∇ u = f in some
2
∂t
∂u
region R, with u (0, t) = u1 and
(0, t) = u2 on the boundary of R.
∂t
An elliptic partial differential equation, for example, is a real 2nd order
partial differential equation of the form:
n

∂ 2u

∑ aij ∂ xi∂ xj

(

+ F x1, … , xn, u,

i,j = 1

)

∂u
∂u
,…,

= 0
∂ x1
∂ xn

n

such that the quadratic form

∑

aijxixj is non-singular and positive definite.

i, j = 1

Typical examples are the Laplace (Dirichlet's problem) and Poisson
equations. We might, in passing, mention the link with potential theory.
21. To show that there always exists a linear differential equation of the
Fuchsian class with given singular points and monodromy group
Several special cases have been solved, for example by H. Röhrl (1957)
and P. Deligne (1970), but a negative solution was found by B. Bolibruch
(1989).
Note: The first indication of a deep relationship between groups and
differential equations emerged in Riemann's investigation of the
hypergeometric differential equation, which belongs to class of equations of
Fuschian type. As it is linear, and of second order, its solutions are
expressible as a sum of basic solutions, the analytic continuation of which
around each singular point gives rise to more branches of the solution that


8

THE MATHEMATICAL GAZETTE

depend linearly on those first chosen. The matrix of constants which
characterises this dependence is called a monodromy matrix and the group
generated by these matrices is called the monodromy group of the equation
[2].
22. Uniformisation of complex analytic functions by means of automorphic
functions
Parametrising all algebraic curves (representing simultaneously their x
and y values by functions of a single parameter) became known as the
uniformisation problem. Poincaré conjectured that all but the simplest of
algebraic curves arise from decompositions of the upper half-plane into a
tessellation by polygons. The problem was resolved by H. Poincaré and P.
Koebe (1907). This result underpins much of modern complex analysis
(and complex dynamical systems).
23. To extend the methods of the calculus of variations
Hilbert, and many others, have made contributions to this area, which
has grown apace, especially since the Second World War, and has been
subsumed under optimisation theory (operations research) which includes
such disciplines as control theory, decision theory, linear programming,
Markov chains and queuing theory.
Acknowledgements
The author wishes especially to thank the referees who read the first
draft of this paper and the editor, Stephen Abbott, for their invaluable help.
References
1. E. T. Bell, The development of mathematics, Dover (1992) p. 340.
2. I. Grattan-Guinness (ed.), Companion encyclopaedia of the history of
mathematical sciences, Routledge (1994) pp. 470-471.
Comprehensive sources

E. J. Borowski and J. H. Borwein, Dictionary of mathematics, Collins
(1989).
Eric W. Weisstein, CRC concise encyclopaedia of mathematics, Chapman
and Hall (1999).
Kiyosi Itô (ed.), Encyclopaedic dictionary of mathematics, MIT Press.
GRAHAM T. Q. HOARE
3 Russett Hill, Chalfont St Peter, Bucks SL9 8JY


LEWIS CARROLL − MATHEMATICIAN AND TEACHER OF CHILDREN

9

Lewis Carroll − mathematician and teacher of
children
CANON D. B. EPERSON
It is well known that Lewis Carroll enjoyed the company of children
and entertained them with fantastic stories, whilst on boating expeditions or
on beaches, but his diaries* also reveal that he enjoyed teaching children
mathematics and other school subjects. On April 16th 1855, he recorded his
concern with the education of his younger sister Louisa, who had an aptitude
for mathematics: ‘Went into Darlington − bought Swale's Chamber's Euclid
for Louisa. I had to scratch out a good deal he had interpolated, (e.g.
definitions of words of his own) and put in some he had left out. An author
has no right to mangle the original writer whom he employs: all additional
matter should be carefully distinguished from the genuine text. N. B. Pott's
Euclid is the only edition worth getting − both Capell and Chamber's are
mangled editions.’ Three days later he recorded: ‘Advanced Louisa's
mathematics to simple Equations (third day of Algebra), and the first 12
propositions of Euclid.’ On the next day, he left his home at the rectory of

Croft, and returned to Oxford for the Easter Term.
At this time his own mathematical education was providing him with
problems. He was studying the monumental works of George Salmon on
conic sections, in which he found deficiencies and inconsistencies that
hindered the compilation of his own Notes on Salmon. New subjects also
worried him; ‘I talked over the calculus of variations with Price (his tutor)
today, but without any effect. I see no prospect of understanding the subject
at all.’ Four days later he wrote: ‘I have spent a good deal of the day
puzzling over a difficulty in Salmon’, and again consultation with Price did
not help.
He was happier a few days later when Price lent him a little book on
finite differences by Knuff, ‘by which all kinds of series can be summed: I
have not yet made it out, but it looks very neat.’
Mr Charles Lutwidge Dodgson was then a 23-year-old undergraduate at
Christ Church, Oxford, who had achieved First Class Honours in
Mathematical Moderations and had been appointed a Student of Christ
Church (equivalent to being elected a Fellow of the college), a post that
initiated him into the teaching profession, as his duties included giving
private tuition to younger men. During the summer vacation he had his first
experience of teaching a class of children at the new National School at
*

I am indebted to my friend Edward Wakeling for his permission to quote
extensively from volumes I and II of his annotated edition of Lewis Carroll's
Diaries (the private journals of Charles Lutwidge Dodgson), published by the
Lewis Carroll Society. In places, the punctuation has been altered and the the
interested reader may wish to consult the originals.


10

THE MATHEMATICAL GAZETTE

Croft-on-Tees: ‘I went to the Boys' School in the morning to hear my father
(Archdeacon Dodgson) teach, as I want to begin trying myself soon. Some
of the boys were much more intelligent than I expected.’ On the following
Sunday ‘I took the first and second class of the Boys' School in the morning
− we did part of the life of St. John, one of the “lessons” on Scripture Lives.
I liked my first attempt in teaching very much.’
On the next day he ‘Took the first class alone in Old Testament (part of
Judges). Mr Hobson (the headmaster) wants some assistance in Latin, and
he also proposes that I should teach Coates (who is trying for a Pupilteachership) some Algebra—we made no definite arrangements’. On the
next day ‘As there was nothing for me to do in the Boys' School, I took the
second class in the Girls', and liked the experiment very much. The
intelligence of the children seemed to vary inversely as their size. They
were a little shy this first time, but answered well nevertheless.’
During the next week ‘I took the first class of boys: besides this I teach
James Coates Euclid and Algebra on Tuesday, Thursday and Friday
evenings, and read Latin with Mr Hobson on Wednesday evening and
Saturday morning − so that I have tolerable practice in teaching.’ Later he
recorded ‘My scheme of teaching now is
Boys' School Sunday morning.
Girls' school, second class, all other mornings except Saturday
and Friday.
Mr Hobson (Latin) Wednesday evening and Saturday morning.
James Coates (Euclid and Algebra) Tuesday, Thursday and
Friday evening.
besides these, I give the first class in the Girls' School a lesson in sums,
every Friday morning, making about 9 hours teaching per week.’
Presumably this continued until the school closed for the summer

holiday, as the last reference to Croft School in the diary was on Sunday
July 29th: ‘Took the Boy's School in texts for the first time − my regular
work now is with the second class girls.’
It is worth noting that the school was built in 1845 by the National
Society with generous help from the Dodgson family; it consisted of two
large rooms, one for boys and one for girls, catering for 120 pupils. There
were only two teachers, Mr Henry Hobson and his wife Sarah, but it is
thought that Archdeacon Dodgson and other members of his family assisted
regularly with the teaching in an honorary capacity.
On January 29th 1856, he breakfasted with the Revd Henry Swabey in
order to arrange about teaching in his school. ‘We settled that I am to come
at ten on Sunday, and at two on Tuesdays and Fridays to teach sums. I gave
the first lesson there today, to a class of 8 boys, and found it much more
pleasant than I expected. The contrast is very striking between town and
country boys: here they are sharp, boisterous, and in the highest spirits, the
difficulty of teaching being not to get an answer, but to prevent all
answering at once. They seem tractable and in good order: I stayed a short


LEWIS CARROLL − MATHEMATICIAN AND TEACHER OF CHILDREN

11

time afterwards to watch: for want of teachers, the master (Charles
Mayhew) had to conduct two lessons at once, while a third (a writing
lesson) went on by itself.’
On Feb. 1st ‘The Master at St Aldate's School asked if I would join the
first class of girls with the boys. I tried it for today, but I do not think they
can be kept together, as the boys are much the sharpest. This made a class
of 15. I went on with “practice” as before.’ Later ‘Gordon (the Senior

Censor) suggested a question in ancient mathematics; viz. how did the
Romans work multiplication? He, Lloyd, and I, tried it, but could not make
much of it.’ On Feb. 5th he ‘varied the lesson at school with a story,
introducing a number of sums to be worked out. I also worked for them the
puzzle of writing the answer to an addition sum, when only one of the five
rows have been written: this, and the trick of counting alternately up to 100,
either putting on no more than 10 to the number last named, astonished them
not a little.’
This shows that Dodgson was well aware of the value of ‘recreational
mathematics’ in the classroom, and he may have been a pioneer in using
puzzles and tricks in order to retain the interest of children in a subject that
they found difficult, although he may have been using methods similar to
those he had experienced when at school. It is thought that he invented
several puzzles and pastimes, and that he planned after his retirement to
publish a book containing some of them, in his series of Curiosa
Mathematica.
On Feb. 8th he found ‘the school class noisy and inattentive, the novelty
of the thing is wearing off, and I find them rather unmanageable. Showed
them the “9” trick of striking out a figure, after subtracting a number from
its reverse’. (This depends upon the fact that the difference between any
number and its reverse is always a multiple of 9, and so has a digit sum that
is 9 or a multiple of 9.)
On the next day he wrote to Swabey, asking ‘what he considered the
best way for my going on at the school: my idea is to form a new class,
consisting only of the bright and attentive boys and girls: the system of
taking the whole of the two first classes does not answer well’ but when he
met him at the school, he ‘agreed to try a little longer taking the whole of
both classes, and set them sums all round, so as to give each something to
do. I taught them a little about fractions, and explained the trick of the
addition sum.’

The struggle to maintain discipline continued for a few more days; one
day he found the ‘school class again noisy and troublesome. I have not yet
acquired the art of keeping order’, though four days later, ‘School class
better, as I threatened to banish those who did not attend from the lesson’.
A week later he recorded: ‘Class again noisy and inattentive, it is most
disheartening, and I almost think I had better give up teaching there for the
present.’ Three days later he ‘Left word at the school that I shall not be able
to come again for the present. I doubt if I shall try again next term: the good
done does not seem worth the time and trouble.’


12

THE MATHEMATICAL GAZETTE

Fortunately his association with the children at Croft School ended on a
happier note. When he returned home during the next vacation, he ‘Heard
the singing lesson in the school, about 50 are learning, and there are many
good voices among them: one piece they sang in full harmony. They are
also learning a choral service, which Mr Baker hopes to introduce in church
on weekdays.’
When Dr Henry George Liddell came to Christ Church to reside in the
deanery with his wife and their four young children, Dodgson was a
frequent visitor and soon made friends with Harry their only boy, whom he
met for the first time whilst watching the torpids* from the Christ Church
barge. On March 6th 1856 he recorded: ‘he is certainly the handsomest boy
I ever saw’. In the following summer term he taught Harry how to row, and
on June 3rd he ‘spent the morning at the deanery, photographing the
children’. Later ‘Frank and I, with Harry Liddell, went down to Sandford in
a gig. We rowed with sculls down with Harry as stroke, and he steered

back.’ On June 5th ‘from half-past four to seven, Frank and I made a
boating excursion with Harry and Ina: the latter (the Dean's eldest daughter),
much to my surprise, having got permission from the Dean to come. We
went down to the island, and made a kind of picnic there, taking biscuits
with us, and buying ginger beer and lemonade there. Harry, as before,
rowed stroke most of the way, and (fortunately, considering the wild spirits
of the children) we got home without accident, having attracted by our
remarkable crew a good deal of attention from almost everyone we met.’
Dodgson regarded this as one of the happiest days of his life.
In the Michaelmas term he soon ‘Fell in with Harry and Ina Liddell, and
took them up to see my book of photographs’, and a few days later, ‘Met
Miss Prickett, the governess at the Deanery, walking with Ina, and settled
that I would come over on Wednesday morning, if fine,’ in order to take
photographs. ‘The morning was fair, and I took my camera over to the
deanery, just in time to see the whole party (except Edith) set off with the
carriage and ponies, a disappointment for me, as it is the last vacant morning
I shall have in the term.’ Five days later, as there was a clear sun, he ‘went
to the deanery to take portraits at two, but the light failed, and I only got one
of Harry. I spent an hour or so afterwards with the children and the
governess, up in the schoolroom, making them paper boats, etc.’
When Mrs Liddell informed Dodgson that she intended to send Harry to
Twyford School after Easter, she ‘took me into the schoolroom to see
specimens of his sums and Latin: in the former he is well on.’ A few days
later he ‘Called at the deanery. The Dean and Mrs L. are going abroad for
four months, for his health. The children are to remain in Oxford: Lloyd has
undertaken to teach Harry his Latin and Greek. I offered to teach him sums
etc. but Mrs L. seemed to think it would take up too much of my time. Two
days before the parents sailed for Madeira, he ‘called at the deanery, and
took Harry a Christmas box, a mechanical tortoise: (I gave Ina one the other
day, Mrs Rutherford's children).’ No mention was made of gifts to Alice

*

A torpid was the second boat of a college.


LEWIS CARROLL − MATHEMATICIAN AND TEACHER OF CHILDREN

13

and Edith, but one can be sure that they were not forgotten. Five days later
he ‘met Harry and Ina in the Quadrangle, coming home from riding, and
went into the deanery with them, and stayed for luncheon (or rather their
dinner).’
Early in February 1857 whilst ‘walking in the afternoon, I fell in with
Ina Liddell and the governess, and returned with them to the deanery, where
I spent about an hour with the young party in the schoolroom. Miss Prickett
showed me a letter the other day from Mrs Reeve (Mrs Liddell's mother) in
which she expressed great alarm at Harry's learning “mathematics” with me!
She fears the effect of overwork on the brain. As far as I can judge, there is
nothing to fear at present on that score, and I sent a message to that effect.’
On February 8th Dodgson ‘went to chapel in surplice for the first time
since the 14th of October 1855. I read the second lesson in the afternoon.
Harry ran up to me afterwards to tell me “you've got your white gown on,
and you read in the church!” Two days later he recorded ‘my pupil Harry
Liddell is beginning to tire of the arithmetic lesson. I talked the subject over
with the governess, and settled that he had better give up coming to me
unless we succeed better in future.’ But on the next day he ‘spent an hour at
the deanery in the afternoon by Harry's invitation,’ and on the following
morning ‘Harry did well today, it is doubtful how long the change will last.’
In the autumn of the following year Dodgson visited Harry at Twyford

School, and, of course, saw him in Oxford during the school holidays, and
enjoyed his company rowing on the Isis. Undoubtedly he had made an
invaluable contribution to the education of the ‘fine young man’ who came
to call on him in November 1862.
On May 5th 1857 Dodgson recorded in his diary ‘I went to the deanery
in the afternoon, partly to give little Alice a birthday present, and stayed for
tea.’ She was then 5 years old, and had already been photographed more
than once. His friendship with Ina, Alice and Edith naturally developed
after Harry had gone to boarding school, and his visits to the deanery were
so frequent that some suspected that the attraction was Miss Prickett! There
is no evidence that Dodgson ever offered to participate in thr education of
the three young maidens, but his imaginative stories and the word puzzles
that he invented must have developed their appreciation of the English
language and increased their vocabularies. This influence upon children and
adults still continues, and, for this reason, Lewis Carroll is still honoured
100 years after his death.
CANON D. B. EPERSON
Hillrise, 12 Tennyson Road, Worthing, Sussex BN11 4BY
An imperfect issue
No feature on perfect numbers for Math Gaz 496? Then there's always
8128!
A comment from J. H. Evans


14

THE MATHEMATICAL GAZETTE

Snubbing with and without eta
H. MARTYN CUNDY

In the recent Gazette [1], John Sharp has given us a most entertaining
account of the number η (eta) which is the only real root of the equation
x3 − x2 − x − 1 = 0, and its connection with the snub cube. I have been
investigating this a bit further, with some results that may be of interest.
More about η
John Sharp suggests that η may be a pervasive number like π or e, but
these are transcendental numbers, while η is an ordinary algebraic number,
although its derivation is interesting. η = 1.8392867552… , but, using
Tartaglia's method − putting x = 13 + u + v, we obtain its value explicitly
as
η =

1
3

[1

+ (19 + 3 33)

1/3

+ (19 − 3 33)

].

1/3

η does have some useful properties: John Sharp gives some, but here is a
short list:
(η − 1) (η2 − 1) = (η + 1) (η − 1)2 = 2,

(η − 1) (η2 + 1) = 2η,
(η + 1)2 (η − 1) = (η2 − 1) (η + 1) = 2η2,
(η + 1)2 = η (η2 + 1) ,
η + 1 / η = 2 / (η − 1) = (1 + 1 / η)2 ,
η4 + 1 = 2η3 and thus η + 1 / η3 = 2.
With a start near 2, y = 2 − 1 / x3 on iteration converges rapidly to η. The
general equation of this type,
xn = xn − 1 + xn − 2 + …

+ x + 1 = (xn − 1) / (x − 1)

gives x = 2 − 1 / xn, which approaches 2 steadily with increasing n.
If we set η − 1 = t , we obtain the equation t 2 (t + 2) = 2, and we
shall find this more meaningful and useful as an aid to the process of
snubbing, to which I now turn.
A more general approach
Measurements of significant features of the snub cube are not too
difficult to make, but most people find themselves baffled by the snub
dodecahedron and wonder how to get started on it. Let me tell two stories
which may help.
(i) Can you discover from local measurements how big the earth is?


SNUBBING WITH AND WITHOUT ETA

15

Eratosthenes had to do some travelling about Egypt to make his
estimate, but I was brought up in a generation (before metrication) that
was taught the following myth. If you go to East Anglia (where there

are lots of zero contours, a spot height of 2 indicates a ‘hill’, and there
is even a trig. point labelled −1) there are long stretches of level water.
Put three posts in a straight stretch, one mile apart. All the posts stick
equally far out of the water, but, due to the curvature of the earth, the
middle one is 8 inches above the line of sight of the other two. From
this and the geometrical theorem in Figure 1, which states that
KA.KB = KC.KD we have at once that 8ins. × earth's diameter =
(1 mile)2. The diameter is therefore given by d = 5280 × 12 / 8 miles
= 7920 miles, which is surprisingly accurate!
A
C

D
K

B
FIGURE 1

(ii) A friend recently was making a model with wooden balls of the carbon
molecule with 60 atoms, called a buckyball. (They are arranged like
the points on a football where hexagons and pentagons meet.) He
could not get the angles right, nor find how big the model would be.
We need the same theorem.
Every uniform polyhedron has equal edges and every vertex alike.
Since the vertices are all alike, they are on a sphere whose centre is their
centroid. So the neighbours of a vertex A, say B1, B2, B3, B4, B5, all lie on two
spheres: the circumsphere of the polyhedron, and a sphere with centre A and
radius equal to the edge-length. Two spheres meet in a plane circle. So the
Bi form a plane cyclic polygon called the vertex figure of A. If we know its
shape and how much the vertex A lies above it, we can deduce all about the

polyhedron.
Snub polyhedra
We will confine our attention to the snub polyhedra that have 4
equilateral triangular faces and a (possibly different) regular polygonal face
surrounding each vertex. There are 6 of these (described in detail in [2]) if


16

THE MATHEMATICAL GAZETTE

we include the icosahedron (which is a snub tetrahedron). The other 5 (in a
notation that is described in the next paragraph) are:
| 2 3 4 snub cuboctahedron
| 2 3 5 snub icosidodecahedron
| 2 3 5/2 great inverted snub icosidodecahedron
| 2 3 5/3 great snub icosidodecahedron
and finally one with its triangles reflexed
| 2 3/2 5/3 great inverted retrosnub icosidodecahedron (!).
These can be seen illustrated in all their glory in [2], nos. 17, 18, 113,
116, 117. Figure 2 shows the snub cube (or, better, snub cuboctahedron).
All of these occur in enantiomorphic pairs, that is, with left-handed or righthanded twists, making mirror-images of one another. As to the names, the
snub cube is related to the octahedron in the same way as it is to the cube; it
contains faces in the planes of both solids. To be fair, then, it is preferred to
call it the snub cuboctahedron, and so on. Some people use other names, but
the numerical code is definitive. All these polyhedra can be generated by
reflections in the sides of spherical triangles, and the numbers give the
angles of these triangles. Thus | 2 3 5/2 indicates the use of a triangle with
angles π / 2, π / 3, 2π / 5, see [2, 3].
The side of a convex regular polygon with p sides (a p-gon) subtends an

angle of 2π / p at its centre K . The pentagram, or regular star-pentagon, has
sides subtending 4π / 5, so it is convenient to label it as a 5/2-gon; an m / ngon will have m vertices equally spaced on a circle, joined by edges to
neighbours n points away.

FIGURE 2

If every edge is of length 2 units, the vertex figures of all these
polyhedra have the form shown here in Figure 3: for i = 1 to 4 the angle
BiKBi + 1 will be 2θ , say, where θ is slightly greater than π / 6. B1B5 will be a
diagonal of a p-gon so that ∠B1AB5 = π(1 − 2/ p) and ∠B1KB5 = 2(π − 4θ).
Let the radius KBi = ρ. Then from B1KB2 we have ρ sin θ = 1, and
from B1KB5 and B1AB5 we have 2c = B1B5 = 2ρ sin 4θ = 4 cos π / p.


SNUBBING WITH AND WITHOUT ETA

17

Therefore
sin 4θ
= c = 2 cos (π / p) .
sin θ

(1)

Let t = 2 cos 2θ , so that t + 2 = 4 cos2 θ . These equations combine to
give
t 2 (t + 2) = c2 = 4 cos2 (π / p) .
(2)
For the six polyhedra listed above, p = 3, 4, 5, 5/2, 5/3 (5/2 reversed), 5/3;

but in the last two the vertex figure is crossed; the theory holds, but it is not
easy to see what. is happening! See [2].
So in equation (2) c2 is equal to 1, 2, τ2, τ−2, where τ = ( 5 + 1) / 2 is
the golden ratio. John Sharp used the letter φ for this number − a
transatlantic name, but the English use of τ is commoner.
B2
B3

A

B1

K
ρ

K

B1

ρ

θ θ
1

1

B2

B4
A

B5
ρ

K

ρ

π − 4θ

B1

π

p

B5
FIGURE 3

Resulting measurements
Well, after all that, what use is t ? This is where Figure 1 comes into its
own. Let h be the height of the vertex A above the plane of the Bi, and let d
be the diameter of the snub polyhedron (Figure 4). Then h (d − h) = ρ2
and h2 + ρ2 = 4. At once we have d = 4 / h (a result we could also get
from the similarity of triangles ABD, AKB), and
1
4 (1 − t)
h2 = 4 − ρ2 = 4 − cosec2 θ = 4 1 −
=
.
2 − t

2 − t
Therefore R, the radius of the circumsphere of the snub polyhedron is given
by
2 − t
R =
.
1 − t

(

)


18

THE MATHEMATICAL GAZETTE

h
B, Bi

ρ

A

K

ρ

d−h

D
FIGURE 4

To find the circumscribing polyhedron
Here, as John Sharp showed, we need the radius of the sphere touching
the faces of the polyhedron we are seeking, the in-radius, which, following
Coxeter [4], we denote by 2R. We are interested in two possible polyhedra:
that with the p-gon as face, and that with the triangle as face belonging to
the dual polyhedron. Which triangle depends on which enantiomorph we
have, but the choice does not affect the measurements. It is immediate that
for a p-gon 2R2 = R2 − cosec2 (π / p) = 1 / (1 − t) − cot 2 (π / p), and for a
triangle 2R2 = 1/ (1 − t) − 13 . If the edge of the circumscribing polyhedron is
2e, we need to find this from our knowledge of the radius 2R of its shared insphere. The polyhedron {p, q} has p-gonal faces and q-gonal vertex figures.
If O is its centre, V a vertex, E the midpoint of an edge, and F the centre of a
face, again following [3], we name the radii OV = 0R, OE = 1R,
OF = 2R. VE = e, sincet 2e is the length of an edge. We need
∠EOF = ψ, where π − 2ψ is the dihedral angle.
First we need
∠VOE = φ. In the diagram (Figure 5), since V, K, O, V′, E are coplanar,
∠VV′K = φ, so V′K = 2e cos φ is the radius of the vertex figure, which is
a q-gon with edge-length 4e cos (π / p), so its radius is
2e cos (π / p) cosec (π / q). Therefore cos φ = cos (π / p) cosec (π / q). The
rest is straightforward, but tiresome.
sin 2 φ = 1 − cos2 (π / p) cosec2 (π / q) = k 2 cosec2 (π / q) ,
where k 2 = sin 2 (π / q) − cos2 (π / p) = sin 2 (π / p) − cos2 (π / q).
Then 0R = e cos φ = (e sin (π / q)) / k, and
2
1R

Finally

= OE2 = e2 (sin 2 (π / q) − k 2) / k 2 = (e2 cos2 (π / p)) / k 2.


SNUBBING WITH AND WITHOUT ETA
2
2R

19

= 0R2 − VF2 = e2 sin 2 (π / q) / k 2 − e2 cosec2 (π / p)
2
= (e2 / k ) cos2 (π / q) cot 2 (π / p) .

[ 1/ (1 − t) − cot2 (π/ p)]

cot 2 ψ = (cos2 (π/ q)) / k 2 = OF2 / FE2 =

So

e2 cot 2 π/ p

,

leading to e2 cos2 (π / p) = k 2 [ tan2 (π / p) / (1 − t) − 1] , or, finally
 sin 2 (π / p)

  tan2 (π / p)
e2 =  2
− 1 

− 1 .

 cos (π / q)
 1 − t
O
φ

K

ψ

V

F
e
φ

E
e

V′

FIGURE 5

Angle of twist
We now know the sizes of the polyhedra which enclose the snub
polyhedron, with faces containing its non-snub faces of a {p, q}. These
faces lie on similar faces of the enclosing polyhedron − the case − but are
twisted in relation to them. We now find the angle of twist. This looks
formidable until we spot the secret. The polygons are forced to twist

because one triangle edge links adjacent p-gons, with its midpoint on OM,
where M is the midpoint of the separating edge of the case. (See Figure 6).
The central plane through this edge bisects PP′, which lies in a plane
perpendicular to it. (If you have made a model of a snub cube, this is seen
very clearly.) If (x, y) as shown in the figure are coordinates of P in its face,
with origin M, we have x2 cos2 ψ + y2 = 1; x = e cot (π/ p) − cosec(π/ p) cosα;
y = cosec(π/ p) sin α, where α is the inclination of 2OP to the x-axis. These
equations lead to the quadratic in cos α:
k 2 cos2 α + 2e cos α cos (π / p) cos2 (π / q) −

(e2 cos2 (π / q)

+ sin 2 (π / p)) cos2 (π / p) = 0.


20

THE MATHEMATICAL GAZETTE

e

P
y

α

M
x

L

π p

2O

P′
e

M
(P′)

L (P)
x cos ψ
2O

ψ

ψ

O
FIGURE 6
2

Inserting the value of e and a little manipulation produces for the positive
root of this quadratic
sin (π / p)
1
− cos (π / q)
− cos2 (π / p),
1 − t

1 − t
i.e. cosα = [ sin (π/ p)/ k] [ 1R′ sin (π/ p)−2 R cos(π/ q)] , where 1R′ = R2 − 1 is
now the mid-radius of the snub polyhedron. This result is so unexpectedly
simple that one wonders if there is a simpler way of getting it!
The angle of twist is π / p − α, and is given in the accompanying table.
k cosec (π / p) cos α =


SNUBBING WITH AND WITHOUT ETA

21

Snubber's kit: [(34, p) only. For others, see [2] and be amazed].
Equations: Snub {p,3}; edge length 2 units.
t 2 (t + 2) = 4 cos2 (π / p). 0R2 = 1 + 1 / (1 − t). 1R = 1 / 1 − t .
If a {p, q} with edges of length 2e contains the p-gonal faces of the snub
(p, q are now interchangeable), then
1
π
π
π
2
− cot 2 .
k 2 = sin 2 − cos2 .
2R =
1 − t
p
p
q
π

π
e = 2R k tan sec .
p
q
Twist of snub on a p-gonal face is β = π / p − α, where
π
π
π
k cos α = sin 1R sin − 2R cos  .
p
p
q
TABLES: numerical results for snub edge = 2
TABLE 1 p = 3, c2 = 1
t

Polyhedron

R

−1
1/τ
−τ

Same tet.
Icosa.
Great icosa.

3/ 2
2+τ

3−τ

{p, q}
{3, 3}
{3, 3}
{3, 3}

e

2R

1

1/ 6
τ2 / 3
3 / τ2

τ2 2
2 / τ2

k
1
2

twist angle
2

0
22·23°
22·23°

TABLE 2 p = 4, c2 = 2
t

Polyhedron

R

0·839287

Snub cube

2·687427

{p, q}

e

2R

k

twist angle

Cube
Octahedron

2·285227
2·972103

2·285227
2·426712

1
2

16·468°
20·315°

k

twist angle

TABLE 3 p = 5, c2 = τ2
t

Polyhedron

0·943151

Snub
dodeca.

{p, q}

R

e

2R

4·311675 Dodecahedron 1·778973 3·961832
Icosahedron
2·748341 4·154179

1 −1
2τ

13·106°
19·518°

TABLE 4 p = 5 / 2, c2 = τ−2
t

Polyhedron

0·399021

(W.113)

−0·505561

(W.116)

−1·893460

(W.117)

R

{p, q}

1·632161 Gt Stellated
dodecahedron
Great
icosahedron
1·290040 Gt Stellated
dodecahedron
Great icosa.
1·160003 Gt Stellated
dodecahedron
Great icosa.

e

2R

6·216529 1·248350

k

twist angle

1
2τ

27·108°

5·230732 1·153524

24·515°

3·721987 0·747415

10·155°

2·608352 0·575214
2·439766 0·489933

4·420°
3·672°

0·551694 0·110786

0·544°


22

THE MATHEMATICAL GAZETTE

Notes on the tables
Tetrahedron. 3 real roots for t . t = −1 means 2θ = 2π / 3 so that B4 will
coincide with B1 and B5 with B2. So triangle KB5B1 will just be triangle
KB1B2 repeated turned over. So at vertex A we have 2 ordinary equilateral
triangles and one covered three times; finally we arrive at the original
tetrahedron covered 5 times—a rather peculiar icosahedron! t = τ−1 gives
us what we expected, but t = −τ reminds us of what we have probably
forgotten. There are in fact two ways of dividing the edges of the core
octahedron of the stella octangula in the ratio τ : 1. It may be divided

eat

Gr

edr

face

on

edr

on

sah

rah

ico

Tet

22.2°

e

fac

icosahedron

face

τ3

1

τ2 2

τ
τ2

l
with ine of i
of s secon ntersec
tella d te
ti
octa trahe on
ngu dron
la

τ4
FIGURE 7

internally, giving the familiar icosahedron, or externally, giving the great
icosahedron, a beautiful polyhedron with star-pointed vertices. Its edges are
parallel to those of the internal icosahedron. See Figure 7 and the fuller
discussion below.
The snub cube. The inverse ratio giving the edge of the snub cube as a
fraction of the encasing cube is 0·4375933…. John's eagle eye found a
similar number in a remote place—Figure 174 in [5]. It would indeed be

miraculous if a number which is the root of a cubic equation should turn up
in the entirely Euclidean process of locating a tetrahedron in a
dodecahedron.
Such has not happened; the ratio here is
1 / (τ 2) = 0·437016…