christopher j bradley - challenges in geometry for mathematical olympians past and present

Challenges in Geometry
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Challenges in Geometry
for Mathematical Olympians
Past and Present
CHRISTOPHER J. BRADLEY
1
3
Great Clarendon Street, Oxford OX2 6DP
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ISBN 0–19–856691–3 9780198566915
ISBN 0–19–856692–1 (pbk) 9780198566922
13579108642
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Preface
This book is written for students who are interested in geometry and number the-
ory, for those involved in Mathematical Olympiads, and for teachers in universities
and schools. It is more of a geometry book than a book about integers and contains,
among other things, a full account of the properties of triangles and circles normally
associated with an advanced course of Euclidean geometry. The restriction to con-
figurations in which various lengths are required to have integer values provides a
natural and appealing link between elementary geometry and interesting problems
inv olving Diophantine equations. Though the content is mostly elementary, some of
the results would appear to be new. The book is not designed for any particular course
of study, but is suitable as additional reading for undergraduates studying these topics,

for students preparing for competitions, and for other mathematically advanced high
school students.
During the last thirteen years I h ave been closely involved in the preparation of
the United Kingdom team for the International Mathematical Olympiad, of which I
was Deputy Leader for three years. The content, therefore, reflects interests developed
during these years.
Though few of the problems treated in the book could ever have been set in
an Olympiad competition (they are mostly too long and detailed), the techniques
inv olved are precisely those suitable for developing the problem-solving skills needed
for competitions. The book also includes a number of topics of a geometrical nature
in which integers appear , that are not normally included in a course primarily devoted
to Euclidean geometry; for e xample, there are chapters on polygonal numbers and on
methods for obtaining rational and integer points on curves.
I ha ve had two most enjo yable careers. My first was as a lecturer at Oxford
University, where my research interests were also in algebra and geometry, and where
I had the good fortune to be associated with Professor C harles Coulson and Dr Simon
Altmann. The latter was my research supervisor when I was a graduate student and I
o we a great deal to his care and enthusiasm. During my years in Oxford I engaged in a
major project with my friend Arthur Cracknell, who later became a professor of Physics
at the University of Dundee. This project resulted in a book entitled The mathematical
theory of symmetry in solids, Clarendon Press, Oxford (1972), a classification of the
irreducible representations of the 230 space groups.
I left Oxford in 1977 to become a schoolteacher, first at Christ’s Hospital, Horsham
and later at Clifton College, Bristol. I am grateful to colleagues at both of these schools
for their help and encouragement. I retired from full-time work in 1998 and since then
there has been time for writing. Since 1990 I have had the privilege to be associated
vi
Preface
with a number of inspiring colleagues, who have encouraged and challenged me
intellectually in ways that I did not anticipate when I became a schoolteacher. These

include the various leaders of the UK International Mathematical Olympiad team,
Dr To ny Gardiner, Professor Adam McBride, Dr Imre Leader , and Dr Geoff Smith.
Perhaps the greatest geometrical influence, however, has been Dr David Monk, who
helped to train the UK team for over thirty years, and whose contrib utions have been
immense. Tony Gardiner has spent much time and effort in helping me prepare the
manuscript for this book, and I am grateful to him for numerous suggestions for
improvements in the style and for the removal of certain ambiguities and conceptual
errors. An y remaining errors are entirely my responsibility.
I should also like to thank Dr Ke vin Buzzard of the Mathematics Department at
Imperial College, London for consultations and assistance with two of the problems
in number theory. Thanks are also due to my nephew, Dr Jeremy Bradley, of the
Department of Computing at Imperial College, London for help with some of the
computational problems in the book and with various pieces of technical help during
the course of preparing the manuscript.
I am indebted to readers of the Oxford Uni versity Press for invaluable comments
and suggestions. I am also extremely grateful to the staff of the Oxford University
Press for their unfailing help and encouragement during production, particularly to
Kate Pullen, the Assistant Commissioning Editor, whose help and courtesy smoothed
my path in the months prior to delivering the final manuscript.
Bristol C. J. B.
July 2004
Contents
Glossary of symbols xi
1 Integer-sided triangles 1
1.1 Integer-sided right-angled triangles 2
1.2 Integer-sided triangles with angles of 60
◦
and 120
◦
4

1.3 Heron triangles 7
1.4 The rectangular box 11
1.5 Integer-related triangles 15
1.6 Other integer-related figures 16
2 Circles and triangles 19
2.1 The circumradius R and the inradius r 20
2.2 Intersecting chords and tangents 22
2.3 Cyclic quadrilaterals and inscribable quadrilaterals 24
2.4 The medians of a triangle 29
2.5 The incircle and the excircles 34
2.6 The number of integer-sided triangles of given perimeter 35
2.7 Triangles with angles
u, 2u,and180
◦
− 3u 38
2.8 Integer r and integer internal bisectors 39
2.9 Triangles with angles
u, nu,and180
◦
− (n +1)u 41
3 Lattices 43
3.1 Lattices and the square lattice 43
3.2 Pick’s theorem 46
3.3 Integer points on straight lines 50
4 Rational points on cur ves 53
4.1 Integer points on a planar curve of degree two 53
4.2 Rational points on cubic c urves with a singular point 58
4.3 Elliptic curves 60
4.4 Elliptic curves o f the form
y

2
= x
3
− ax −b 65
viii
Contents
5 Shapes and numbers 71
5.1 Triangular numbers 71
5.2 More on triangular numbers 75
5.3 Pentagonal and
N-gonal numbers 78
5.4 Polyhedral numbers 83
5.5 Catalan numbers 86
6 Quadrilaterals and triangles 89
6.1 Integer parallelograms 89
6.2 Area of a cyclic quadrilateral 92
6.3 Equal sums of squares on the sides of a triangle 96
6.4 The integer-sided equilateral triangle 98
7 Touching circles and spheres 107
7.1 Three circles touching each other and all touching a line 107
7.2 Four circles touching one another externally 109
7.3 Five spheres touching each other externally 112
7.4 Six touching hyperspheres in four-dimensional space 115
7.5 Heron triangles revisited 117
8 More on triangles 123
8.1 Transversals of integer-sided triangles 123
8.2 The pedal triangle of three Cevians 126
8.3 The pedal triangle of a point 131
8.4 The pivot theorem 134
8.5 The symmedians and other Cevians 136

8.6 The Euler line and ratios
2:1in a triangle 137
8.7 The triangle of excentres 141
8.8 The lengths of
OI and OH 142
8.9 Feuerbach’s theorem 144
9 Solids 145
9.1 Tetrahedrons with integer edges and integer volume 145
9.2 The circumradius of a tetrahedron 149
9.3 The five regular solids and six regular hypersolids 153
Contents
ix
10 Circles and conics 157
10.1 Sequences of intersecting circles of unit radius 157
10.2 Simson lines and Simson conics 159
10.3 The nine-point conic 161
11 Finite geometries 163
11.1 Finite projective and affine geometries 163
Appendix
A Areal co-ordinates 167
A.1 Preliminaries 167
A.2 The co-ordinates of a line 168
A.3 The vector treatment of a triangle 169
A.4 Why the co-ordinates
(l, m, n) are called areal co-ordinates 171
A.5 The area of a triangle PQRand the equation of the line PQ 173
A.6 The areal co-ordinates of key points in the triangle 174
A.7 Some examples 175
A.8 The areal metric 177
A.9 The condition for perpendicular displacements 179

A.10 The equation of a circle 180
Answers to exercises 185
References 201
Index 203
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Glossary of symbols
The following symbols are repeatedly used in connection with a triangle ABC.
A, B, C The vertices of the triangle ABC.
∠BAC, ∠CBA,
∠ACB or



The angles of the triangle ABC.
∠A, ∠B, ∠C
a, b, ca= BC, b = CA,andc = AB are the side lengths of the triangle
ABC.
L, M, N The midpoints of BC, CA,andAB, respectively.
D, E, F The feet of the altitudes from A, B,andC, respectively.
O The circumcentre of the triangle ABC.
G The centroid of the triangle ABC.
H The orthocentre of the triangle ABC.
I The incentre of the triangle ABC.
I
1
, I
2
, I
3
The excentres opposite A, B,andC, respectively.

U, V , W The points where the internal angle bisectors meet BC, CA,and
AB, respectively.
X, Y , Z The points where the incircle touches BC, CA,andAB, respec-
tively.
T The nine-point centre.
S The symmedian point of the triangle ABC.
J The centre of mass of a uniform wire framework in the shape of
the triangle ABC.
R The radius of the circumcircle.
r The radius of th e incircle.
ss=
1
2
(a + b + c) is the semi-perimeter of the triangle ABC.
r
1
, r
2
, r
3
The radii of the excircles opposite A, B,andC, respectively.
[ABC], [PQR] The areas of the triangles ABC and PQR, respectively.
[X
1
X
2
···X
n
] The area of the polygon X
1

X
2
···X
n
.
Occasionally, we use some of these symbols to denote other points or quantities,
bu t when this happens it is always made clear in the text. For example, L, M ,andN
are also used as the points on BC, CA,andAB where the Cevians through a general
xii
Glossary of symbols
point P meet BC, CA,andAB, respectiv ely. L, M ,andN are also used to denote
the feet of the perpendiculars from an arbitrary point P onto the sides BC, CA,and
AB, respectively. U , V ,andW are also used to denote the midpoints of AH, BH,
and CH, respectively, where H is the orthocentre.
1 Integer-sided triangles
To say that a triangle has one side of integer length or that a circle has integer radius is
not mathematically significant, as the unit of length can always be adjusted so that this
is the case. To say that a triangle has all its sides of integer length is mathematically
significant. It means that, whatever the unit of length, the ratio of any pair of side
lengths is a rational number. However, even if significant, it is scarcely interesting.
This is because, if you are given three positive integers such that the greatest is smaller
than the sum of the other two, then you can always construct a triangle having sides
with these integer lengths. Integer-sided triangles become mathematically interesting
only when some further condition is imposed.
In this first chapter we treat a number of basic problems involving integer-sided
triangles, when an additional property is introduced.
A Babylonian tablet confirms that geometers of that era, about 3500 years ago, were
aware of the existence of right-angled triangles having inte ger sides, and may well
have had some method for constructing them based on the sexagesimal arithmetic the y
used. Problems of all sorts involving integers have always been regarded as fascinating

and not only by professional m athematicians. Witness the general interest aroused by
the solution of Fermat’s last theorem.
As seems fitting, since it is such an ancient problem, we start with an account of
those integer-sided triangles that have an angle of 90
◦
. Next we show how to obtain
all integer-sided triangles with angles of either 60
◦
or 120
◦
. It would be possible to
consider integer-sided triangles in which the angles have cosines that are equal to
rational numbers other than 1,
1
2
,or−
1
2
. Ho wever, the angles 60
◦
, 90
◦
,and120
◦
are
special as they feature in the rectangular and hexagonal lattices.
We then consider triangles with integer sides and integer area, and towards the end
of the chapter we investigate geometrical figures in which the area and perimeter are
related. There is also a section on the rectangular box, which is appropriate to consider
at an early stage because of its connection with integer-sided right-angled triangles.

For the most part we deal with configurations in which lengths have integer values,
b ut occasionally we relax this condition and just require them to ha ve rational values.
When this is done, it is done simply as a matter of convenience. It is not a restriction
because a figure with a finite number of lengths that are rational may always be
magnified so that the y become integers, the enlargement factor being the least common
multiple of all the denominators.
2
Integer-sided triangles
1.1 Integer-sided right-angled triangles
Theorem 1.1.1 (Pythagoras) ABC is a triangle with ∠BCA =90
◦
if and only if
a
2
+ b
2
= c
2
,wherea = BC, b = CA, and c = AB. ✷
No proof is needed here.
Theorem 1. 1.2 Suppose that a, b, and c ar e positive integer s with no common factor,
that a
2
+ b
2
= c
2
, and that a and b are coprime. Then a and b have opposite parity, so
b may be chosen to be even; and with this choice there exist coprim e positive integers
u and v of opposite parity with u>vsuch that

a = u
2
−v
2
,b=2uv , c = u
2
+ v
2
. (1.1.1)
Proof The integers a and b cannot both be even, for then a, b,andc would hav e a
common factor of 2. The integers a and b cannot both be odd, since all odd squares
are equal to 1 (mod 4) and c
2
=2(mod 4) is impossible. Suppose then that a is odd
and b is even. Then c is also odd. We hav e b
2
= c
2
−a
2
=(c −a)(c + a).Now,since
a and c are both odd, c −a and c + a have a f actor of 2 in common and cannot both be
divisible by 4. But a and c themselves have no factor in common, so c − a and c + a
have no factor other than 2 in common. Hence each must be twice a perfect square,
the squares having no factor in common. Writing c + a =2u
2
and c − a =2v
2
,we
find b

2
=4u
2
v
2
and b =2uv, c = u
2
+ v
2
,anda = u
2
− v
2
, where, since a and c
are odd and coprime, u and v must be of opposite parity and coprime. Also, u and v
may be chosen to be both positive, since b is positive and u>v,sincea is positiv e.
Note that (u
2
−v
2
)
2
+(2uv)
2
=(u
2
+ v
2
)
2

, so that condition (1.1.1) is sufficient
as well as necessary for such an integer triple (a, b, c) to exist. ✷
These integer triples (a, b, c) are called primitive Pythagorean triples.Theyare
Pythagorean because a, b,andc may then be chosen to be the integer sides of a
right-angled triangle and primitive because a, b,andc have no common factor. The
general solution of a
2
+ b
2
= c
2
in integers is then found by enlargement by a scale
factor k:
a = k(u
2
− v
2
) ,b=2kuv , c = k(u
2
+ v
2
) , (1.1.2)
where k is a positive integer. The simplest e xample with k =1, u =2,andv =1is
illustrated in Fig. 1.1.
Integer-sided triangles
3
3
4
5
A

BC
Fig. 1.1 Pythagoras’ theorem.
Exercises 1.1
1.1.1 Generalise the patterns appearing in Table 1.1 when (a) v =1,and(b)
u = v +1.
1.1.2 Prove that one member of a Pythagorean triple is always divisible by 5 and
that the area of an inte ger-sided right-angled triangle is always divisible by
6.
1.1.3 Find all integer -sided right-angled triangles with hypotenuse 145.
1.1.4 Prove that if c is the hypotenuse of a primitive integer-sided right-angled
triangle then c
2
is also.
1.1.5 Let (a, b, c) be a primiti ve Pythagorean triple. Prove that there exists an
infinite number of sets of integers l, m,andn such that
a = −lb + mc , b = la −nc , c = ma − nb (1.1.3)
and
m
2
+ n
2
=1+l
2
. (1.1.4)
Conversely, prove that if l, m,andn are integers satisfying eqn (1.1.4) then
there exists a primitive Pythagorean triple (a, b, c) satisfying eqn (1.1.3).
1.1.6 Prove that there are an infinite number of primiti ve Pythagorean triples in
which |a − b| =1. Explain how this result is related to finding rational
approximations of
√

2.
1.1.7 Is it possible to find an infinite set of points in the plane, not all on the same
straight line, such that the distance between every pair of points is rational?
4
Integer-sided triangles
Tab l e 1 .1 Primitive Pythagorean triples with c<100.
uva= u
2
− v
2
b =2uv c = u
2
+ v
2
21 3 4 5
32 5 12 13
41 15 8 17
43 7 24 25
5 2 21 20 29
54 9 40 41
6 1 35 12 37
6 5 11 60 61
7 2 45 28 53
7 4 33 56 65
7 6 13 84 85
8 1 63 16 65
8 3 55 48 73
8 5 39 80 89
(8 7 15 112 113)
9 2 77 36 85

9 4 65 72 97
1.2 Integer-sided triangles with angles of 60
◦
and 120
◦
Suppose now that a, b,andc are the side lengths of an integer-sided triangle with
angles of either 60
◦
or 120
◦
. Note that cos 60
◦
=
1
2
and cos 120
◦
= −
1
2
. It follows
from the cosine rule that if angle C =60
◦
then c
2
= a
2
− ab + b
2
, and if angle

C = 120
◦
then c
2
= a
2
+ ab + b
2
.
We first show how to obtain all solutions in nonzero integers a, b,andc of the
equation
c
2
= a
2
− ab + b
2
, (1.2.1)
without regard to the geometrical application. In the analysis that follows it turns out
that we can always ensure that c is positiv e. Then a solution of eqn (1.2.1) in which
precisely one of a or b turns out to be negativ e is a solution with positive a and b of
the equation
c
2
= a
2
+ ab + b
2
(1.2.2)
by a change of sign of a or b, as appropriate. If both of a and b turn out to be negative,

then a change of sign of both of them gives a solution of eqn (1.2.1) in which all of
a, b,andc are positive. In this way, the positive solutions of both equations may be
obtained simultaneously.
The method is to find a non-singular unimodular linear transformation from a, b,
and c to new integer variables u, v,andw to provide an equation that is linear in each
of the variables u, v,andw. The transformed equation can then be solved to fin d w in
terms of u and v. Finally, by eliminating w, the variables a, b,andc can be expressed
Integer-sided triangles
5
in terms of the two-parameter system u and v. The method works with most of the
homogeneous quadratic Diophantine equations that I have encountered and it is one
that I use several times in this text. There are many non-singular linear transformations
that will do the job.
Here a suitable transformation is a = u − w, b = v −w,andc = u + v −w,and
then substitution into eqn (1.2.1) gives 3uv = w(u + v). Hence w =3uv/(u + v).
Now, substituting back for w and multiplying up by u+v, we obtain the two-parameter
solution
a = u
2
− 2uv ,
b = v
2
− 2uv ,
c = u
2
− uv + v
2
.
(1.2.3)
Certain points need to be made about the method. Firstly, it does not always lead to a

solution in which a, b,andc are coprime. For e xample, u =5and v =1gives a =15,
b = −9,andc =21. This provides the primitive solution a =5, b =3,andc =7to
eqn (1.2.2). Secondly, solutions get repeated. For example, u =3and v =1leads to
the solution a =3, b =5,andc =7of eqn (1.2.2). To get all solutions one has to
multiply each of the expressions above for a, b,andc by any positive integer k.The
cases u =2v and v =2u give only trivial solutions and must be excluded. The case
u =1and v =1gives a = −1, b = −1,andc =1, which on changing the signs of
both a and b gives the equilateral triangle solution a = b = c =1of eqn (1.2.1). The
case u =1and v = −1 gives the solution a = b = c =3of eqn (1.2.1).
The important question is whether all solutions arise using the method. The answer
is ‘yes’. To see this, note that the inverse transformation is u = c −b, v = c −a,and
w = c − a − b, so that for any a, b,andc (c>0) satisfying eqn (1.2.1) values of u,
v,andw always e x ist. However, since we multiply up by u + v the solution for a, b,
and c may be a multiple of the specified solution. For example, consider the solution
a =8, b =15,andc =13of eqn (1.2.1). From the inverse transformation we have
u = −2, v =5,andw = −10. Substituting back into our expressions for a, b,and
c in terms of u and v we get a =24, b =45,andc =39. It is easy to see from the
transformation that a, b,andc are coprime provided that u and v are coprime and that
3 is not a factor of u + v. When 3 is a factor of u + v all that happens is that a, b,and
c have a common factor of 3, so the difficulty is a minor one. In Table 1.2 we give
the first few solutions to eqns (1.2.1) and (1.2.2) for triples {a, b, c} for triangles with
C =60
◦
and [a, b, c] for triangles with C = 120
◦
. See Figs 1.2 a nd 1.3 f or examples
of each of these cases.
6
Integer-sided triangles
Tab l e 1. 2 The following triangles have C =60

◦
or 120
◦
and corresponding triples are
labelled {a, b, c} or [a, b, c], respectively.
{1, 1, 1};
{8, 5, 7}, {8, 3, 7}, { 5, 8, 7}, [3, 5, 7];
{15, 7, 13}, {15, 8, 13}, {7, 15, 13}, [8, 7, 13];
{24, 9, 21}, {24, 15, 21}, {9, 24, 21}, [15, 9, 21];
{21, 16, 19}, {21, 5, 19}, {16, 21, 19}, [5, 16, 19];
{35, 11, 31}, {35, 24, 31}, {11, 35, 31}, [24, 11, 31];
{48, 13, 43}, {48, 35, 43}, {13, 48, 43}, [35, 13, 43];
{45, 24, 39}, {45, 21, 39}, {24, 45, 39}, [21, 24, 39];
{40, 33, 37}, {40, 7, 37}, {33, 40, 37}, [7, 33, 37]
87
5
A
BC
60
Fig. 1.2 A triangle with C =60
◦
and integer sides.
57
3
A
BC
120
Fig. 1.3 A triangle with C = 120
◦
and integer sides.

Integer-sided triangles
7
Exercises 1.2
1.2.1 Explain algebraically the repeats in the list of triangles with angle C =60
◦
in Table 1. 2.
1.2.2 Explain geometrically why, if [a, b, c] is a triple with angle C = 120
◦
,then
{a + b, b, c} is a triple with angle C =60
◦
.
1.2.3 Find all integer-sided triangles with angle C = 120
◦
and c =91.
1.2.4 Prove the statement in the text, that a, b,andc given by eqn (1.2.3) are
coprime provided that u and v are coprime and that 3 is not a factor of u + v.
1.3 Heron triangles
A Heron triangle is often defined to be one with rational side lengths and rational area.
Clearly, a triangle similar to a Heron triangle is also a Heron triangle, provided that
the scale f actor is rational, and indeed any Heron triangle is similar to a Heron triangle
with integer side lengths and area. We shall adopt this more restrictive definition and
insist that all of our Heron triangles have integer side lengths and integer area. Indeed,
since the altitudes of a Heron triangle must have rational length, we can, if needed, find
a Heron triangle similar to a giv en one, for which one (or even all) of the altitudes has
integer length. In what follows we insist that the altitude from A is of integer length.
It is our purpose, in this section, to give a parametric classification of Heron
triangles as defined above, by means of the theory of Section 1.1. In Section 7.5
we describe an alternative and more subtle approach to the classification of Heron
triangles.

An integer-sided right-angled triangle is trivially a Heron triangle and these have
already been dealt with in Section 1.1.
From the cosine rule, cos C =(a
2
+b
2
−c
2
)/2ab, etc., it follows that for an integer-
sided triangle the cosine of each angle is rational. From this observation alone, it is
ev ident that each acute-angled Heron triangle ABC is the union of two integer-sided
right-angled triangles ABD and ACD or, if angle B (or C) is obtuse, the difference
of two integer-sided right-angled triangles ACD and ABD. It would be possible to
dispense with Heron triangles that are the difference of two right-angled triangles if
we were to insist that any obtuse angle is placed at the vertex A, but we choose not to
do this.
The following formulae hold for the area [ABC] of triangle ABC:
[ABC]=
1
2
bc sin A =
1
2
ca sin B =
1
2
ab sin C
=
abc
4R

= {s(s − a)(s − b)(s − c)}
1/2
=
1
2
ad =
1
2
be =
1
2
cf = rs.
(1.3.1)
8
Integer-sided triangles
Here R is the circumradius, r is the inradius, s =
1
2
(a + b + c) is the semi-perimeter,
and d, e,andf are the lengths of the altitudes AD, BE,andCF, respectively. It
follo ws, in particular, that for a Heron triangle the sine of each angle is rational.
As a first example, tak e AB =13, AC =15, BC =14, AD =12, BD =5,
CD =9,and[ABC]=84. This is the union of a (5, 12, 13) triangle and a threefold
enlargement of a (3, 4, 5) triangle. The triangle with BC =4, AB =13, AC =15,
and [ABC]=24is the difference of these triangles. These examples, illustrated in Fig.
1.4, involve a component right-angled triangle that is an enlargement of a primitive
right-angled triangle. However, Heron triangles exist that are the union or difference
of two primitive Pythagorean triangles, and we now obtain formulae for their side
lengths.
Theorem 1.3.1 (Heron triangles with even height and primitiv e components) Let

ABC be a Heron triangle, as defined above , with integer sides, integ er ar ea, and
integer altitude AD.IfAD is even and the Heron triangle is built from two primitive
right-angled triangles, then positive integer parameters w, x, y, and z exist with
wx > yz and wy > xz such that AB = w
2
x
2
+ y
2
z
2
, AC = w
2
y
2
+ x
2
z
2
, and
either BC =(w
2
− z
2
)(x
2
+ y
2
) and [ABC]=wxyz(x
2

+ y
2
)(w
2
− z
2
) when
ABC is acute or BC =(w
2
+ z
2
)|x
2
−y
2
| and [ABC]=wxyz(w
2
+ z
2
)|x
2
−y
2
|
when ABC is obtuse.
Proof Take the altitude AD =2uv =2pq; then we have integers w, x, y,andz such
that u = wx, v = yz, p = wy,andq = xz,whereu>vand p>q, u and v are
Area 84
5
A

BC
9
13 15
14
Area 24
5
A
B
C
4
1312 15
9
(a) (b)
Fig. 1.4 (a) The acute-angled Heron triangle, and (b) the obtuse-angled Heron triangle, with
shortest side lengths.
Integer-sided triangles
9
coprime and of opposite parity, and p and q are coprime and of opposite parity. Then
AB = u
2
+v
2
, AC = p
2
+q
2
, BD = u
2
−v
2

, CD = p
2
−q
2
,andBC = BD+CD
in the acute case, and BC = |BD −CD| in the obtuse case. ✷
As an example, take w =2, x =3, y =5,andz =1.Thenu =6, v =5, p =10,
and q =3.SoAB =61, AC = 109,andBC = 11 + 91 = 102 in the acute case and
BC =91− 11 = 80 in the obtuse case. In both cases AD =60. The acute case is
illustrated in Fig. 1.5.
Theorem 1.3.2 (Heron triangles with odd height and primitive components) With
the same hypotheses as Theorem 1.3.1, but with the altitude AD being an odd in-
teger, then positive inte ger parameters w, x, y, and z exist with wx > yz and
wy > xz such that AB =
1
2
(w
2
x
2
+ y
2
z
2
), AC =
1
2
(w
2
y

2
+ x
2
z
2
), and either
BC =
1
2
(w
2
− z
2
)(x
2
+ y
2
) and [ABC]=
1
4
(w
2
− z
2
)(x
2
+ y
2
)wxyz when ABC
is acute, or BC =

1
2
(w
2
+ z
2
)|x
2
− y
2
| and [ABC]=
1
4
(w
2
+ z
2
)|x
2
− y
2
|wxyz
when ABC is obtuse.
Proof Taking the altitude AD = u
2
− v
2
= p
2
− q

2
with u>v, p>q, u and v
coprime and of opposite parity, and p and q coprime and of opposite parity, then we
have odd integers w, x, y,andz such that u =
1
2
(wx + yz), v =
1
2
(wx − yz), p =
1
2
(wy + xz),andq =
1
2
(wy −xz).ThenAB = u
2
+ v
2
, AC = p
2
+ q
2
, BD =2uv,
CD =2pq,andBC = BD + CD in the acute case and BC = |BD − CD| in the
obtuse case. ✷
Area 3060
11
A
B C

91
61 60 109
102
Fig. 1.5 A primitive Heron triangle.
10
Integer-sided triangles
As an example, take w =3, x =5, y =7,andz =1.Thenu =11, v =4,
p =13,andq =8.SoAB = 137, AC = 233,andBC = 88 + 208 = 296 in the
acute case and BC = 208 − 88 = 120 in the obtuse case. In both cases AD = 105.
Heron triangles with component right-angled triangles that are not primiti ve are
even more common. They can be constructed in a similar manner by choosing integers
h, k, u, v, p,andq such that either 2huv =2kpq or k(p
2
−q
2
) or h(u
2
−v
2
)=2kpq
or k(p
2
−q
2
). We give an example of a Heron triangle with non-primitive components
in which h =3, u =4, v =1, k =2, p =3,andq =2,sothat2huv =2kpq.The
v alues of the parameters h and k show that one component triangle is an enlargement
by a factor of 3 of an (8, 15, 17) triangle, and that the other component triangle is an
enlargement by a f actor of 2 of a (12, 5, 13) triangle. Hence AB =51, AC =26,
AD =24,andBC =55when ABC is acute and BC =35when ABC is obtuse.

An alternative and instructiv e method of characterising Heron triangles is as fol-
lows. Suppose that AD = h, AB = c, AC = b, BD = n,andCD = m,whereh, b,
c, m,andn are integers. We have h
2
= c
2
−n
2
= b
2
−m
2
,soc
2
+ m
2
= b
2
+ n
2
.It
follo ws that integers p, q, r,ands exist so that
b =
1
2
(pr + qs) ,
c =
1
2
(pq + rs) ,

m =
1
2
(pr − qs) ,
n =
1
2
(pq − rs) .
(1.3.2)
Here, either p, q, r,ands are all odd, or p and s are both ev en, or q and r are both
even (or three of p, q, r,ands are ev en). Then h
2
= c
2
− n
2
= pqrs. Now it is very
easy to choose four integers so that their product is a perfect square, and ev ery such
choice leads to two Heron triangles, an acute one with BC = m + n or an obtuse
one with BC = |m − n|.Asanexample,letp =6, q =3, r =4,ands =2.Then
AB =13, AC =15, AD =12,andBC =14or 4. More generally, one can choose
p = klmw
2
, q = ktux
2
, r = ltvy
2
,ands = muvz
2
.

Note from formulae (1.3.1) that [ABC]=abc/4R = rs,whereR and r are the
circumradius and inradius of the triangle ABC, respectively, and it follows that in a
Heron triangle both R and r are rational.
Integer-sided triangles
11
Exercises 1.3
1.3.1 Show that if cos B = n/c and cos C = m/b,whereb, c, m,andn are defined
in terms of p, q, r,ands as in the penultimate paragraph of Section 1.3, then
cos A =
ps(q + r)
2
− qr(p − s)
2
(pq + rs)(pr + qs)
.
1.3.2 Prove that there are an infinite number of Heron triangles whose side lengths
are consecutive integers.
1.3.3 Find all Heron triangles with an altitude of 40.
1.3.4 Show that an infinite number of Heron triangles exist in which two side
lengths are perfect squares.
1.3.5 Prove that, in a Heron triangle in which the sides have no common factor,
two of the sides are odd and one is even.
1.4 The rectangular box
We consider three interesting possibilities that arise in connection with a rectangular
parallelepiped with integer side lengths a, b,andc.
The first problem, and one to which a complete answer can be given, is whether
a, b,andc can be chosen so that the main diagonals are integers. That is, for which
positive integers a, b,andc does an integer d exist such that a
2
+b

2
+c
2
= d
2
?Thisis,
of course, nothing more than the three-dimensional generalisation of the right-angled
triangle problem, and its solution leads to Pythagorean quartets,suchas(1, 2, 2, 3)
and (2, 3, 6, 7). See below for a complete solution to this problem.
The second problem is a more sophisticated one and asks whether a, b,andc can
be chosen so that all three face diagonals are integers. T hat is, for w hich inte gers a, b,
and c do three integers A, B,andC exist so that b
2
+ c
2
= A
2
, c
2
+ a
2
= B
2
,and
a
2
+ b
2
= C
2

? One might imagine that cases would be rare or even non-existent, but
the surprising result is that a two-parameter system of solutions exists, which is in 1–1
correspondence with the primiti ve Pythagorean triples.
Evidence from similar problems (see Sections 2.4 and 6.4) indicates that other
parametric systems of solutions probably exist and that sporadic solutions may also
exist. T h e triple with least positive a, b,andc is (44, 117, 240). See below f or further
details about these results.
The third problem is whether a, b,andc exist so that simultaneously all face
diagonals and leading diagonals are integers. This is an open question and it is a v ery
difficult problem, not only because of our incomplete knowledge of the solutions of
the second problem, but also because the existence of a fourth equation raises the
problem to a higher le vel of difficulty.
12
Integer-sided triangles
Integer-sided main diagonal
Theorem 1. 4.1 The g eneral solution in positive integers of the equation a
2
+b
2
+c
2
=
d
2
is
a = k(p
2
+ q
2
− u

2
−v
2
) ,
b =2k(pu − qv) ,
c =2k(qu + pv) ,
d = k(p
2
+ q
2
+ u
2
+ v
2
) ,
(1.4.1)
where p, q, u, and v have no common factor, one or three of p, q, u, and v are odd,
p
2
+ q
2
>u
2
+ v
2
, pu > qv, qu > −pv, and k is a positive integer.
Outline pr oof As the equation is homogeneous the factor k accounts for a ny common
factor, so we need only consider the case in which a, b, c,andd have no common
factor. This means that a, b,andc cannot all be even. Neither can two or three of them
be odd, since d

2
cannot equal 2 or 3 (mod 4). H ence we may suppose that a is odd, b
and c are even, and d is odd.
Writing the equation as (b +ic)(b −ic)=(d − a)(d + a) and working over the
Gaussian integers, we may factorise into Gaussian primes and use the property of
unique factorisation to obtain b +ic =2wz, b − ic =2w
∗
z
∗
, d + a =2zz
∗
,and
d −a =2ww
∗
. Putting z = p +iq and w = u +iv we now obtain the solution given.
In order that a and d should be odd when k =1, we must choose precisely one or
three of p, q, u,andv to be odd. The other conditions ensure that a and b are positive.
For those unfamiliar with the Gaussian integers I recommend Silverman (1997). In
chapters 33 and 34 he discusses the units and primes of the Gaussian integers and the
property of unique f actorisation necessary for the above argument to be made fully
complete. ✷
When a, b, c,andd hav e no common factor we call the Pythagorean quartet
primitive. We note that, as every positiv e integer is the sum of four squares, there is a
primitive quartet for almost all odd values of d. Any exception results solely from the
fact that at least three of p, q, u,andv must be nonzero and one or three of them odd.
Sometimes a selection leads to a quartet that is not primitive. For example, p =3,
q =1, u =2,andv =1leads to a =5, b =10, c =10,andd =15. (Cases such as
this may arise when p
2
+ q

2
and u
2
+ v
2
have a common factor, and it is more trouble
than it is worth to exclude these cases.) In Table 1.3 we list the primitive Pythagorean
quartets with d<20 and in Fig. 1.6 we illustrate one case.
Integer-sided face diagonals
A two-parameter set of solutions is given below i n eqn (1.4.2), in which a, b, c, A,
B,andC are integers satisfying the equations b
2
+ c
2
= A
2
, c
2
+ a
2
= B
2
,and
a
2
+ b
2
= C
2
. A derivation is not possible since other solutions exist. It is left to the

reader to verify that the above three equations are satisfied.