TrianglesOn

with rational altitudes,angle bisetors or medians

Ralph H. BuhholzNovember 1989 (PhD Thesis, University of Newastle)November 1999 (Seond orreted, LaTeXed edition)

ContentsI Prependa

vii

Abstrat

ix

Aknowledgements

xi

II Thesis

xiii

Introdution

0.1 Historial Perspetive . . . . . . . . . . . . . . . . . . . . . . . .0.2 Personal Perspetive . . . . . . . . . . . . . . . . . . . . . . . . .1 Three Integer Altitudes

1.11.21.31.41.51.6

Dening Equations . . . .Isoseles Restrition . . .Pythagorean Restrition .Area of Alt Triangles . . .Related Alt Triangles . . .General Parametrization .

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. 5. 6. 7. 7. 9. 13

Dening Equations . . . . . . . . .Isoseles Restrition . . . . . . . .Pythagorean Restrition . . . . . .Sides in an Arithmeti ProgressionRoberts Triangles . . . . . . . . . .Area of Bis Triangles . . . . . . . .General Parametrization . . . . . .

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3 Three Integer Medians

3.13.23.33.4

13

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2 Three Integer Angle Bisetors

2.12.22.32.42.52.62.7

1

Dening Equations .Even SemiperimeterPaired Triangles . .Negative Results . .3.4.1 Isoseles . . .

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17

17182021212224

27

2728292930

ii

CONTENTS

3.53.63.73.8

3.4.2 Pythagorean . . . . . . . . . . . . . . .3.4.3 Arithmeti Progression . . . . . . . . .3.4.4 Automedian . . . . . . . . . . . . . . . .3.4.5 Roberts Triangles . . . . . . . . . . . . .Euler's Parametrizations . . . . . . . . . . . . .Two Median General Parametrization . . . . .Med Triangles . . . . . . . . . . . . . . . . . . .3.7.1 Tiki Surfae . . . . . . . . . . . . . . . .3.7.2 Plotting Med Triangles on Tiki Surfae3.7.3 Generating New Med Triangles . . . . .Euler's Parametrization Revisited . . . . . . . .

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Two Median Heron Triangles . . . . . . . . . . . . .Even Semiperimeter . . . . . . . . . . . . . . . . . .Area Divisibility . . . . . . . . . . . . . . . . . . . .Shubert's Oversight . . . . . . . . . . . . . . . . . .Dening Surfae . . . . . . . . . . . . . . . . . . . .Plotting Heron-2-Median Triangles on the -plane .Ellipti Curves on Surfae . . . . . . . . . . . . . . .Rational Points on the Ellipti Curves . . . . . . . .

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4 Heron Triangles with Integer Medians

4.14.24.34.44.54.64.74.8

5 General Rational Conurrent Cevians

5.15.25.35.45.55.65.7

Renaissane Results . . . . . . . . . . . . .Rational Cevians . . . . . . . . . . . . . . .Generating New Ceva points from Old . . .Allowable Base Segments . . . . . . . . . .Rational Cevians through the CirumentreVinents of a Triangle . . . . . . . . . . . .Tiling a Triangle with Rational Triangles .

III Addenda

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333435363839424346495255

5556575860646569

75

75777881848589

93

Epilogue

95

Bibliography

97

A Searh Program

99

B Coordinatised Triangle

105

List of Figures1231.11.21.32.13.13.23.33.43.53.64.14.24.34.44.54.65.15.25.35.45.55.6B.1

Rhind Papyrus { Problem 51 . . . . . . . . . . .Euler's Related Triangles . . . . . . . . . . . . .Pitorial Overview . . . . . . . . . . . . . . . . .Alt Triangle . . . . . . . . . . . . . . . . . . . . .Four related alt triangles . . . . . . . . . . . . . .Generating a Heron triangle . . . . . . . . . . . .Bis Triangle . . . . . . . . . . . . . . . . . . . . .Med Triangle . . . . . . . . . . . . . . . . . . . .Level urves of Tiki surfae . . . . . . . . . . . .Three dimensional view of Tiki surfae . . . . . .Med triangles in the -plane . . . . . . . . . . .Med triangles in the positive quadrant . . . . . .The ellipti urve t = 4s 150444s + 9595800 .Heron Parallelogram . . . . . . . . . . . . . . . .Heron Triangles with two integer medians . . . .Heron-2-median Contours . . . . . . . . . . . . .Heron-2-median Surfae . . . . . . . . . . . . . .Heron-2-median triangles in the -plane . . . .Torsion subgroup of y = x(x 8)(x 9) . . . .Cevians of a triangle . . . . . . . . . . . . . . . .New Ceva Point . . . . . . . . . . . . . . . . . . .Ceva Points of an equilateral triangle . . . . . . .Cevian through the irumentre . . . . . . . . .Vinent of a triangle . . . . . . . . . . . . . . . .An integer-tiled equilateral triangle . . . . . . . .The oordinatised triangle . . . . . . . . . . . . .2

3

2

iii

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1246111218284747495052596165666772768081858690106

iv

LIST OF FIGURES

List of Tables1.11.22.13.13.24.14.24.34.44.55.1

Alt Triangles . . . . . . . . . . . . . . . . . . . . . . . .-sets of the (13; 14; 15) triangle . . . . . . . . . . . . .Triangles with three rational Angle Bisetors . . . . . .Triangles with three integer medians . . . . . . . . . . .Critial Points of the Tiki surfae . . . . . . . . . . . . .Terms in equation 4.4 ongruent to zero modulo 5 . . .Heron triangles with two integer medians . . . . . . . .Heron-2-median triangles w.r.t. Shubert's Parameters .Critial Points of Heron-2-median surfae . . . . . . . .Points (; ) orresponding to Heron-2-med triangles . .Isoseles triangles with two integer vinents . . . . . . .

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913243945585960646488

vi

LIST OF TABLES

Part I

Prependa

vii

AbstratThis thesis is prinipally onerned with integer-sided triangles. Suh a triangleis alled a Heron triangle if its area is also an integer. The main theme of thethesis is a study of the relationships between Heron triangles and triangles withvarious types of evians of integer length. Cevians are lines joining the vertiesof a triangle to any points on the opposite sides.The rst hapter overs the ase of three integer altitudes and initially onsiders isoseles and Pythagorean examples. We then examine two transformations, the hinge and the pivot, whih an be used to lassify integer-sided triangles with three integer altitudes. The general parametrization emerges froman analogous one for the set of Heron triangles.The seond hapter deals with the ase of three angle bisetors of integerlength. Again we searh for isoseles and Pythagorean examples but this timeit turns out that Pythagorean triangles with three integer angle bisetors donot exist. Next we show that, somewhat surprisingly, any integer-sided trianglewith three integer angle bisetors must neessarily have integer area. This leadsto a general parametrization of suh triangles, by the same tehnique as wasused in Chapter 1.The third hapter deals with the ase of three integer medians. Previouslyknown results are that the semiperimeter must be even and that the primitivetriangles o

ur in related pairs. The searh for simple examples leads to negativeresults, for example, suh triangles annot be isoseles, Pythagorean or havesides in an arithmeti progression.One of the unsolved problems in this area asks whether there is a nonemptyintersetion between the set of Heron triangles and the set of triangles withthree integer medians. To this end we note that Euler's parametrization ofa proper subset of triangles with three integer medians does not inlude anytriangle with integer area.A omplete parametrization is given for triangles with three rational sides andtwo rational medians. This leads to a method for generating triangles with threerational medians by appliation of the tangent-hord proess to various ellipti

urves.In the fourth hapter we onvert the parametrization of Chapter 3 into a

omputer searh routine. This leads to the disovery of several examples ofHeron triangles with two integer medians. These all lie on a surfae dened by apolynomial funtion of two variables. The problem of disovering rational pointsix

xon this surfae turns out to be diult to attak with any known tehniques.The fth and nal hapter deals with rational onurrent evians througharbitrary points inside and outside a triangle. The dening equations are onsidered along with some interesting speial ases e.g. evians through the irumentre. The relationship between three rational evians and a rational tilingof the triangle is also disussed.

AknowledgementsI would like to thank my supervisor Roger B. Eggleton for his enthusiasti enouragement when I was rst exploring diophantine problems about triangles and for his subsequent expert guidane during the preparation of this thesis. Imust also thank my ating supervisors; Mike Hayes who supplied some ritialinsights into Chapter 3; Brue Rihmond whose help was greater than he antiipated and Professor Warren Brisley for his skillful assistane during the nalstages.I am grateful to Kevin Sharp, my high shool mathematis teaher of sixyears duration, for instilling in me a desire to explore and enjoy mathematis.Finally I must thank Judy whose onstant gentle nudges helped push this thesisfrom my mind into the real world.Any errors that remain in this thesis are the sole responsibility of the author.

xi

xii

Part II

Thesis

xiii

Introdution0.1 Historial PerspetiveHistorially, triangles are among the rst objets doumented to have attratedmathematial attention. This attention has frequently emphasised triangleswith integer dimensions.One of the oldest mathematial douments in existene is the Rhind papyrus[13, pp. 169-178. It was written by the Egyptian sribe A h-mose (pronounedAah-mes) in 4th month of the ood season in the 33rd year of the reign of oneof the Apophis kings sometime between 1585 B.C. and 1542 B.C. [1, p. 36.Furthermore A h-mose writes, rather matter-of-fatly, that this is just a opyof an earlier work (whih he presumably has in front of him) made for kingAmenemhet III who reigned from 1844 B.C. to 1797 B.C.In Problem 51, as translated by J. Peet [14, pp. 91-94, the sribe showshow to alulate the area of a triangular setion of land (see Figure 1). It hasa base length of 4 khet and a height of 10 khet (where 1 khet = 100 ubits and1 ubit = 523mm) whih makes the blok about 209 metres by 523 metres.While the hierati sript in Figure 1 is unreadable exept to an egyptologist,the a

ompanying gure, drawn by A h-mose three and a half millenia ago, anbe desribed by a hild. (My ve year old daughter reassured me that it isindeed a triangle).

Figure 1: Rhind Papyrus { Problem 511

2131

12720485

68127

12785

6813187

131261

255

15887

158

158

Figure 2: Euler's Related TrianglesThe anient Egyptians were pratial people who invariably solved pratialproblems. The anient Greeks on the other hand were probably the rst to studythe triangle (amongst other gures) just for its own sake. Reall Pythagorasand his theorem, Eulid's Elements and Diophantus' Arithmetia. Book 1 of theElements devotes over half of its propositions to triangles. While in Arithmetia,for example, Problem 16 of Book VI [10, pp. 240-241, asks for a Pythagoreantriangle in whih the length of the bisetor of one of the aute angles is rational.Diophantus obtained the triangle 4(7; 24; 25) whih has an angle bisetor of 35to the side 96. In Chapter 2 we will derive the general solution to this problem.(Notationally we will speify triangles by just the three sides, in inreasing order,with any nontrivial gd of the sides alone expliitly shown as a sale fator).In more reent times Giovanni Ceva (1648-1734) beame interested in theproperties of internal side dividers of triangles. The line segments joining eahvertex of a triangle to any point on the opposite side have sine beome known as

evians [4, pp. 4-5. We will denote their lengths by d ; d ; d : this thesis we willnormally onsider only primitive triangles in whih gd(a; b; ; d ; d ; d ) = 1.When he was thirty years old Ceva published a remarkably onise onditionfor the onurreny of the three evians of a triangle. Though he apparentlyproved only the neessity of the ondition, in more reent times it has also beenproved to be suient. So we have the formulation: If three evians divide thethree sides of a triangle in the ratios a : a , b : b and : in a lokwisesense, the evians are onurrent if and only if aa21 bb21

12 = 1.The rst mathematiian known to have worked on the problem of ndinginteger-sided triangles with three integer medians seems to have been Euler [6,p. 282 in 1773. Working with simplifying assumptions he produed severalsuh triangles inluding the two smallest examples namely, 2(68; 85; 87) and2(127; 131; 158) (see Figure 2).These two triangles are related in that the side lengths of the seond triangleare twie the median lengths of the rst. Euler went on [7, p. 290, in 1778, toinvestigate triangles with rational vertex-to-entroid lengths (whih neessarily1

2

3

1

1

2

1

2

1

2

2

3

0.2.

3

PERSONAL PERSPECTIVE

implies rational medians) and the following year [7, p. 399 he produed aparametrization to the original problem:a = m(9m + 26m n + n ) n(9m 6m n + n )b = m(9m + 26m n + n ) + n(9m 6m n + n )

= 2m(9m 10m n 3n ):Unfortunately this parametrization turned out to be inomplete as we will showin Chapter 3. Later [8 he produed still more parametrizations of triangles withthree integer medians but these all turn out to be essentially the same as theone shown above.In 1813 N. Fuss [5, p. 210 produed the rst examples of triangles withthree rational angle bisetors and neessarily rational area e.g. (14; 25; 25): In1905 H. Shubert [15 thought that he had parametrized all Heron triangles withone integer median and from his parametrization he dedued that no Heron triangle ould have two or more integer medians. Dikson showed [5, p. 208 thatShubert's parametrization does not over all solutions. But the ounterexamples given by Dikson only have one integer median, so the existene of Herontriangles with more than one integer median remained open. We take up andadvane this partiular problem in Chapter 3.4

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0.2 Personal PerspetiveMy earliest reolletion of triangles, at least of a mathematial nature, is ofthe delightful proofs of many of the Eulidean propositions during high shool.The organised and self-ontained nature of these proofs was (and still is) veryappealing and likely steered my areer into that of a mathematiian.I rst looked at a opy of Rihard Guy's Unsolved Problems in NumberTheory [9 early in 1985 and beame interested in several problems suh as F25,the largest persistene of a number and F4, the no-three-in-line problem beforedeiding to make a onerted eort on the innouous looking D21.While D21 itself seemed diult to solve diretly there was a rih \sea"of sub-problems in whih a mathematiian ould easily \surf" away the hours.However, following disussions with a olleague, Dr. Roger Eggleton, a moreorganised and aggressive approah was undertaken. As the work proeededit beame inreasingly obvious that the mathematial researh was more likean experimental siene than the phrase \pure mathematis" would lead us tobelieve. Constant testing of hypotheses against observations (from the omputersearhes) and searhing for patterns in the data to reveal new hypotheses beamethe order of the day.The hapters of this thesis follow similar paths. From the onsiderationof simple ases, they proeed to the more general formulation of the problem.This often leads to greater generality whih subsumes the earlier ases. Butthe simpler ases are inluded expliitly beause they give an instrutive and

onrete introdution to the more general problems and solutions presented.

4

INTEGER SIDEDTRIANGLES

2(68,85,87)

3

2

2

3 .7.11.13 .29.37 .53(248,1183,1369)

2

2.7.11 .19.31(91,250,289)

Euler(1779)

2(233,255,442)

3.13(14,25,25)

Med

2

3.5 .13(14,25,25)

Bis

Heron

AutoEqu.mdianIso.

Roberts(1893)

Alt

(31,41,49)5(5,5,6)5(7,15,20)

Pythagorean5(3,4,5)

(9,10,17)

Int

AP

(13,14,15)

Figure 3: Pitorial Overview

Chapter 1

Three Integer AltitudesThis hapter will mainly onsider the subset of integer-sided triangles whihhave three integer length altitudes. After obtaining the dening equations ofthe altitudes in terms of the sides the main theme is to derive parametrizationsfor isoseles triangles, Pythagorean triangles and nally for a general triangle.

1.1 Dening EquationsThere are at least two possible methods that an be used to obtain expressionsfor the lengths of the altitudes of a triangle in terms of the sides. One methoduses the expression for area in terms of base and height and then appeals toHeron's formula for the area in terms of the sides. An alternative path is asfollows: Denote the lengths of the sides of the triangle by a; b; and the orresponding altitudes by ; ; as in Figure 1.1. Expressing the sine and osine of\ABC in terms of the sides and one altitude and then eliminating \ABC gives4 = 4a 2 2

(a

2 2

2

b2 + 2 )2 :

By expanding the right hand side to replae the term 4a by 4a b oneobtains a more symmetri equation. Hene the three altitudes satisfy2 2

(a b )4a4a (b a )(1.1) =4b4a b ( a b ) :

=4Using the relations a = b =

= 24, where 4 is the area of the triangle,one nds the inverse system of equations, expressing the sides in terms of the2 =2

2

4b

2 2

2 2

2

2

2 2

2

2 2

2

2 2

2

2 2

2

2 2

2

2

2

5

6

CHAPTER 1.

THREE INTEGER ALTITUDES

CD

Eb

a

FB

A

c

Figure 1.1: Alt Trianglealtitudes:a2 =

4 2

4

4

4

( )4 b =(1.2)4 ( )4

=4 ( ) :Having found these expressions let us now restrit our attention to integersided triangles and analyse some simple ases.Denition : An alt triangle is a triangle with three integer sides and three integeraltitudes. The set of all alt triangles is denoted by Alt.4

2

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2 2

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2 2

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2

2

2 2

1.2 Isoseles RestritionThe simplest nontrivial ase to attak is that of the isoseles triangle. Withoutloss of generality, onsider the ase a = b. Then equations (1.1) show = and they redue to two equations for and ,4 a = (4a )4 = 4a :Combining them redues the rst to a =

, and the seond shows that mustbe even. Setting 2C := leads to C + = a , whih has the primitive solutiona = u + v , C = 2uv, = u v where u; v 2 , 2 j uv and gd(u; v) = 1. Nowsine =

=a and gd(u + v ; 2uv) = 1 and gd(u + v ; u v ) = 1, salingup the parametri solution by ensures that 2 , so the general solution in2 2

2

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N

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N

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2

1.3.

7

PYTHAGOREAN RESTRICTION

this ase is

a = b = (u2 + v2 )2

= 4uv(u2 + v2 ) = = 4uv(u2 v2 )

= u4 v 4 :

Notie that eah of the triangles produed by this parametrization generates arelated but dissimilar isoseles triangle whih also has three integer altitudes. Inpartiular the triangle 5(5; 5; 8) generates the triangle 5(5; 5; 6). This reets thealgebrai fat that the parameters C and an be interhanged. Alternatively,it reets the geometri fat that eah of the isoseles solutions is omposed oftwo opies of the same Pythagorean triangle joined along the same leg. Sinethis an be done in two possible ways we obtain two sets of solutions.

1.3 Pythagorean RestritionConsider any integer-sided right-angled triangle referred to as a Pythagoreantriangle. It already has two integer altitudes so it is neessary only to oerethe third altitude to beome an integer to obtain an alt triangle. For anyprimitive Pythagorean triangle, (where a + b = ), the sides are expressibleas a = u v , b = 2uv, = u + v where u and v are relatively prime integersof opposite parity. Then the altitudes are = b, = a and = a=. Now asbefore gd(a; ) = 1 and gd(; ) = 1 so saling up by yields the solutiona==u vb = = 2uv(u + v )

= (u + v )

= 2uv(u v ):2

2

2

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2

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2

4

4

2

2

2

2 2

2

2

1.4 Area of Alt TrianglesAt this stage one might onjeture that any alt triangle must have integer area.In fat, this turns out to be true but is not entirely obvious sine at this pointit is oneivable that there are instanes in whih all the sides and altitudesare odd. To show that this is never the ase, via Theorem 1, we need twopreliminary lemmas.Lemma 1 If a; b; ; 2 and and are the distanes from the foot of thealtitude to the endpoints of the side then ; 2 .Proof : By denition = a and = b . So there exists e ; e 2suh that = e and = e . But + = , sope + pe = :N

1

2

1

21

1

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21

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1

1

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2

2

2

1

2

N

8

CHAPTER 1.

Squaring leads to 2pe e = that1 2

THREE INTEGER ALTITUDES

e1 e2 2 N .

2

Hene there exists an n 2 suhN

4e e = n :So n = 2N for some N 2 and e e = N . If gd(e ; e ) = g > 1 thenthere exists E ; E 2 suh that e = gE , e = gE and gd(E ; E ) = 1.Substituting these gives g E E = N . Hene g divides N , so E E = mfor some m 2 . By the Fundamental Theorem of Arithmeti, there exists" ; " 2 suh that E = " and E = " . Sine g an be written as g = k swhere k; s 2 and s is squarefree, substituting into + = givespk(" + " ) s = whene s = 1 and ; 2 .

Lemma 2 Let a; b; ; ; ; 2 . Then gd(a; b; ; ; ; ) 0 (mod 2) iff gd(a; b; ) 0 (mod 2).Proof : =) If gd(a; b; ; ; ; ) 0 (mod 2) then 2 divides eah of a; b; and so gd(a; b; ) 0 (mod 2).(= If gd(a; b; ) 0 (mod 2) then there exist A; B; C 2 suh that a =2A, b = 2B , = 2C . Using the same notation as in Lemma 1, Pythagoras'sTheorem gives

+ = (2A)where 2 by Lemma 1 and 2 by assumption. Consequently and have the same parity. But from the form of Pythagorean triplets at least one of

and must be even { hene both are even. By symmetry and are evenand so gd(a; b; ; ; ; ) is even.

Theorem 1 If a; b; ; ; ; 2 and gd(a; b; ; ; ; ) = 1 then exatly one2

1 2

N

1

N

2

1

2

1

2

1 2

1

2

2

1

2

2

2

1

1

2

2

2

N

1

2

N

21

1

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2

N

1

1

1

2

2

2

N

N

N

21

1

N

2

2

N

1

1

N

of a; b and is even.

Proof : The equation in (1.1) yields the modulo 2 ongruene(a b ) 0 (mod 2):Hene a + b + 0 (mod 2) and so an odd number of the sides is even. Butthe possibility that gd(a; b; ) is even ontradits our assumption about theprimitivity of the six parameters, by Lemma 2. Hene exatly one side is even.

2

2

2 2

Corollary 1 All triangles in Alt have integer area.

Proof : Without loss of generality let the even side be . Then

=2 2 . Corollary 2 If gd(a; b; ; ; ; ) = 1 then at least two of ; ; are even.Proof : Sine 4 = a=2 = b=2 =

=2 2 then if any two sides are oddthe orresponding two altitudes must be even.

N

N

1.5.

9

RELATED ALT TRIANGLES

1.5 Related Alt TrianglesA program written to generate all primitive solutions to equations (1.1) withlargest side less than or equal to 1000 turned up 17 suh triangles. Srutinisingthe results in Table 1.1 leads one to the onlusion that the two related isoseles alt triangles mentioned earlier ould generate a third distint, salene alttriangle.By Lemma 1 any altitude of an alt triangle deomposes that triangle intotwo Pythagorean triangles. So an alt triangle has assoiated with it at most 6Pythagorean triangles whih will be referred to as the -set. The triangles ADBand CF B in Figure 1.1 belong to the same similarity lass, so they must eahbe integer-saled versions of the same primitive Pythagorean triangle. Sinethe same is true for other pairs of right triangles in Figure 1.1 the -set of anyalt triangle ontains at most three distint primitive Pythagorean triangles. To

larify the onnetion between alt triangles alluded to in the above paragraphimagine a hinge at the foot of eah altitude (i.e. D, E and F ). Rotate eahpair of Pythagoreans either side of an altitude about the hinge, keeping them

oplanar, until the old bases meet and beome a new altitude (after appropriateresaling). This leads to three potentially new alt triangles eah of whih havethe same -set. With the original triangle this yields a set of four relatedtriangles.Semiperimeters

3040451051952343404253255447008008251015104013651300

a

1525253565130136272169289175350275580260845625

Sidesb

Altitudes

20 25 201525 30 242425 40 242475 100 6028156 169 15665169 169 156120255 289 255136289 289 255240169 312 120120289 510 240240600 625 600175625 625 600336625 750 600264609 841 609580845 975 780240910 975 840780975 1000 936600Table 1.1: Alt Triangles

122015216012012024065136168336220420208728585

Area41503003001050507010140173403468010140346805250010500082500176610101400354900292500

Let f(a; b; )g represent the similarity lass of a triangle with sides (a; b; )and altitudes , , respetively. Also let f denote the rotation and appro-

10

CHAPTER 1.

THREE INTEGER ALTITUDES

priate resaling about the altitude to produe a related triangle and similarlyfor f and f . Clearly, `hinging' about the altitude transforms f(a; b; )g intothe related f(a; ba ; a )g. The new altitudes of the triangle (a; ba ; a ) are0 = a a , 0 = a , 0 = a . These transformations an be written expliitlyasf f(a; b; )g = f(a; ba ; a )gf f(a; b; )g = f(ab ; b; b )gf f(a; b; )g = f(a ; b ;

)gwhere a , a , b , b , , are the base segments as in Lemma 1. For example,hinging (a; b; ) = 5(5; 5; 6) about the altitude = 20 leads to the triangle5(5; 5; 8) as before. But hinging it about = 24 and resaling leads to thealt triangle 5(15; 20; 7), while hinging and resaling about = 24 leads to5(20; 15; 7). Now hinging a new triangle about the same altitude leads bak tothe original triangle sinef ff f(a; b; )gg = f f(a; ba ; a )g= f(aa a ; ba a0 ; a a0 )gwhereqpa0 = (ba ) (a a ) = a b a = a qpa0 = (a ) (a a ) = a a = a :Sof fff(a; b; )gg = f(aa a ; ba a ; a a )g= f(a; b; )g:Hene (f ) = 1. Similarly (f ) = 1 and (f ) = 1. Next onsider the

omposition f f . From the denitionsf fff(a; b; )gg = f f(a; ba ; a )g= f(ab0 ; ba a ; a b0 )g:whereppb0 = (a ) (a ) = a = a bppb0 = (a) (a ) = a (b ) (a ) = :Sof ff f(a; b; )gg = f(aa b ; ba a ; a )g

ab ba = f a;;

g

2

1 2

1

2

1

2

2

1

1

2

2

1

2

1

2

1

1

2

2

1

1 2

1

2

2

1

2

1 2

2

1 2

2

2

2

1

1 2

2

2

21

2

2

22

1

2 1

2

1 2

2

2

2

1

2

2

1

1

2

2

1

2

1

1

2

1

1

1 1

2 2

1 2

2

1 1

2

1

1 1

1

2

1

1

1

1

2

= f a(a ; b ;

)g

= f f(a; b; )g:1

2

1

2

1

1.5.

11

RELATED ALT TRIANGLES

fba 2ca 1

ab 1a

cb 2

bf

fa

bcf

f

fc

ac

2

bc 1

Figure 1.2: Four related alt trianglesHene f f = f . Similarly f f = f , and f f = f f = f while f f =f f = f . So under omposition these transformations form a group ating onsimilarity lasses of alt triangles. The group table is 1 f f f1 1 f f ff f 1 f ff f f 1 ff f f f 1

and the group struture is isomorphi to C C . The relationship between thesimilarity lasses of triangles is shown in Figure 1.2.Allowing the hinge at the foot of eah altitude to beome a three dimensionalpivot inreases the number of related alt triangles from 4 to innity. Now thereare eight ways in whih any pair of Pythagorean triangles about a given altitude

an meet at their \legs" to form, after appropriate resaling, a potentially newalt triangle. For the \hinge" transformation the -sets of the new alt triangleswere idential to the original -set whih led to the \early losure" of the set ofrelated alt triangles. However, the \pivot" transformation generates alt triangleswith dierent -sets and does not lose the related set of alt triangles - in fatit beomes innite in size.Begin with two arbitrary Pythagorean triangles (a; b; ) and (d; e; f ) where 2

2

12

CHAPTER 1.

na

nc

md

THREE INTEGER ALTITUDES

mf

mfncme-nb

me

nb

nc

me-nbs2

s1mf

Figure 1.3: Generating a Heron triangleand f are the hypotenuses. If they are dissimilar the legs of eah an be pairedin four ways then either adding or subtrating the areas of the Pythagoreansresults in the aforementioned eight alt triangles. For example mathing sidea with side d means that there exist integers m and n suh that na = md.Supposing, without loss of generality, that nb < me leads to an alt triangle of(n; mf; me nb) by subtrating areas and (n; mf; me + nb) by adding areas(see Figure 1.3).In the \subtration ase" the altitude to the side mf is given by 24=mfwhih is just (me nb)d=f . If mf is split into the two base segments s and s ,whih are integers by Lemma 1, then the two Pythagoreans either side of thisaltitude are similar to (s f; d(me nb); nf ) and (s f; d(me nb); f (me nb)).Using na = md and Pythagoras' Theorem shows that these are similar to (ad +be; ae bd; f ) and (e; d; f ) i.e. one \new" and one \old". Interhanging a withb leads to the two Pythagoreans (ad + be; ae bd; f ) and (bd + ae; be ad; f ).It turns out that any other pairing of the two original Pythagoreans will onlyprodue alt triangles whih have as -sets any three of f(a; b; ); (d; e; f ); (ad +be; ae bd; f ); (bd + ae; be ad; f )g. As a spei example begin with the alttriangle 65(13; 14; 15) whih has the -set of f(3; 4; 5); (5; 12; 13); (33; 56; 65)g.Pairing up the rst two Pythagorean triangles in the above eight ways leads toeight alt triangles with their assoiated -sets (see Table 1.2).Clearly the appearane of the Pythagorean triangle (16; 63; 65) whih is nota member of the original -set shows that the proess beomes self-propagating.Indeed, the union of the -sets of all the alt triangles produed from eah pairof the three Pythagoreans in the -set of 65(13; 14; 15) isf(3; 4; 5); (5; 12; 13); (33; 56; 65); (16; 63; 65); (36; 323; 325); (116; 837; 845)g:Finally sine the same proess an be applied to any pair of the Pythagoreansin a reursive manner the union of the -sets ontain only those Pythagoreanswith hypotenuses of the form 5 13 for non-negative integers and . For the1

1

2

2

1.6.

13

GENERAL PARAMETRIZATION

two arbitrary initial Pythagorean triangles the hypotenuses in the -set are ofthe form f .Legs Alt Triangle-set4,12 (13; 14; 15) f(3; 4; 5); (5; 12; 13); (33; 56; 65)g(4; 13; 15) f(3; 4; 5); (5; 12; 13); (16; 63; 65)g3,12 (13; 20; 21) f(3; 4; 5); (5; 12; 13); (16; 63; 65)g(11; 13; 20) f(3; 4; 5); (5; 12; 13); (33; 56; 65)g4,5 (25; 52; 63) f(3; 4; 5); (5; 12; 13); (16; 63; 65)g(25; 33; 52) f(3; 4; 5); (5; 12; 13); (33; 56; 65)g3,5 (25; 39; 56) f(3; 4; 5); (5; 12; 13); (33; 56; 65)g(16; 25; 39) f(3; 4; 5); (5; 12; 13); (16; 63; 65)gTable 1.2: -sets of the (13; 14; 15) triangle

1.6 General ParametrizationBy virtue of the equations a = b =

= 24, all Heron triangles must havethree rational altitudes while by Theorem 1, all alt triangles have integer area.Consequently we an sale up Heron triangles to produe alt triangles and besure that no alt triangle is exluded by this proess.Carmihael [3, p. 12 showed that all Heron triangles have sides proportionalto a, b and wherea = n(m + k )b = m(n + k )(1.3)

= (m + n)(mn k )for some integers m; n; k suh that mn > k . Note that dierent sets of triples(m; n; k) an produe the same primitive Heron triple e.g.(m; n; k) = (2; 1; 1) =) (a; b; ) = (5; 4; 3)(m; n; k) = (3; 1; 1) =) (a; b; ) = 2(5; 3; 4)(m; n; k) = (3; 2; 1) =) (a; b; ) = 5(4; 3; 5):To eliminate this degeneray it is useful to invert equations (1.3) to obtain m,n, k in terms of a, b, .Lemma 3 For any Heron triple (a; b; ) the parameters (m; n; k ) of equations2

2

2

2

2

2

1.3 are given by

m = (a b + )(a + b + )n = (b a + )(a + b + )k = 44:

14

CHAPTER 1.

THREE INTEGER ALTITUDES

Proof : Three useful linear ombinations of equations (1.3) area + b + = 2mn(m + n)a + b = 2k (m + n)a b = (m n)(mn k ):Dividing the rst two of these yields

a+b+mn = ka+b and substituting this into the third equation leads to(a + b )(a b) :m n=2kThe seond equation yields m + n = a kb2 hene solving these last two for mand n leads to(a + b )(a b + )m=4k(a + b )( a + b + ):n=4k2

2

2

2

+2

2

2

Substituting these two expressions into the expression for the produt mn yieldsk in terms of a, b and only i.e.(a + b ) (a b + )( a + b + ) :k =16 (a + b + )Using this to eliminate k from the expressions for m and n one obtains(a b + ) (a + b + )m =4( a + b + )( a + b + ) (a + b + ) :n =4(a b + )3

6

2

2

3

2

3

Multiplying the expression for k by (a + b + )=(a + b + ) and rearranging leadsto4(a + b ) :k =

(a + b + )Saling up these last three expressions by the fator 4(a + b + ) (a b + )(b +

a) to make them integers leads tom = (a b + ) (a + b + )n = (b a + ) (a + b + )k = (44)6

3

2

3

3

3

3

3

3

3

3

1.6.

15

GENERAL PARAMETRIZATION

from whih we obtain the required result.

Seleting any partiular Heron triangle (a; b; ) it is possible to permutethe sides in 6 ways to produe 6 potentially distint orresponding sets of theparameters (m; n; k). To selet exatly one member from this set of six one needsimply hoose a b . This inequality leads toa b + b a + :So by Lemma 3 we have m n. Furthermore b implies by equations (1.3)that k m n=(2m + n). To exlude any sale multiples of smaller Herontriangles reappearing a nal onstraint is that gd(m; n; k) = 1. Hene we haveproved:2

2

Theorem 2 Exatly one member of eah similarity lass of Heron triangles is

obtained from Carmihael's parametrization by applying the onstraints

gd(m; n; k) = 1; m n 1mn > k2

m2 n2m + n :

Now to obtain a general parametrization for alt triangles substitute (1.3)into the equations a = b =

= 24 whih results in2km(m + n)(mn k )=m +k2kn(m + n)(mn k )=n +k

= 2kmn:To ensure that and are both integers sale up the triangle by the fator(m + k )(n + k ). So the Heron triangle dened by equations (1.3) gives riseto the alt triangle dened bya = n(m + k ) (n + k )b = m(m + k )(n + k )

= (m + n)(mn k )(m + k )(n + k ) = 2km(m + n)(mn k )(n + k ) = 2kn(m + n)(mn k )(m + k )

= 2kmn(m + k )(n + k ):A question that now arises is the following. Is the sale fator from primitiveHeron triangle to primitive alt triangle an integer? Consider all alt trianglessuh that gd(a; b; ; ; ; ) = 1 and let g := gd(a; b; ). Then g 1 (mod 2)by Lemma 2. Also by Theorem 1 exatly one of a, b and is even so without lossof generality let 2A := a hene g j A. Now 4 = a=2 = A while from equations2

2

2

2

2

2

2

2

2

2

2

2 2

2

2

2

2

2 2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

16

CHAPTER 1.

THREE INTEGER ALTITUDES

(1.1) g j so that g j 4. Consequently the greatest ommon divisor of a, b andis also a divisor of the area and an be removed, leaving a primitive Herontriangle of the form (a=g; b=g; =g; 4=g ). In summary, there is a one-to-one

orrespondene between the similarity lasses of alt triangles and the similarity

lasses of Heron triangles given by the parametrization above. Furthermore alttriangles are related by their deomposition into related sets of Pythagoreantriangles.2

2

Chapter 2

Three Integer AngleBisetors2.1 Dening EquationsIn this hapter we will onsider integer-sided triangles with integer length anglebisetors. Denote the bisetors by p, q, r and, as before, denote the sides by a,b, as in Figure 2.1. The sine rule applied to triangles BF C and AF C leads toBFsin C=2

= sinr B

BFsin C=2

and

= sinr A :

Eliminating the length BF from these two expressions yieldsr

11C= sin B + sin A 1 os2 :Sine the sines and osines of the angles an be expressed as funtions of thesides (via the osine rule) the angle bisetor r an be expressed in terms of thesides of the triangle. By analogous reasoning on the other two pairs of trianglesone obtains the three dening equations for the angle bisetorsb(b + a)(a + b + )p =(b + )a(a + b)(a + b + )q =(2.1)(a + )ab(a + b )(a + b + ):r =(a + b)Denition : A bis triangle is a triangle with three integer sides and three integerangle bisetors. The set of all bis triangles will be denoted by Bis.

r

2

2

2

2

2

2

17

18

CHAPTER 2.

THREE INTEGER ANGLE BISECTORS

Cra

b

D

E

qB

p

**

Fc

A

Figure 2.1: Bis Triangle

2.2 Isoseles RestritionAs in Chapter 1, it is onvenient to begin with the relatively simple task ofdeiding the existene or otherwise of isoseles bis triangles. If we require that

ase a = b, then p = q by equations (2.1), whih redue to4r = 4a p = (2a + a)=(a + ) :2

2

2

2

2

2

2

From the rst of these must be even so let 2C := . This leads to thePythagorean equation r +C = a , whih has the general solution of a = u +v ,C = u v , r = 2uv where u; v 2 , 2 j uv and gd(u; v) = 1. Substitutingthese into the seond equation of the above pair leads to2u(2u 2v ) pu + v :p=(3u v )Sine we require p to be an integer the expression under the radial must be thesquare of an integer, say w, and so u + v = w . Again resorting to the generalsolution of the Pythagorean equation leads to u = 2xy, v = x y , w = x + ywhere x; y 2 , 2 j xy and gd(x; y) = 1, or the orresponding equations with uand v interhanged. Saling up the triangle by the fator 3u v gives us thegeneral parametrization of isoseles bis triangles:2

2

2

2

2

2

2

N

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

N

a = b = (u2 + v2 )(3u2 v2 )

= (2u2 2v2 )(3u2 v2 )p = q = 2uw(2u2 2v2 )r = 2uv(3u2 v2 )

(2.2)

2.2.

19

ISOSCELES RESTRICTION

where (u; v; w) = (2xy; x y ; x + y ) or (x y ; 2xy; x + y ). Notie that Cand r ould have been interhanged earlier whih would have led to a dierentexpression for p, namely4uv(u + v) p2u + 2v :p=(u + 4uv + v )Now the requirement that p be an integer leads to 2u + 2v = w . Clearly wmust be even, so dene W by 2W := w. Then u + v = 2W and u and v havethe same parity. Dening U and V by U + V := u and U V := v leads toU + V = W . Solving this in the usual way yieldsu = x + 2xy yv = y + 2xy x or x 2xy yw = 2x + 2y :This time sale up by the fator u + 4uv + v , to give the parametrizationa = b = (u + v )(u + 4uv + v )

= 4uv(u + 4uv + v )(2.3)p = q = 4uvw(u + v)r = (u v )(u + 4uv + v ):Substituting the expressions for u, v and w into equations (2.3) shows that thisparametrization is just four times the one in equations (2.2). So nally theprimitive isoseles bis triangles are given expliitly by(1) If x y < 2xy thena = b = (x + y ) (14x y x y )

= 2(6x y x y )(14x y x y )p = q = 8xy(x + y )(6x y x y )r = 4xy(x y )(14x y x y )(2) and if x y > 2xy thena = b = (x + y ) (3x 10x y + 3y )

= 2(x 6x y + y )(3x 10x y + 3y )p = q = 4(x y )(x 6x y + y )r = 4xy(x y )(3x 10x y + 3y ):Note that only one of the above pair will produe a bis triangle for any partiularx; y 2 . This leads to no obvious lassiation for bis triangles, as happenedfor alt triangles (and as will happen for triangles with three integer medians).2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2 2

2 2

2 2

2

4

4 4

2

2

2

2

2 2

4

4

2 2

2 2

4

4

2 2

4

4

4

4

2

2

4

2 2

4

4

2

N

2

2

2

2

2

2

2

2 2

4

4

2

4

2 2

4

4

2 2

4

2 2

4

4

20

CHAPTER 2.

THREE INTEGER ANGLE BISECTORS

2.3 Pythagorean RestritionDo there exist any Pythagorean bis triangles? The answer in the negative isobtained from Theorem 3. However in the proess of proving this we obtain ageneral parametrization of Pythagorean triangles with one integer angle bisetor, whih turns out to be a solution to Diophantus' Problem 16 Book VI.Theorem 3 If a2

+b =

then at most one of the angle bisetors of thetriangle (a; b; ) an have rational length.2

2

Proof : Substituting a + b = into the dening equations (2.1) gives2

2

2

p2 = 2b2 =(b + )q2 = 2a2 =(a + )r2 = 2a2 b2 =(a + b)2 :

But the third equation leads tor2 (a + b)2a2 b2

=2

pwhih annot be solved in integers sine 2 is irrational. Hene (a; b; ) 62 Bis.Now sine (a; b; ) form a Pythagorean triangle let us assume without loss ofgenerality that a = u v , b = 2uv, = u + v where 2 j uv and gd(u; v) = 1.Then p and q are given byp(u v ) u +vp=p u2uv 2u + 2v :q=u+vFor both p and q to be integers we require that u + v and 2u + 2v besimultaneously squares of integers. But this is impossible as u + v = w and2u + 2v = x imply that 2w = x . So nally letting u = 2UV , v = U V ,w = U + V where 2 j UV and gd(U; V ) = 1 leads to the parametrization ofPythagorean triangles with one integer angle bisetor,2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

a = 2UV (U 2 V 2 )2b = 8U 2 V 2 (U 2 V 2 )

= 2UV (U 2 + V 2 )2q = (U 2 + V 2 )(6U 2 V 2 U 2 V 2 )

providing the required result.

2.4.

21

SIDES IN AN ARITHMETIC PROGRESSION

2.4 Sides in an Arithmeti ProgressionFor ompleteness regarding the set intersetions of Figure 2.1 we will now onsider bis triangles in whih the sides are in an arithmeti progression. Let(a; b; ) := (e d; e; e + d). Then from equations (2.1) we get4e q = (e d )f(e d) + (e + d) + 2(e d ) e g4q = 3(e d ):Clearly 3 j q so set 3Q := q to give the equation d +3(2Q) = e whose generalsolution isd = u 3ve = u + 3v2Q = 2uv:Now onsider the p and r equations from (2.1). They give(2e + d) p = e(e + d)(3e + 6ed)(2e d) r = e(e d)(3e 6ed):Substituting for d and e in terms of u and v leads to(3u + 3v ) p = 32(u + 3v ) (2u )(u v )(u 9v )2r = 32(u + 3v ) (2v )( u + 9v ):So for p and q to be rational we require that 2(u v ) and 2(9v u ) aresimultaneously squares. So letting x := 2(u v ), and y := 2(9v u ), weobtain after rearrangementx + y = (4v)9x + y = (4u) :By a result of Collins [5, p. 475 this last pair of quadrati forms annot besquares together. Note that this was also proved by Diophantus and beamehis Proposition 15. We also give an independent proof in Lemma 4.2 2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2 2

2

2 2

2

2

2 2 2

2

2 2

2

2

2

2

2 2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2.5 Roberts TrianglesIn 1893, C.A. Roberts [5, p. 198 showed that any triangle with sides given by(a; b; ) = (x + 2y ; x + 4y ; 2x + 2y ) has integer area. We shall refer tothem as Roberts triangles. As we shall show in Chapter 3, the motivation for

onsidering Roberts' triangles is that no suh triangle has three integer medians.It turns out that a Roberts triangle annot have three integer angle bisetorseither.2

2

2

2

2

2

22

CHAPTER 2.

THREE INTEGER ANGLE BISECTORS

Substitute Roberts' parametrization into equations (2.1) to give9p = 16(x + 4y )(x + y )(3x + 4y )q = 16x (x + 2y ) (x + y )(2x + 6y )r = 16y (x + 2y ) (x + 4y ):Now rearranging these and taking the square root we nd thatpp4x + 4y x + yp=3p4x(x + 2y ) x + yq=3x + 4py2y(x + 2y ) x + 4yr=:2x + 6ySo if we require p, q and r to be rational we must have x + 4y and x + ysimultaneously squares of integers. This is also impossible, again by Collins'result [5, p. 475.2

2

2

2

2

2

2

2

2

2

2 2

2

2

2

2

2

2

2 2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2.6 Area of Bis TrianglesW. Rutherford [5, p. 212 showed that triangles in whih all three angle bisetorshave rational length must also have rational area. Here we will review andextend this result to show that bis triangles have integer area. Of ourse thisdoes not mean that all Heron triangles have rational angle bisetors - the set ofPythagorean triangles is an obvious ounterexample.Theorem 4 The area of any bis triangle is rational.

Proof : If we rewrite equations (2.1) in terms of the semiperimeter s thenwe getp (b + ) = 4bs(s a)q (a + ) = 4as(s b)r (a + b) = 4abs(s ):The produt of these three expressions givesp (b + ) q (a + ) r (a + b) = 64a b s 42

2 2

2

2

2

2

2

2

2 2

2

2 2 2 2

2

so 4 = pqr bab

aa b a b . Hene the area is learly rational when a; b; ; p; q; r 2.

( + )( + )( + )4( + + )

N

Theorem 5 If (a; b; ) 2 Bis then the perimeter of the triangle must be even.

2.6.

23

AREA OF BIS TRIANGLES

Proof : Let P := a + b + . Assuming that P is odd leads to two ases.Case (i): a; b; 1 (mod 2). From equations (2.1) we nd thatp(b + ) b mod 2:But this is impossible sine b + is even while b is odd.Case (ii): a; b 0 (mod 2), 1 (mod 2). Now suppose 2m k a (that is,the largest power of two dividing a is 2m ) and 2n k b, where without loss ofgenerality we take m n 1. Then let := a=2m and := b=2n, where and are both odd. Substituting into equations (2.1) givesp (2n + ) = 2n(2m + 2n + )( 2m + 2n + )q (2m + ) = 2m(2m + 2n + )(2m 2n + ):Sine is odd these imply that 2n k p and 2m k q so that 2 j n and 2 j m.Dividing out the powers of two from both equations leads to1 (mod 4)1 (mod 4):Now let A := a=2n where A an be even or odd. Then the last of equations(2.1) implies thatr (A + ) = A (2n (A + ) + )(2n (A + ) )r (A + ) A (mod 4):This last equation will turn out to be impossible by onsideration of the dierent

ases of parity of r and A + . If both r and A + are odd then A must be even,sine is odd. This leads to a ontradition as r (A + ) 1 (mod 4) whilethe right hand side is A 0 or 2 (mod 4). Next if r is odd and A + is eventhen A must be odd so r (A + ) 0 (mod 4) and A 1 or 3 (mod 4).These two ases show that r must be even so A must be divisible by 4. In fat,if 2k k r then 2 k k A and m = n + 2k where k 1. Now set := r=2k andusing and as dened earlier we obtain from equations (2.1) (2 k + ) = (2m + 2n + )(2m + 2n )1 (mod 4):Reall that 1 (mod 4) whih implies that (mod 4) so that wehave 1 (mod 4), the nal ontradition for Case (ii).

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Corollary 3 The area of any bis triangle is an integer.

Proof: Sine the semiperimeter of any bis triangle must be an integer byTheorem 5 and the area is rational, by Rutherford's result, Heron's formula forthe area of a triangle yields the required result.

It is not diult to implement a program to searh for bis triangles and usingCorollary 3 one an disregard any triangles without integer area. The results of

24

CHAPTER 2.

THREE INTEGER ANGLE BISECTORS

one suh searh are listed in Table 2.1, and sine the numbers are so large bistriangles are listed with the greatest ommon divisor of the sides removed. Theyseem to o

ur less frequently than alt triangles did. For example the smallestbis triangle, (546; 975; 975) whih is just a saled version of N. Fuss' examplementioned in the Introdution, is the only one with all sides less than 1000.Semiperimeters

32189224288297315385450352483

a

Sidesb

Angle Bisetors

p

q

r

560392620825326002174256407316817464138117815216422061148048415101745344

560391260020930800279

14 25 25 2484 125 169125 154 169120169 169 238150 169 27591 250 289231 250 289240289 289 32277 289 338289 338 399Table 2.1: Triangles with three rational Angle Bisetors9757480483237425640715015742040077170071642206111768057248976737

2340292730034192400481

10472183265211

Area4168504092401428011088109202772038640924047880

2.7 General ParametrizationSine all bis triangles have integer area we an use Carmihael's parametrization,as was done in Chapter 1, to obtain a parametrization of bis triangles. Rewritingequations (1.3) here, for onveniene of referene, we havea = n(m + k )b = m(n + k )

= (m + n)(mn k ):From equations (2.1) we have2

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1 (b +a )

bq = a 1(a + )

p2 = b

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2.7.

25

GENERAL PARAMETRIZATION

Meanwhile equations (1.3) imply

m +k=b + 2mn + m kb= n +ka + 2mn + n k

= mn k :a + b mn + ka

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Substituting the latter set of equations into the former leads to2m(m + n)(mn k ) pn + kp=2mn + m k2n(m + n)(mn k ) pm + kq=2mn + n k2mnk p(n + k )(m + k ):r=mn + kSo p, q and r are rational if and only if n + k and m + k are simultaneouslyperfet squares. This leads to four potentially dierent ases depending onwhih leg of a Pythagorean triple is represented by eah of the parameters m,n and k. These ases are now onsidered in turn.Case (i): Assume thatm = u v , k = 2uv, where u; v 2 , 2 j uv and gd(u; v) = 1;n = x y , k = 2xy, where x; y 2 , 2 j xy and gd(x; y) = 1.This requires that uv = xy := say, where u = , v = , x = , y = .Substituting these relations into Carmihael's equations (1.3) yieldsa = ( )( + )b = ( )( + )(2.4)

= ( + )( )[( )( ) (2) whih produes a triangle with three rational angle bisetors for any hoie of; ; ; 2 .Case (ii): Assume thatm = 2uv, k = u v , where u; v 2 , 2 j uv and gd(u; v) = 1;n = 2xy, k = x y , where x; y 2 , 2 j xy and gd(x; y) = 1.This requires that u v = x y . But the transformationsu = U + V; v = U V; x = X + Y; y = X Ylead to m, n and k having the same form as in Case (i), so this leads to thesame parametrization.Case (iii): Assume that2

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26

CHAPTER 2.

THREE INTEGER ANGLE BISECTORS

m = 2uv, k = u2 v2 , where u; v 2 N ,

2 j uv and gd(u; v) = 1;n = x y , k = 2xy, where x; y 2 , 2 j xy and gd(x; y) = 1.Now equating the two k's requires that u v = 2xy, so u and v must have thesame parity. Let u = U + V , v = U V and x = 2X . Then UV = Xy := say, where U = , V = , X = , y = . Consequently, u = + ,v = , X = 2 , and y = . Substituting these relations bak intoequations (1.3) gives the seond parametrizationa = 4(4 )( + )b = 2( )(4 + )(2.5)

= [2( ) + 4( ) [2( )(4 ) (4) :Case (iv): Assume thatm = u v , k = 2uv, where u; v 2 , 2 j uv and gd(u; v) = 1;n = 2xy, k = x y , where x; y 2 , 2 j xy and gd(x; y) = 1.Sine equations (1.3) are symmetri in m and n, this ase turns out to be thesame as ase (iii) by interhanging m and n. To produe bis triangles fromthese parametrizations it only remains to sale up equations (2.4) and (2.5) by(mn + k )(2mn + n k )(2mn + m k ).2

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Chapter 3

Three Integer Medians3.1 Dening EquationsIn this hapter we onsider integer-sided triangles in whih the evians bisetingthe opposite sides are of integer length. In other words, this hapter treats the

ase of three integer medians. To obtain the dening equations we note thatthe osine rule for triangle ADC (see Figure 3.1) givesAD = CD + AC 2AC CD os C;while for triangle ABC we getAB = BC + AC 2BC AC os C:Eliminating os C between these two equations produes an expression for themedian in terms of the sides of the triangle, 4AD = 2AC + 2AB BC .Hene if the lengths of the sides and median are integers then BC must beeven. It follows similarly that all three sides have even length. For the rest ofthis hapter we will denote the lengths of the sides by 2a, 2b, 2 and the lengthsof the medians by k, l, m. By repeating the above proess we nd that themedians are dened byk = 2b + 2 al = 2 + 2a b(3.1)m = 2a + 2b :These equations an be rearranged to give the sides in terms of the medians9a = 2l + 2m k9b = 2m + 2k l(3.2)9 = 2k + 2l m :Denition : A med triangle is an integer-sided triangle with three integer medians. We will use Med to denote the set of all med triangles.2

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CHAPTER 3.

THREE INTEGER MEDIANS

C

m2a

D

l

E

2b

kF2c

B

A

Figure 3.1: Med Triangle

3.2 Even SemiperimeterOne of the rst patterns I notied in searhing for med triangles was that thesemiperimeter always seemed to be even. I assumed that this was always the

ase and proeeded to inorporate this pattern into the omputer searhes formed triangles. It was not until some time later that I thought I should try toprove this \fat". It turned out as follows.Theorem 6 If a; b; ; k; l; m 2 N andone of the half-sides is even.

gd(a; b; ; k; l; m) = 1 then one and only

Proof : Sine there are only four distint ases for the parity of the half-sideswe will eliminate three of them leaving the one we want as follows:Case (i) : If a; b 0 (mod 2) and 1 (mod 2) then by equation (3.1)k 2(b + ) a 2 (mod 4)whih is impossible.Case (ii) : If a; b; 1 (mod 2) then as beforek 2(b + ) a 3 (mod 4)whih is also impossible.Case (iii) : Finally if all three half-sides are even then by equations (3.1) allthe medians are even, ontraditing the assumption that we have a primitivetriangle.

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Corollary 4 The semiperimeter of a med triangle is even.

Proof : This is immediate from Theorem 6, sine the semiperimeter is givenby s = (BC + AC + AB )=2 = a + b + .

3.3.

29

PAIRED TRIANGLES

Corollary 5 Exatly one of the medians is even.

Proof : Without loss of generality suppose that a is even. Then k must beeven, by equation (3.1).

Corollary 6 Exatly one of a, b, , k, l, m is divisible by four.

Proof : If we assume that a = 2, b = 2 + 1, = 2 + 1 while k = 2,l = 2 + 1, m = 2 + 1 then substitution into the rst of equations (3.1) gives

us

4 + 4 = 8 + 8 + 2 + 8 + 8 + 2:So + = 2 + 2 + 2 + 2 + 1 i.e. + 1 (mod 2). So and have opposite parity and the result follows.

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3.3 Paired TrianglesAs mentioned in the introdution Euler was already aware that there is a natural

orrespondene between pairs of med triangles. Suppose that we already have amed triangle (2a; 2b; 2) with medians k; l; m. Now if we onsider a new trianglewith sides (2k; 2l; 2m) then its medians k0 ; l0 ; m0 are given byk0 = 2l + 2m kl0 = 2m + 2k lm0 = 2k + 2l m :But sine k, l and m are expressible in terms of the sides of the original trianglea, b and we havek0 = 2(2 + 2a b ) + 2(2a + 2b ) (2b + 2 a ) = 9al0 = 2(2a + 2b ) + 2(2b + 2 a ) (2 + 2a b ) = 9bm0 = 2(2b + 2 a ) + 2(2 + 2a b ) (2a + 2b ) = 9 :So the new medians k0 , l0 , m0 are respetively 3a, 3b, 3 and hene are integers.So (2a; 2b; 2) 2 Med implies that (2k; 2l; 2m) 2 Med. If we repeat the aboveproess on the triangle (2k; 2l; 2m) we obtain another med triangle, namely(6a; 6b; 6), whih is just a saled version of the original triangle. Hene thisproess generates only one new dissimilar med triangle.2

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3.4 Negative ResultsWhile the various programs I had written were searhing for med triangles withinteger area I onsidered the possibility that some sublasses of Heron trianglesmight not ontain any med triangles at all. The results of this approah tosolving D21 [9 are ontained in the next few setions.

30

CHAPTER 3.

THREE INTEGER MEDIANS

3.4.1 Isoseles

Do there exist isoseles med triangles? Beause the urrent setion deals withnegative results it would be foolish to maintain any air of suspense. The answeris No! This is an interesting result sine referring to Figure 3, just before theIntrodution, we see that examples of isoseles alt and isoseles bis triangles areknown. In fat in Chapters 1 and 2 we found the general parametrizations forall suh triangles. My rst proof of the non-existene of isoseles med trianglesrelied on muh ase work and the method of innite desent. Subsequently Ifound the following more onise method.Theorem 7 No med triangle an be isoseles.

Proof : If a = b in equations (3.1) then k = l, and we have just two equationsfor the medians i.e.k = a + 2m = 4a :The seond equation shows that m and have the same parity while Theorem 6shows that must be even (otherwise two sides would be even). So let := 2Cand m := 2M whih leads tok = a + 8CM =a C :If we onsider the seond of this pair together with the sum of the two equationswe have, in the terminology of Dikson [5, p. 475, a \disordant" pair ofquadrati forms,k = M + 9Ca =M +Ci.e. have no ommon solutions.

In our attempt to make this thesis as self-ontained as possible the following lemma will display an independent proof that these two equations have no

ommon solution in integers.Lemma 4 If x; y 2 then the expressions x + y and x + 9y annot both be2

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simultaneously squares of integers.

Proof : Let us take x + y = z and x + 9y = t . These two diophantineequations an be solved by the standard Pythagorean parametrization. If u andv are the parameters for the rst solution set while r and s are the parametersfor the seond solution set then z = u + v and t = r + s but x and y antake four distint ombinations of the parametri forms. We now onsider these

ases and eliminate them one by one.Case (i): Assume that2

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3.4.

31

NEGATIVE RESULTS

x = 2uv, y = u2 v2 , where u; v 2 N

and gd(u; v) = 1, 2 j uv; andx = 2rs, 3y = r s , where r; s 2 and gd(r; s) = 1, 2 j rs.Then equating x's gives uv = rs := say, where u := , v := , r := and s := . Thus omparing the expressions for y shows that we require3( ) = (3 ) = (3 ):Now gd(u; v) = 1 implies that gd(; ) = 1 whih means that there is someinteger " for whih we have3 = "3 = ":(3.3)Combining these two we nd that8 = ( + 3 )"8 = (3 + )":(3.4)So " j 8 otherwise " j and " j would fore " = 1. Now " = 1, 2, 4 or 8 willeah lead to a ontradition hene eliminating this rst ase.If " = 1 then by equation (3.4) and are both odd but then the equation + 3 0 (mod 8) is insoluble.If " = 2 then by equations (3.3) and (3.4) , , and are all odd whihimplies that u, v, r and s are all odd. But this ontradits the originalassumptions.If " = 4 then by equation (3.3) and are both odd but then, as before,the equation 3 0 (mod 4) is insoluble.If " = 8 then by equation (3.3) and are both odd. So that nally theequation 3 0 (mod 8) is insoluble.Case (ii): Assume thatx = u v , y = 2uv, where u; v 2 and gd(u; v) = 1, 2 j uv; andx = 2rs, 3y = r2 s2, where r; s 2 and gd(r; s) = 1, 2 j rs.Now the requirement 2rs = u v is a ontradition as u and v must haveopposite parity.Case (iii): Assume thatx = 2uv, y = u v , where u; v 2 and gd(u; v) = 1, 2 j uv; andx = r s , 3y = 2rs, where r; s 2 and gd(r; s) = 1, 2 j rs.2

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32

CHAPTER 3.

THREE INTEGER MEDIANS

As before we obtain an easy ontradition by equating the x's.Case (iv): Assume thatx = u v , y = 2uv, where u; v 2 and gd(u; v) = 1, 2 j uv; andx = r s , 3y = 2rs, where r; s 2 and gd(r; s) = 1, 2 j rs.So 3uv = rs := say, where either3u := ; v := ; r := ; s := oru := ; 3v := ; r := ; s := :Note that if we interhange with and with that the seond identiationis equivalent to the rst. So onsidering the rst set leads to( 9 ) = 9 9 ( 9 ) = (9 9 ):Now, just as in the rst ase, gd(u; v) = 1 implies that gd(; ) = 1 whihmeans that there is some integer " for whih 9 = "9 9 = ":(3.5)Combining these two we nd that8 = ( + )"72 = ( + 9 )":(3.6)Now gd("; ) = 1 and the seond of (3.6) implies that " j 72. We an nowuse innite desent to eliminate the two subases 3 j and 3 . If 3 then equations (3.6) imply that 9 j " so substituting " := 9 into(3.5) leads to 9 = 9

= :(3.7)If 2 j then and are both odd and sine at least one of or must be oddwe see that 0 (mod 8). Together with 9 j 72 this shows that = 8.Substitution into (3.6) gives us the pair of equations + =B + 9 = where B := =9 < ( =9 + ) = z . This last pair of equations isequivalent to the rst pair in the statement of the lemma. So any solution2

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3.4.

33

NEGATIVE RESULTS

would imply the existene of a smaller solution (as B < z ) and hene by innitedesent there an be no non-trivial solution.If = 1 then eliminating from equations (3.7) gives + 9 = 8 whihmeans that and are both odd. But then + 9 2 (mod 8) leads to a

ontradition. If 3 j then equation (3.6) implies that 3 " and so " j 8. Substituting = 3 into equations (3.5) redues them to 9 = "

= ":(3.8)Now as before if 2 j " then and are both odd and sine at least one of or must be odd we see that " 0 (mod 8) hene " = 8. Substitution into (3.8)gives us the familiar pair of equations + = + 9 = :This time < ( =9 + ) = z and so we have no solution again byinnite desent.If " = 1 then eliminating from equations (3.8) gives us + = 8whih means that and are both odd. But then + 2 (mod 8) leadsto a ontradition whih nally eliminates ase (iv).

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3.4.2 Pythagorean

The next subset of Heron triangles I hose to onsider was the set of integersided triangles with a right angle i.e. Pythagorean triangles (sometimes referredto as Pythagorean triples). The following theorem proves that no suh triangle

an have three integer medians.Theorem 8 If a; b; 2 suh that a + b = then the triangle (2a; 2b; 2)2

N

2

2

has exatly one integer median.

Proof : If we set a + b = in equations (3.1) we nd thatk = 4b + al = 4a + bm =a +b :So learly the median m is an integer for any Pythagorean triangle. But if wetake, without loss of generality, a = u v , b = 2uv, and = u + v then theequations for k and l beomek = u + 14u v + vl =u u v +v :Mordell shows [12, pp. 20-21 that these pair of equations have only the solutions(u ; v ) = (1; 0); (1; 1) and (0; 1) whih all lead to degenerate triangles.

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CHAPTER 3.

THREE INTEGER MEDIANS

3.4.3 Arithmeti Progression

The set of integer-sided triangles with sides in an arithmeti progression alsoturns out to have no member in ommon with the set of Med triangles. Notiethat in the following proof of Theorem 9 no mention is made of the rationalityor otherwise of the area of the triangle. This explains why, in Figure 3, theset Int is not wholly ontained within the set of Heron triangles and remainsdisjoint from Med.AP

Theorem 9 No med triangle an be have its sides in an arithmeti progression.

Proof : If (a; b; ) = (p q; p; p + q) where gd(p; q) = 1, p is even and q isodd then equations (3.1) beomek = 3p + 6pq + ql = 3p + 4q(3.9)m = 3p 6pq + qSolutions to the seond of these are(i) p = 2xy, 2q = x 3y , gd(x; y) = 1, 2 xy(ii) p = 2xy, 2q = 3x y , gd(x; y) = 1, 2 xySubstituting set (i) into (3.9) leads to4k = x + 24x y + 42x y 72xy + 9y4m = x 24x y + 42x y + 72xy + 9yAny rational solution (x ; y ) to the rst equation orresponds to two rationalsolutions ( x ; y ) and (x ; y ) of the seond equation and vie versa. So

ommon solutions to both quarti equations an only o

ur when x = x ory0 = y0 i.e. x y = 0. This leads to only degenerate triangles. Substitutingset (ii) into (3.9) leads to4k = 9x + 72x y + 42x y 24xy + y4m = 9x 72x y + 42x y + 24xy + yAs above, the only ommon rational solutions to both equations have xy = 0whih result in degenerate triangles.

Note that the equation 4k = x +24x y +42x y 72xy +9y does have rational solutions e.g. (x; y; k) = (1; 1; 1), (3; 1; 15), (7; 5; 97), or (1; 13; 163). UsingMordell's transformation (as on p.47) this is equivalent to the ubi equationT = s 147s + 610with orresponding rational solutions(s; T ) = (11; 18); (41; 252); (419=52; 6678=53); (833=132; 5004=133):2

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3.4.

35

NEGATIVE RESULTS

Using the tangent-hord proess the third rational point generates an innitenumber of rational solutions on the ellipti urve and hene there exists aninnite number of triangles with sides in an arithmeti progression and twointeger medians.3.4.4 Automedian

If the medians of a triangle are proportional to the sides then the triangle isdened to be automedian. Shortly, we will prove a neessary and suient

ondition for a triangle to be automedian. This will then be used to show thatthe set of isoseles automedian triangles is equivalent to the set of equilateraltriangles hene larifying Figure 3 a little more. The above ondition will alsoshow that no integer-sided automedian triangle an have an integer median orinteger area whih will justify our onsideration of this type.Theorem 10 A triangle with sides 2a, 2b, 2 is automedian if and only if one

of a2 + 2 = 2b2, a2 + b2 = 22 , b2 + 2 = 2a2 is true.

Proof : If a + = 2b then substitution into equations (3.1) gives the threeexpressions k = 3 , l = 3b , m = 3a and so the medians are proportionalto the sides. Similarly so for the other two onditions. On the other hand ifk= = l=b = m=a = x then equations (3.1) beomex = 2b + 2 ax b = 2 + 2a bx a = 2a + 2b :(Note that solving these equations for x leads to x = 9 and so the denitionof automedian triangles is muh more restritive than rst impressions wouldimply.) Eliminating x from the rst two of these equations gives

(2 + 2a b ) = b (2b + 2 a ):Expanding and rearranging this one obtains(2 + b )( + a 2b ) = 0:Clearly this implies that + a = 2b . If k=a = l= = m=b then b + = 2aand if k=b = l=a = m= then a + b = 2 . While the other three permutationslead to equilateral triangles or degenerate triangles (zero side), both of whihidentially satisfy the onditions as stated in the theorem.

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2

2

2

2

2

2

Corollary 7 All isoseles automedian triangles are equilateral.

Proof: If a = then from Theorem 10 we have 2a = 2b so that a = b aswell. The remaining two ases are similar.

2

2

36

CHAPTER 3.

THREE INTEGER MEDIANS

Corollary 8 An automedian triangle with integer sides annot have an integer

median.

Proof:p If a; b; are integers then k; l; m are all irrational sine k= = l=b =3.

m=a =

Corollary 9 An automedian triangle with integer sides a; b; annot have in-

teger area.

Proof: The area of a triangle is given by Heron's formula164 = 2a + 2b + 2a b a b :Eliminating by Theorem 10 gives164 = 4a b 2a b :First assume that a solution exists with gd(a; b) = 1. From the last expressionwe must have 2 j b so letting b = 2B then84 = 8a B a 8B :But now 2 j a whih is a ontradition.

2

2 2

2 2

2

2 2

4

4

4

2 2

2

2

2

4

4

4

4

3.4.5 Roberts Triangles

Reall, from Chapter 2, that a Roberts triangle whih has sides (a; b; ) = (x +2y ; x + 4y ; 2x + 2y ) for some integers x and y also has integer area. Moreexpliitly, using Heron's formula for the area of a triangle one nds that 4 =2xy(x +2y ). Notie that the above expressions for the sides lead to a + b = 3.In fat it turns out that all Heron triangles whose sides satisfy this relationshipmust be of Roberts' form. The main result in this setion is that no Roberts'triangle an belong to the set of Med triangles.2

2

2

2

2

2

2

2

Theorem 11 A Roberts' triangle annot have three integer medians.

Proof : Considering only the similarity lasses of Roberts' triangles let thesides be (2a; 2b; 2) = (X + 2Y ; X + 4Y ; 2X + 2Y ) where gd(a; b; ) = 1.Then learly X must be even. Setting X = 2x and Y = y, say and dividing outthe fator of 2 shows that the half-sides have the same parametri form as thesides, (a; b; ) = (2x + y ; 2x + 2y ; 4x + y ) where gd(x; y) = 1. Note thaty must be odd by Theorem 6. Substituting this into equations (3.1) we obtaink = (6x 3y ) + (8xy)l = 4x (9x + 4y )m = y (16x + 9y ):2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2 2

2

2

2

2

2

3.4.

37

NEGATIVE RESULTS

Now fousing attention on the latter two equations, the requirement that allthree medians be integers means there exist integers p and q whih satisfyp = (3x) + (2y)q = (4x) + (3y) :As usual, applying the solution for Pythagorean triples and the additional onstraint that y is odd leads to only one possible simultaneous solution:3x = r s , 2y = 2rs, where r; s 2 and gd(r; s) = 1, 2 rs;4x = 2uv, 3y = u v , where u; v 2 and gd(u; v) = 1, 2 j uv.Sine r and s have the same parity let r = R + S and s = R S whih leads to3x = 4RS , 2y = R S , gd(R; S ) = 1, 2 j RS ,4x = 2uv, 3y = u v , gd(u; v) = 1, 2 j uv.Now we must have 8RS = 3uv so setting both to gives(i) 8R = , S = , 3u = , v = ,(ii) R = , 8S = , 3u = , v = ,(iii) 8R = , S = , u = , 3v = ,(iv) R = , 8S = , u = , 3v = .Without loss of generality one need only onsider ase (i). Substituting this into3(R S ) = u v leads to (27 64 ) = (1728 576 ):Now gd(R; S ) = 1 implies that gd(; ) = 1 so there must exist an integer, "say, suh that1728 576 = "27 64 = ":Thus solving these two equations for and leads to153 = ( + 27 )"1088 = (3 + 64 )":Sine gd("; ) = 1 implies that " j 1088 = 2 17 while gd("; ) = 1 impliesthat " j 153 = 3 17 we have " j 17. If " = 1 or 17 then the latter equationimplies that 8 j . In all four ases the former equation redues to 3(mod 8). But is odd as gd(; ) = 1 and is even. Hene 1 (mod 8)implies that 3 (mod 8) whih is impossible.

2

2

2

2

2

N

2

2

2

-

N

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

6

2

2

2

2

2

38

CHAPTER 3.

THREE INTEGER MEDIANS

3.5 Euler's ParametrizationsIn a paper written in 1779 Euler showed that if the half-sides of a triangle aregiven by the expressions2a = 2(e f )2b = (d + e) + (d e)2 = (d + e) (d e)where d = 16 , e = ( + )(9 + ) and f = 2(9 ) for integers and then the medians are integers. This parametrization appeared severaltimes in Euler's works. In a posthumous paper of 1849 he noted thata = (m + n)p (m n)qb = (m n)p + (m + n)q(3.10)

= 2mp 2nqwhere p = (m + n )(9m n ) and q = 2mn(9m + n ) for integers m and nis suient to make the three medians integers as well. With a little algebraimanipulation it is not hard to show that these two formulations are essentiallythe same. Interhanging a with m and b with n yields the orrespondenes(a; b; )f g = 2(b; ; a)f g.To show that the parametrization (3.10) does not generate every med triangleone needs to solve for the ratio m=n in terms of a, b and .a + b + = 4mpa + b = 4nqa b + = 2(p q)(m + n):Eliminating the parameters p and q in these expressions leads to a quadrati inm and n(a + b )m + 2(a b)mn (a + b + )n = 0whenepm b a 2a + 2b =:na+b The med triangle 2(233; 255; 442) yields m=n = 5 or 93=23. For the rstvalue substitute m = 5k and n = k, where k 2 , into the equation in set(3.10). This leads to = 53720k but then k = 442=53720 is not solvablefor integer k. Similarly, substituting m = 93k and n = 23k results in =116557658136k and as before k = 442=116557658136 is not solvable forinteger k. To omplete the urrent line of argument the related med trianglealso needs to be heked. If (a; b; ) = 2(208; 659; 683) then m=n = 25=4 or31=23. Substituting m = 25k and n = 4k leads to k = 683=170742850 whih,as expeted, is insoluble for integer k. Similarly m = 31k and n = 23k leadsto k = 683=148085512. Hene neither of these two med triangles an be2

2

2

2

2

1849

2

2

2

2

4

2

2

2

1779

2

2

2

2

2

N

5

5

5

5

5

5

4

3.6.

39

TWO MEDIAN GENERAL PARAMETRIZATION

Semiperimeters

24064668079893011221200176422623248349641004350440851526240

Sides

a

b

Medians

k

l

Eulerian

m

136 170 174 158131127226 486 580 523367244290 414 656 529463142318 628 650 619404377466 510 884 683659208654 772 818 725632587554 892 954 881640569932 982 1614 1252 1223 5151162 1548 1814 1583 1312 10251620 2198 2678 2312 1921 13911018 2646 3328 2963 2075 11181754 2612 3834 3161 2680 11292446 3048 3206 2879 2410 22511678 3280 3858 3481 2482 17512802 3556 3946 3485 2924 25213810 3930 4740 3915 3825 3060Table 3.1: Triangles with three integer medians

yesnoyesyesnonoyesnonoyesnononononono

generated by Euler's parametrization. Eulerian and non eulerian med trianglesare listed in Table 3.1.Another negative result is that no med triangle from Euler's parametrization

an have integer area. If the triangle is a primitive one, i.e. gd(a; b; ; k; l; m) =1, then a and b are odd by Theorem 6. This implies that gd(m; n) = 1 andthat m and n have opposite parity. From Heron's formula and equations (3.10)the area is given by4 = 4mp [2(m n)p + 2(m n)q [2(m + n)p 2(m + n)q 4nq= 64mnpq(m n )(p q )= 128m n (34m n )(m n )(p q ):Hene if the area is to be an integer then 2(34m n )(m n )(p q ) = zfor some integer z . Note that gd(2; 34m n ; m n ; p q ) = 1 sine thelatter three terms are all odd. This learly leads to a ontradition sine z 2(mod 4) is impossible.2

2

2

2

2

4

2

4

2

4

4

2

2

4

4

4

4

4

4

4

2

4

2

2

2

2

2

3.6 Two Median General ParametrizationTo date no med triangle has been disovered whih also has integer area sothe approah used in Chapters 1 and 2 of applying Carmihael's equations isunlikely to produe a parametrization of even a subset of med triangles.p Adierent approah is needed. Fatorizing equations (3.1) over the eld ( 2)Q

40

CHAPTER 3.

THREE INTEGER MEDIANS

gives the following for k:pppp(k b 2)(k + b 2) = (a 2)(a + 2)ppp(a 2)(k + b 2)k + bp2=(k 2b )a+ 2 ppr sp k + b 2 = + 2 (a + 2)t t2

2

pwherepr = 2b ak, s = k ab, t = k 2b . Now the norm of any a := x + y 2in ( 2) is pgiven by Norm(a) := x 2y . It turns out that the norm of thefator rt + st 2 is just -1 sine

p) 2(k ab)Norm rt + st 2 = (2b ak(k 2b )= a (k 2(bk ) 22b )(k 2b )= ka 22b= 1:pppIn other words k + b 2 is just some unit of ( 2) times a + 2. To nd allunits whih have a norm of 1 we simply solve the equation:r sp+ 2 = 1:t t2

2

2

Q

2

2

2

2

2

2

2 2

2

2

2

2

2

2

2

2

2

2 2

Q

Thus r + t = 2s and so r and t have the same parity. Letting 2R := r + t and2T := r t leads to R + T = s . Using the general Pythagorean triple givesr = 2p q + p qs=p +q(3.11)t = (2p q p + q )where gd(p ; q ) = 1 and 2 j p q . Sine the same analysis an be applied tothe l and m equations from (3.1) then dening u; v; w and x; y; z analogously tor; s; t givespp r sp k + b 2 = + 2 (a + 2)p tu tv p p+l+ 2=2(b + a 2)(3.12)p wx ywp pm + a 2 = + 2 ( + b 2)z zppp

where Norm rt + st 2 = Norm wu + wv 2 = Norm xz + yz 2 = 1. So asin equations (3.11) the variables u; v; w and x; y; z are expressible in terms of the2

2

2

2

2

2

21

1 1

21

21

1 1

1

1

1 1

21

21

21

3.6.

41

TWO MEDIAN GENERAL PARAMETRIZATION

parameters p ; q and p ; q . Equating the irrational parts of equation (3.12)leads tor + sa = tbua + vb = w(3.13)xb + y = za:Solving the rst two of these three for the ratios a=b and =b gives2

2

3

3

ab

b

tw rv= sw+ rutu + sv= sw + ru :So any integer-sided triangle with two integer medians has sides proportional toa = tw rvb = sw + ru

= tu + sv:Substituting for the parameters r; s; t and u; v; w as given by equations like(3.11) the sides area = 4p q p + 2(2p q p + q )p q + 2(q p )q(3.14)b = (2p q 2q )p + 2(2p q + 2p )p q + 2(q p q )q

= (2p q + 2q )p + 2(2p q p + q )p q + 2(p p q )qwhere p ; q ; p ; q 2 . Sine a; b; an be either positive or negative by equations (3.1) one may take, without loss of generality, the positive sign in equations(3.11). However u; v; w an also take both positive and negative values. Comparing the result of the eight possible permutations of the signs of u; v; w on theform of a; b; leads to only four distint types, namely The last three forms are21 1 2

1

1

2

21

1 1

1 1

21

22

1 1

1 1

21

22

1 1

2

N

u

v

v

+ + ++ ++++

atw rvtw rvtw + rvtw + rv

21

21

2 2

21

21

22

21

2 2

21

21

1 1

21

2 2

22

1 1

22

b

sw + ru tu + svsw + ru tu + svsw + ru tu svsw + ru tu sv.

all equivalent to the rst form i.e. to (3.14) by the respetive interhanges(p ; q ; p ; q ) =) ( q ; p ; q ; p )(p ; q ; p ; q ) =) (p ; q ; q ; p )(p ; q ; p ; q ) =) ( q ; p ; p ; q ):Considering the more general ase of rational sided triangles with rational medians the arguments above readily lead to1

1

2

2

1

1

2

2

1

1

2

2

1

1

1

1

1

1

2

2

2

2

2

2

42

CHAPTER 3.

THREE INTEGER MEDIANS

Theorem 12 If the rational numbers a; b and are the lengths of sides of a

triangle with at least two rational medians then

a = t 22 + ( 2 + 2 + 1) + ( 2 + 1)

b = t ( 1)2 + 2(2 + ) + ( + 1)

= t ( + 1)2 + ( 2 + 2 + 1) + (2 )where ; ; t 2 Q .

Proof : Sine any pair of the equations (3.13) an substitute for any otherpair by symmetry, set := p =q and := p =q in equations (3.14) to give theabove result.

1

1

2

2

3.7 Med TrianglesUsing Theorem 12 as a starting point one ould proeed in at least two relevantdiretions given a parametrization of rational-sided triangles with two rationalmedians. Either onstrain the third median of a triangle to be rational lengthor onstrain the area to be rational. The seond ase is taken up in Chapter 4while here the rst ase leads to med triangles.Reverting the disussion bak to integer-sided triangles: fore the \m" median of equations (3.12) to be of integer length. Begin with the third equationof set (3.13) namelyxb + y = za:pThe other onstraint on the median is that Norm xz + yz 2 = 1 whih hassolutionsx = 2p q + p qy =p +qz = (2p q p + q ):Substituting these into the above gives(a + b + )p + 2(b a)p q + ( a b + )q = 0:Finally substituting the parametrization (3.14) into this last equation leads top + p q + q = 0where := (a + b + )=4 = (3p q + q )p q + (q p q )q := (2b 2a)=4 = (3p q q )p + (3p q )p q + (p p q )q

:= ( a b + )=4 = (p q + q )p + ( p p q )p q + (p q )q :23

3 3

23

23

23

3 3

23

23

23

23

3 3

23

21

1 1

1 1

1 1

23

3 3

21

22

21

21

2 2

21

22

1 1

21

21

22

21

2 2

1 1

2 2

1 1

22

21

21

22

3.7.

43

MED TRIANGLES

Now the ratio p =q is rational if and only if the disriminant of the quadrati inthis ratio is the square of some integer, say. Hene expanding 4 = interms of the parameters p ; q ; p ; q and then rewriting these in terms of and leads to a sixth degree polynomial in two variables whih must be a rationalsquare, namely = (3 1) + 2 (9 9 11 1) + 3 (3 + 6 + 2 + 2 1)+ 2( 1)(3 8 11 2) + ( 1) ( + 2) :(3.15)Thus any rational solutions of equation (3.15) will provide rational numbers , , and hene integers p ; q ; p ; q whih when substituted into the parametrization (3.14) will produe a triangle with three integer medians. For examplethe rational solution (; ; ) = (11=2; 2; 765=4) orresponds to (p ; q ; p ; q ) =(2; 1; 11; 2) whih in turn orresponds to 2(127; 131; 158) 2 Med.Notie that substituting = 1 in equation (3.15) leads to = 4 ( +3) so that is trivially rational for any rational hoie of . And similarlysubstituting = 1 leads to = 4 ( 3) whih means is rational forany rational . These two ases lead to degenerate triangles sine substitutingp = q into the parametrization (3.14) leads to a + = b while p = q leadsto b + = a.3

3

2

1

2

4

2

3

1

2

3

3

2

2

2

2

1

1

2

2

2

4

3

2

2

2

1

1

2

2

2

2

2

2

1

2

2

1

2

3.7.1 Tiki Surfae

2

Sine the equation (3.15) is so losely assoiated with med triangles it makessense to investigate its properties a little more fully. Consider the funtion oftwo variables f (; ) dened byf (; ) = (3 1)+ 2 (9 9 11 1)+ 3 (3 + 6 + 2 + 2 1)+ 2( 1)(3 8 11 2)+ ( 1) ( + 2) :then the problem of nding rational triples whih satisfy (3.15) is equivalent tonding rational , and whih make f (; ) the square of some rational number.Firstly note that f ( ; ) = f (; ) whih implies that f (; ) is symmetriabout the line = in the -plane. Now if = then the funtionsimplies tog() := f (; ) = 24 + 52 16 24 + 8 + 4:Hene g0 () = 120 + 208p 48 48 + 8 = 8( 1) (15 + 4 1)whih is zero when = 1; . The points orresponding to these values of in the -plane turn ou