Heron triangles - Department of Mathematical Sciences - Fau.edu

Heron triangleswhich cannot be decomposedinto two integer right triangles

Paul YiuDepartment of Mathematical Sciences,

Florida Atlantic University,Boca Raton, Florida 33431

[email protected]

41st Meeting of Florida Section ofMathematical Association of America

Flordia Southern College,Lakeland, Florida

February 15-16, 2008

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2

AbstractA Heron triangle is one whose sides and area are integers.

While it is quite easy to construct Heron triangles by joiningtwo integer right triangles along a common side, there are somewhich cannot be so obtained. For example, the Heron triangle(25, 34, 39) has integer area420 but no integer altitude. In thistalk, a systematic construction of such indecomposable Herontriangles will be presented.

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CONTENTS

1. Heron’s formula for the area of a triangle 12. Construction of primitive Heron triangles 143. Decomposability of primitive Heron triangles 174. Triple of simplifying factors (for the similarity class

of a Heron triangle) 195. Gaussian integers 236. Heron triangles and Gaussian integers 267. Construction of Heron triangles

with given simplifying factors 308. Decomposability of Heron triangles

in terms of triple of simplifying factors 339. Orthocentric Quadrangles and

triples of simplifying factors 3710. Examples of indecomposable Heron triangles

as lattice triangle 42

Indecomposable Hereon triangles 1

1. Heron’s formula for the area of a triangle

△ =√

s(s − a)(s − b)(s − c),

wheres := 12(a + b + c).

r

r

r

s − b s − c

s − c

s − a

s − a

s − b

I

X

Y

Z

A

B C

2 P. Yiu

s − c s − as − b

s

s

Ia

YY ′

I

A

B

C

ra r

• △ = rs.• From the similarity of trianglesAIY andAIaY

′,r

ra

=s − a

s.

• From the similarity of trianglesCIY andIaCY ′,

r · ra = (s − b)(s − c).


s − c s − as − b

s

s

Ia

YY ′

I

A

B

C

ra r

Fromr

ra

=s − a

s, r · ra = (s − b)(s− c),

we obtain

r =

√

(s − a)(s − b)(s − c)

s,

△ =√

s(s − a)(s − b)(s− c).

4 P. Yiu

Examples(1)

a b c s s − a s − b s − c △3 4 5 6 = 2 · 3 3 2 1 2 · 3 = 6

(2)

a b c s s − a s − b s − c △13 14 15 21 = 3 · 7 8 = 23 7 6 = 2 · 3 22 · 3 · 7 = 84

(3)a b c s s − a s − b s − c △25 34 39 49 = 72 24 = 23 · 3 15 = 3 · 5 10 = 2 · 5 22 · 3 · 5 · 7 = 420


How can one construct Heron triangles?(1) Putu = s − a, v = s − b, andw = s − c. Thens =

u + v + w.We require

uvw(u + v + w) = �.

s − c s − as − b

s

sIa

YY ′

I

A

B

C

ra r

(2) From three positive rational numberst1, t2, t3 satisfyingt1t2 + t2t3 + t3t1 = 1. (Section 2 below).

6 P. Yiu

(3) A more naıve approach is to put two integer right trianglestogether along a common side:

13 12

5 9

15

(13, 14, 15; 84) = (12, 5, 13; 30)∪ (12, 9, 15; 54).


By joining 4(3, 4, 5) and(5, 12, 13) along the common side12, we also get(13, 20, 21; 126):

13 12

5 16

20

(13, 20, 21; 126) = (12, 5, 13; 30)∪ (12, 16, 20; 96).

8 P. Yiu

We may also cut out a small Pythagorean triangle from alarger one. For example,

1312

5 11

20

(11, 13, 20; 66) = (12, 16, 20; 96) \ (12, 5, 13; 30).


Does every Heron triangle arise in this way?

We say that a Heron triangle isdecomposable if it can beobtained by joining two Pythagorean triangles along a commonside, or by excising a Pythagorean triangle from a larger one.

Clearly, a Heron triangle is decomposable if and only if it hasaninteger height (which is not a side of the triangle).

The Heron triangle(25, 34, 39; 420) is not decomposable be-cause it does not have an integer height. Its heights are

840

25=

168

5,

840

34=

420

17,

840

39=

280

13,

none an integer.

10 P. Yiu

This example was first obtained by Fitch Cheney.

W. F. Cheney, Heronian triangles, AMER. MATH . MONTHLY,36 (1929) 22–28.

Cheney was perhaps more famous for his card trick describedin

Michael Kleber, The best card trick, THE MATHEMATICAL

INTELLIGENCER, 24 (2002) Number 1, 9–11.

Kleber wrote[William Fitch Cheney, Jr.] was born in San Fran-cisco in 1894, . . . . After receiving his B.A. andM.A. from the University of California in 1916 and1917, . . . [i]n 1927 he earned the first math Ph.D.ever awarded by MIT. . . . Fitch [taught] at the Uni-versity of Hartford . . . until his death in 1974.


Cheney’s second example of indecomposable Heron triangle:

(39, 58, 95; 456).

a b c s s − a s − b s − c △39 58 95 96 = 25 · 3 57 = 3 · 19 38 = 2 · 19 1 23 · 3 · 19 = 456

Heights:304

13,

456

29,

48

5.

12 P. Yiu

Nowadays, it is much easier to do a computer search.

Smallest indecomposable Heron triangle:

(5, 29, 30; 72).

a b c s s − a s − b s − c △5 29 30 32 = 25 27 = 33 3 2 23 · 32 = 72

Smallest indecomposableacute Heron triangle:

(15, 34, 35; 252).

a b c s s − a s − b s − c △15 34 35 42 = 2 · 3 · 7 27 = 33 8 = 23 7 22 · 32 · 7 = 252


How can one construct examples of indecomposable Herontriangles?

Restrict toprimitive ones in which the sides do not havecommon divisors.

We give a systematic construction of primitive Heron tri-angles and examine the condition for the indecomposabil-ity.

14 P. Yiu

2. Construction of primitive Heron trianglesThe similarity class of a Heron triangle is determined by

three positiverational numbers

t1 = tanA

2, t2 = tan

B

2, t3 = tan

C

2.

SinceA2 + B

2 + C2 = π

4 , these numbers satisfy

t1t2 + t2t3 + t3t1 = 1.

1

1

1

1

t2

1

t3

1

t3

1

t1

1

t1

1

t2

I

X

Y

Z

A

B C


By clearing denominators, we obtain Heron triangles.

Puttingti = ni

di, i = 1, 2, 3, with gcd(ni, di) = 1, and

magnifying byn1n2n3 times, we have the Heron triangle

n1n2n3

n1n2n3

n1n2n3

n1d2n3 n1n2d3

n1n2d3

d1n2n3

d1n2n3

n1d2n3

I

X

Y

Z

A1

A2 A3

Here,

n1n2d3 + n1d2n3 + d1n2n3 = d1d2d3. (1)

16 P. Yiu

This is a Heron triangle with sides

a = n1(d2n3+n2d3), b = n2(d3n1+n3d1), c = n3(d1n2+n1d2),

semiperimeters = n1n2d3 + n1d2n3 + d1n2n3 = d1d2d3

and area△ = n1d1n2d2n3d3.

n1n2n3

n1n2n3

n1n2n3

n1d2n3 n1n2d3

n1n2d3

d1n2n3

d1n2n3

n1d2n3

I

X

Y

Z

A1

A2 A3

A primitive Heron triangleΓ0 results by dividing by thesides byg := gcd(a, b, c).


3. Decomposability of primitive Heron triangles

Theorem 1. A primitive Heron triangle can be decomposedinto two Pythagorean components in at most one way, i.e., itcan have at most one integer height.

Proof. This follows from three propositions.(1) A primitive Pythagorean triangle is indecomposable.1

(2) A primitive, isosceles Heron triangle is decomposable,the only decomposition being into two congruent Pythagoreantriangles.2

(3) If a non-Pythagorean Heron triangle has two integer heights,then it cannot be primitive.3

1Proof of (1). We prove this by contradiction. A Pythagorean triangle, if decomposable, ispartitioned by the altitude on the hypotenuse into two similar but smaller Pythagorean triangles.None of these, however, can have all sides of integer length by the primitivity assumption on theoriginal triangle.

2Proof of (2). The triangle being isosceles and Heron, the perimeter and hence the base must beeven. Each half of the isosceles triangle is a (primitive) Pythagorean triangle,(m2 −n2, 2mn, m2 +n2), with m, n relatively prime, and of different parity. The height on each slant side of the isoscelestriangle is

2mn(m2 − n2)

m2 + n2,

which clearly cannot be an integer. This shows that the only way of decomposing a primitive isosce-les triangle is into two congruent Pythagorean triangles.

3Proof of (3). Let(a, b, c;△) be a Heron triangle, not containing any right angle. Supposetheheights on the sidesb andc are integers. Clearly,b andc cannot be relatively prime, for otherwise,

18 P. Yiu

�

the heights of the triangle on these sides are respectivelych and bh, for some integerh. This isimpossible since, the triangle not containing any right angle, the height onb must be less thanc,Suppose thereforegcd(b, c) = g > 1. We writeb = b′g andc = c′g for relatively prime integersb′ andc′. If the height onc is h, then that on the sideb is ch

b= c′h

b′. If this is also an integer, then

h must be divisible byb′. Replacingh by b′h, we may now assume that the heights onb andc arerespectivelyc′h andb′h. The sidec is divided intob′k and±(c − b′k) 6= 0, whereg2 = h2 + k2.It follows that

a2 = (b′h)2 + (c′g − b

′

k)2

= b′2(h2 + k

2) + c′2

g2 − 2b

′

c′

gk

= g[g(b′2 + c′2) − 2b

′

c′

k]

From this it follows thatg dividesa2, and every prime divisor ofg is a common divisor ofa, b, c.The Heron triangle cannot be primitive.


4. Triple of simplifying factors (for the similarity class of aHeron triangle)

Unless explicitly stated otherwise, whenever the three in-dicesi, j, k appear altogether in an expression or an equation,they are taken as apermutation of the indices1, 2, 3.

Note that from

t1t2 + t2t3 + t3t1 = 1,

any one ofti, tj, tk can be expressed in terms of the remainingtwo.

i

jk

20 P. Yiu

In the process of expressingti = ni

diin terms of

tj =nj

djandtk = nk

dkfrom

t1t2 + t2t3 + t3t1 = 1,

we encounter certaincancellations. Namely,

ti =1 − tjtk

tj + tk=

djdk − njnk

njdk + djnk

is simplified by canceling thegcd

gi := gcd(djdk − njnk, njdk + djnk)

from the numerator and denominator. We call thisgi thesimplifying factor of ti from tj andtk:

gini = djdk − njnk,

gidi = djnk + njdk.

Likewise, there are simplifying factorsgj andgk.(g1, g2, g3) is called thetriple of simplifying factors for the

numbers(t1, t2, t3), or of the similarity class of triangles theydefine.


Example 1. For the(13, 14, 15; 84), we havet1 = 12, t2 = 4

7

andt3 = 23 .

6

7

8

7

C A

B

6 8

4

4

4

t1 =1 − t2t3

t2 + t3=

7 · 3 − 4 · 27 · 2 + 4 · 3 =

13

26=

1

2=⇒ g1 = 13,

t2 =1 − t3t1

t3 + t1=

3 · 2 − 2 · 13 · 1 + 2 · 2 =

4

7= 1 =⇒ g2 = 1,

t3 =1 − t1t2

t1 + t2=

2 · 7 − 1 · 42 · 4 + 7 · 1 =

10

15=

2

3=⇒ g3 = 5.

22 P. Yiu

Example 2. For the indecomposable Heron triangle(25, 34, 39; 420)(Cheney’s example),

a b c s s − a s − b s − c △25 34 39 49 = 72 24 = 23 · 3 15 = 3 · 5 10 = 2 · 5 22 · 3 · 5 · 7 = 420

we haver = △s

= 607 ,

t1 =r

s − a=

5

14, t2 =

r

s − b=

4

7, t3 =

r

s − c=

6

7.

t1 =1 − t2t3

t2 + t3=

7 · 7 − 4 · 64 · 7 + 6 · 7 =

25

70=

5

14=⇒ g1 = 5,

t2 =1 − t3t1

t3 + t1=

7 · 14 − 6 · 56 · 14 + 5 · 7 =

68

119=

4

7=⇒ g2 = 17,

t3 =1 − t1t2

t1 + t2=

14 · 7 − 5 · 45 · 7 + 14 · 4 =

78

91=

6

7=⇒ g3 = 13.

The simplifying factors are(g1, g2, g3) = (5, 17, 13).


5. Gaussian integers

We shall associate with each positive rational number

t =n

d, gcd(n, d) = 1,

a primitive, positive Gaussian integer

z(t) := d + n√−1 ∈ Z[

√−1].

Here, we say that a Gaussian integerx + y√−1 is

• primitive if x andy are relatively prime, and• positive if both x andy are positive.

The norm of the Gaussian integerz = x + y√−1 is the

integerN(z) := x2 + y2.

The argument of a Gaussian integerz = x + y√−1 is the

unique real numberφ = φ(z) ∈ [0, 2π) defined by

cos φ =x

√

x2 + y2, sin φ =

y√

x2 + y2.

24 P. Yiu

We also consider thecomplement of z = x +√−1y:

z∗ := y + x√−1 =

√−1 · z.

0

z = x + y√

−1

z∗ = y + x√

−1

z = x − y√

−1

φ(z) + φ(z∗) =π

2.


Some basic facts about the Gaussian integers

(1) The units ofZ[√−1] are precisely±1 and±

√−1.

(2) Multiplicativity of norm: N(z1z2) = N(z1)N(z2).

(3) An odd (rational) prime numberp ramifies into twononassociate primesπ(p) andπ(p) in Z[

√−1], namely,

p = π(p)π(p) if and only if p ≡ 1 mod 4.

This is a consequence of Fermat’s 2-square theorem: A odd prime numberp is a sum

of two squares in a unique way if and only ifp ≡ 1 mod 4.

(4) Letg > 1 be an odd number. There is aprimitive Gauss-ian integerθ satisfyingN(θ) = g if and only if each primedivisor ofg is congruent to 1 (mod 4).

26 P. Yiu

6. Heron triangles and Gaussian integers

Corresponding to the rational numbersti = ni

di, we consider

the Gaussian integerzi = di +√−1ni.

The relations

gini = djdk − njnk, gidi = djnk + njdk,

can be rewritten as

gizi =√−1 · zjzk = (zjzk)

∗.


Lemma 2. N(zi) = gjgk.

Proof. Fromgizi = (zjzk)∗, we have

g2i N(zi) = N((zjzk)

∗) = N(zjzk) = N(zj)N(zk).

Similarly,

g2jN(zj) = N(zk)N(zi) and g2

kN(zk) = N(zi)N(zj).

Multiplying these latter two, and simplifying, we obtain

g2jg

2k = N(zi)

2 =⇒ N(zi) = gjgk.

�

28 P. Yiu

Lemma 2. N(zi) = gjgk.

Proposition 3.(1) gi is a common divisor of N(zj) and N(zk).(2) At least two of gi, gj, gk exceed 1.(3) gi is even if and only if all nj, dj, nk and dk are odd.(4) At most one of gi, gj , gk is even, and none of them is

divisible by 4.(5) gi is prime to each of nj, dj, nk, and dk.(6) Each odd prime divisor of gi, i = 1, 2, 3, is congruent to

1 (mod 4).

Proposition 4. gcd(g1, g2, g3) = 1.


Proof. (1) follows easily from Lemma 2.(2) Supposeg1 = g2 = 1. Then,N(z3) = 1, which is clearly impossible.(3) is clear from the relation (??).(4) Supposegi is even. Thennj , dj , nk, dk are all odd. This means thatgi, being a divisor of

N(zj) = d2

j + n2

j ≡ 2 (mod 4), is not divisible by 4. Also,djdk − njnk andnjdk + djnk areboth even, and

(djdk − njnk) + (njdk + djnk)

= (dj + nj)(dk + nk) − 2njnk

≡ 2 (mod 4),

it follows that one of them is divisible by 4, and the other is 2(mod 4). After cancelling the commondivisor 2, we see that exactly one ofni anddi is odd. This means, by (c), thatgj andgk cannot beodd.

(5) If gi andnj admit a common prime divisorp, thenp divides bothnj andn2

j + d2

j , and hencedj as well, contradicting the assumption thatdj + nj

√−1 be primitive.

(6) is a consequence of Proposition??. �

Proof of Proposition 4. We shall derive a contradiction by assuming a common rational prime

divisor p ≡ 1 (mod 4) ofgi, gj , gk, with positive exponentsri, rj , rk in their prime factorizations.

By the relation (??), the productzjzk is divisible by the rational prime powerpri . This means that

the primitive Gaussian integerszj andzk should contain in their prime factorizations powers of the

distinct primesπ(p) andπ(p). The same reasoning also applies to each of the pairs(zk, zi) and

(zi, zj), so thatzk andzi (respectivelyzi andzj) each contains one of the non - associate Gaussian

primesπ(p) andπ(p) in their factorizations. But then this means thatzj andzk are divisible by the

same Gaussian prime, a contradiction.

30 P. Yiu

7. Construction of Heron triangleswith given simplifying factors

Example 3. g1 = 17, g2 = 13, g3 = 5.N(z1) = 13·5 = 65, N(z2) = 5·17 = 85, N(z3) = 17·13 =

221.

z1 : 1 + 8√−1 4 + 7

√−1 7 + 4

√−1 8 +

√−1

z2 : 2 + 9√−1 6 + 7

√−1 7 + 6

√−1 9 +

√−1

z3 = 1

g3

(z1z2)∗ 2 + 9

√−1 6 + 7

√−1 7 + 6

√−1 9 +

√−1

1 + 8√−1 5 − 14

√−1 11 − 10

√−1

4 + 7√−1 10 − 11

√−1 14 − 5

√−1

7 + 4√−1 14 + 5

√−1 10 + 11

√−1

8 +√−1 11 + 10

√−1 5 + 14

√−1

There are four classes of rational triangles with triple of sim-plifying factors(17, 13, 5):

(t1, t2, t3) =

(

4

7,

6

7,

5

14

)

,

(

4

7,

2

9,

11

10

)

,

(

1

8,

6

7,

10

11

)

,

(

1

8,

2

9,

14

5

)

.


Primitive Heron triangles with triple of simplifying factors(17, 13, 5):

t1 t2 t3 (a, b, c;△)47

67

514 (34, 39, 25; 420)

18

29

514 (68, 117, 175; 2520)

18

67

1011 (68, 273, 275; 9240)

47

29

1110 (238, 117, 275; 13860)

32 P. Yiu

Theorem 5. Let g1, g2, g3 be odd numbers satisfying the fol-lowing conditions.

(i) At least two of g1, g2, g3 exceed 1.(ii) The prime divisors of gi, i = 1, 2, 3, are all congruent

to 1 (mod 4).(iii) gcd(g1, g2, g3) = 1.

Suppose g1, g2, g3 together contain λ distinct rational (odd)prime divisors. Then there are 2λ−1 distinct, primitive Herontriangles with simplifying factors (g1, g2, g3).

Proof. Suppose(g1, g2, g3) satisfies these conditions. By (ii), there are primitive Gaussianintegersθi, i = 1, 2, 3, such thatgi = N(θi). Sincegcd(g1, g2, g3) = 1, if a rational primep ≡ 1 (mod 4) dividesgi andgj, then, in the ringZ[

√−1], the prime factorizations ofθi

andθj contain powers of the same Gaussian primeπ or π.Therefore, ifg1, g2, g3 together containλ rational prime divisors, then there are2λ

choices of the triple of primitive Gaussian integers(θ1, θ2, θ3), corresponding to a choicebetween the Gaussian primesπ(p) andπ(p) for each of these rational primes. Choose unitsǫ1 andǫ2 such thatz1 = ǫ1θ2θ3 andz2 = ǫ2θ3θ1 are positive.

Two positive Gaussian integersz1 and z2 define a positive Gaussian integerz3 viag3z3 = (z1z2)

∗ if and only if

0 < φ(z1) + φ(z2) <π

2. (2)

Sinceφ(z∗1) + φ(z∗

2) = π − (φ(z1) + φ(z2)), it follows that exactly one of the two pairs

(z1, z2) and (z∗1, z∗

2) satisfies condition (2). There are, therefore,2λ−1 Heron triangles

with (g1, g2, g3) as simplifying factors. �


8. Decomposability of Heron trianglesin terms of triple of simplifying factors

Proposition 6. A Heron triangle is Pythagorean if and only ifits triple of simplifying factors is of the form (1, 2, g), for anodd number g whose prime divisors are all of the form 4m+1.

Proof. If the Heron triangle contains a right angle, we may taket3 = tan π

4= 1 so thatg1g2 = N(1 +

√−1) = 2. From this

the numbersg1 andg2 must be 1 and 2 in some order.Conversely, ifg1 = 1 andg2 = 2, thenN(z3) = 2 =⇒

z3 = 1 +√−1, t3 = 1, andC = 2 arctan1 = π

2. �

Here is the main theorem on indecomposable Heron trian-gles.

34 P. Yiu

Consider the Heron triangleΓ again.

n1n2n3

n1n2n3

n1n2n3

n1d2n3 n1n2d3

n1n2d3

d1n2n3

d1n2n3

n1d2n3

I

X

Y

Z

A1

A2 A3

a = n1(d2n3 + n2d3) = g1n1d1,

b = n2(d3n1 + n3d1) = g2n2d2,

c = n3(d1n2 + n1d2) = g3n3d3.

Proposition 7. gcd(a, b, c) = gcd(n1d1, n2d2, n3d3).

Proof. For i = 1, 2, 3, ai = ginidi. �


Proposition 8. The Heron triangle Γ (assumed non-Pythagorean)is indecomposable if and only if each of the simplifying factorsgi, i = 1, 2, 3, contains an odd prime divisor.

Proof. We first consider the triangleΓ:

n1n2n3

n1n2n3

n1n2n3

n1d2n3 n1n2d3

n1n2d3

d1n2n3

d1n2n3

n1d2n3

I

X

Y

Z

A1

A2 A3

SinceΓ has area△ = n1d1n2d2n3d3, the height on the sideai = ginidi is given by

hi =2njdjnkdk

gi

.

Since the triangle does not contain a right angle, it is indecom-posable if and only if none of the heightshi, i = 1, 2, 3, is aninteger. By Proposition 3(4), this is the case if and only if eachof g1, g2, g3 contains an odd prime divisor. �

36 P. Yiu

Theorem 9. The primitive Heron triangle Γ0 with half-tangents(t1, t2, t3) (assumed non-Pythagorean) is indecomposable if andonly if each of the simplifying factors gi, i = 1, 2, 3, containsan odd prime divisor.

Proof. The sides (and hence also the heights) ofΓ0 are 1g

timesthose ofΓ, where

g := gcd(a1, a2, a3) = gcd(n1d1, n2d2, n3d3).

The heights ofΓ0 are therefore

h′i =

2njdjnkdk

gi · g=

2

gi

· njdjnkdk

gcd(n1d1, n2d2, n3d3).

Note that njdjnkdk

gcd(n1d1,n2d2,n3d3)is an integer prime to gi. If h′

i isnot an integer, thengi must contain an odd prime divisor, byProposition 3(4) again. �


9. Orthocentric Quadrangles andtriples of simplifying factors

A

B C

H

Γ2

Γ1

Γ3

Triangle Orthocenter Half tangents Simplifying factorsΓ = ABC H t1, t2, t3 g1, g2, g3(assumed odd)

Γ1 = HBC A 1t1, 1−t3

1+t3, 1−t2

1+t22g1, g3, g2

Γ2 = AHC B 1−t31+t3

, 1t2, 1−t1

1+t1g3, 2g2, g1

Γ3 = ABH C 1−t21+t2

, 1−t11+t1

, 1t3

g2, g1, 2g3

38 P. Yiu

Proposition 10. The simplifying factors for the four (ratio-nal) triangles in an orthocentric quadrangle are of the form(g1, g2, g3), (2g1, g2, g3), (g1, 2g2, g3) and (g1, g2, 2g3), with g1,g2, g3 odd integers.

Corollary 11. Let Γ be a primitive Heron triangle. Denote byΓi, i = 1, 2, 3, the primitive Heron triangles in the similarityclasses of the remaining three rational triangles in the ortho-centric quadrangle containing Γ. The four triangles Γ and Γi,i = 1, 2, 3, are either all decomposable or all indecomposable.


Example 4. The orthocentric quadrangle from Cheney’s inde-composable(25, 34, 39; 420):

(a, b, c) (t1, t2, t3) (g1, g2, g3)(25, 34, 39; 420) 5

14,47 ,

67 (5, 17, 13)

(700, 561, 169; 30030) 145 , 3

11,113 (10, 17, 13)

(855, 952, 169; 62244) 919,

74,

113 (5, 34, 13)

(285, 187, 364; 26334) 919,

311,

76 (5, 17, 26)

40 P. Yiu

Summary: To constructindecomposable primitive Heron tri-angles, one begins with a triple ofodd integers(g1, g2, g3), eachgreater than 1, such that

(i) the prime divisors ofgi, i = 1, 2, 3, are all congruent to1 (mod 4),

(ii) gcd(g1, g2, g3) = 1.The primitive Heron triangle withN(zi) = gjgk is decom-

posable.


(g1, g2, g3) (d1, n1) (d2, n2) (d3, n3) (a, b, c;△)(5, 13, 17) (14, 5) (7, 6) (7, 4) (25, 39, 34; 420)

(5, 14) (9, 2) (8, 1) (175, 117, 68; 2520)(11, 10) (7, 6) (8, 1) (275, 273, 68; 9240)(10, 11) (9, 2) (7, 4) (275, 117, 238; 13860)

(5, 13, 29) (4, 19) (12, 1) (8, 1) (95, 39, 58; 456)(16, 11) (8, 9) (8, 1) (110, 117, 29; 1584)(11, 16) (12, 1) (7, 4) (220, 39, 203; 3696)(19, 4) (8, 9) (7, 4) (95, 234, 203; 9576)

(5, 17, 29) (22, 3) (12, 1) (2, 9) (55, 34, 87; 396)(18, 13) (9, 8) (9, 2) (65, 68, 29; 936)(18, 13) (12, 1) (6, 7) (195, 34, 203; 3276)(22, 3) (9, 8) (7, 6) (55, 204, 203; 5544)

(13, 17, 29) (22, 3) (16, 11) (10, 11) (39, 136, 145; 2640)(22, 3) (19, 4) (5, 14) (429, 646, 1015; 87780)(18, 13) (19, 4) (11, 10) (1521, 646, 1595; 489060)(18, 13) (16, 11) (14, 5) (1521, 1496, 1015; 720720)

Further examples can be obtained by considering the ortho-centric quadrangle of each of these triangles.

42 P. Yiu

10. Examples of indecomposable Heron trianglesas lattice triangle

Theorem 12 (Y, M ONTHLY, 108 (2001) 261–263).Every Heron triangle is a lattice triangle.

39

25

34

(0, 0)

(15, 36)

(30, 16)

The indecomposable Heron triangle(25, 34, 39; 420) as a lattice triangle


30

5

29

(0, 0)

(18, 24)

(21, 20)

The smallest indecomposable Heron triangle(5, 29, 30; 72) as a lattice triangle

44 P. Yiu

35

15

34

(0, 0)

(21, 28)

(30, 16)

The smallest acute indecomposable Heron triangle(15, 34, 35; 252) as a lattice triangle

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