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Heron triangleswhich cannot be decomposedinto two integer right triangles
Paul YiuDepartment of Mathematical Sciences,
Florida Atlantic University,Boca Raton, Florida 33431
yiu@fau.edu
41st Meeting of Florida Section ofMathematical Association of America
Flordia Southern College,Lakeland, Florida
February 15-16, 2008
1
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.
3
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).
Indecomposable Hereon triangles 3
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
Indecomposable Hereon triangles 5
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).
Indecomposable Hereon triangles 7
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).
Indecomposable Hereon triangles 9
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.
Indecomposable Hereon triangles 11
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
Indecomposable Hereon triangles 13
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
Indecomposable Hereon triangles 15
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).
Indecomposable Hereon triangles 17
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.
Indecomposable Hereon triangles 19
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.
Indecomposable Hereon triangles 21
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).
Indecomposable Hereon triangles 23
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.
Indecomposable Hereon triangles 25
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)
∗.
Indecomposable Hereon triangles 27
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.
Indecomposable Hereon triangles 29
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
)
.
Indecomposable Hereon triangles 31
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. �
Indecomposable Hereon triangles 33
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. �
Indecomposable Hereon triangles 35
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. �
Indecomposable Hereon triangles 37
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.
Indecomposable Hereon triangles 39
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.
Indecomposable Hereon triangles 41
(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
Indecomposable Hereon triangles 43
30
5
29
(0, 0)
(18, 24)
(21, 20)
The smallest indecomposable Heron triangle(5, 29, 30; 72) as a lattice triangle
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