Heron triangles
which cannot be decomposed
into two integer right triangles
Paul Yiu
Department of Mathematical Sciences,
Florida Atlantic University,
Boca Raton, Florida 33431
[email protected]
41st Meeting of Florida Section of
Mathematical Association of America
Flordia Southern College,
Lakeland, Florida
February 15-16, 2008
1
2
Abstract
A Heron triangle is one whose sides and area are integers.
While it is quite easy to construct Heron triangles by joining
two integer right triangles along a common side, there are some
which cannot be so obtained. For example, the Heron triangle
(25, 34, 39) has integer area 420 but no integer altitude. In this
talk, a systematic construction of such indecomposable Heron
triangles will be presented.
3
C ONTENTS
1. Heron’s formula for the area of a triangle
2. Construction of primitive Heron triangles
3. Decomposability of primitive Heron triangles
4. Triple of simplifying factors (for the similarity class
of a Heron triangle)
5. Gaussian integers
6. Heron triangles and Gaussian integers
7. Construction of Heron triangles
with given simplifying factors
8. Decomposability of Heron triangles
in terms of triple of simplifying factors
9. Orthocentric Quadrangles and
triples of simplifying factors
10. Examples of indecomposable Heron triangles
as lattice triangle
1
14
17
19
23
26
30
33
37
42
Indecomposable Hereon triangles
1
1. Heron’s formula for the area of a triangle
where s :=
1
2 (a
p
△ = s(s − a)(s − b)(s − c),
+ b + c).
A
s−a
s−a
Y
r
Z
I
r
s−c
s−b
B
r
s−b
X
s−c
C
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P. Yiu
B
s
Ia
I
ra
r
Y′
s−b
C
s−c
Y
s−a
A
s
• △ = rs.
• From the similarity of triangles AIY and AIaY ′ ,
s−a
r
=
.
ra
s
• From the similarity of triangles CIY and Ia CY ′ ,
r · ra = (s − b)(s − c).
Indecomposable Hereon triangles
3
B
s
Ia
I
ra
r
Y′
s−b
C
s−c
Y
s−a
s
From
s−a
r
=
,
ra
s
r · ra = (s − b)(s − c),
we obtain
r
(s − a)(s − b)(s − c)
,
s
p
△ = s(s − a)(s − b)(s − c).
r=
A
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P. Yiu
Examples
(1)
(2)
a b c s
s−a s−b s−c △
3 4 5 6=2·3 3
2
1
2·3=6
a b c s
s−a s−b s−c
△
3
13 14 15 21 = 3 · 7 8 = 2 7
6 = 2 · 3 22 · 3 · 7 = 84
(3)
a b c s
s−a
s−b
s−c
△
2
3
25 34 39 49 = 7 24 = 2 · 3 15 = 3 · 5 10 = 2 · 5 22 · 3 · 5 · 7 = 420
Indecomposable Hereon triangles
5
How can one construct Heron triangles?
(1) Put u = s − a, v = s − b, and w = s − c. Then s =
u + v + w.
We require
uvw(u + v + w) = .
B
s
Ia
I
ra
r
Y′
s−b C
s−cY
s
s−a
A
(2) From three positive rational numbers t1 , t2 , t3 satisfying
t1 t2 + t2 t3 + t3 t1 = 1. (Section 2 below).
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P. Yiu
(3) A more naı̈ve approach is to put two integer right triangles
together along a common side:
13
15
12
5
9
(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 side
12, we also get (13, 20, 21; 126):
13
12
5
20
16
(13, 20, 21; 126) = (12, 5, 13; 30) ∪ (12, 16, 20; 96).
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P. Yiu
We may also cut out a small Pythagorean triangle from a
larger one. For example,
12
20
13
5
11
(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 is decomposable if it can be
obtained by joining two Pythagorean triangles along a common
side, or by excising a Pythagorean triangle from a larger one.
Clearly, a Heron triangle is decomposable if and only if it has
an integer height (which is not a side of the triangle).
The Heron triangle (25, 34, 39; 420) is not decomposable because it does not have an integer height. Its heights are
840 168 840 420 840 280
=
,
=
,
=
,
25
5
34
17 39
13
none an integer.
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P. Yiu
This example was first obtained by Fitch Cheney.
W. F. Cheney, Heronian triangles, A MER . M ATH . M ONTHLY,
36 (1929) 22–28.
Cheney was perhaps more famous for his card trick described
in
Michael Kleber, The best card trick, T HE M ATHEMATICAL
I NTELLIGENCER, 24 (2002) Number 1, 9–11.
Kleber wrote
[William Fitch Cheney, Jr.] was born in San Francisco in 1894, . . . . After receiving his B.A. and
M.A. from the University of California in 1916 and
1917, . . . [i]n 1927 he earned the first math Ph.D.
ever awarded by MIT. . . . Fitch [taught] at the University 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 △
5
39 58 95 96 = 2 · 3 57 = 3 · 19 38 = 2 · 19 1
23 · 3 · 19 = 456
Heights:
304 456 48
,
,
.
13 29 5
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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
3
5 29 30 32 = 2 27 = 3 3
2
23 · 32 = 72
Smallest indecomposable acute 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 Heron
triangles?
Restrict to primitive ones in which the sides do not have
common divisors.
We give a systematic construction of primitive Heron triangles and examine the condition for the indecomposability.
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P. Yiu
2. Construction of primitive Heron triangles
The similarity class of a Heron triangle is determined by
three positive rational numbers
t1 = tan
Since
A
2
+
B
2
+
A
,
2
C
2
t2 = tan
B
,
2
t3 = tan
= π4 , these numbers satisfy
t1 t2 + t2 t3 + t3 t1 = 1.
A
1
t1
1
t1
Y
1
Z
1
t2
B
I
1
t3
1
1
1
t2
X
1
t3
C
C
.
2
Indecomposable Hereon triangles
15
By clearing denominators, we obtain Heron triangles.
Putting ti = ndii , i = 1, 2, 3, with gcd(ni, di) = 1, and
magnifying by n1 n2n3 times, we have the Heron triangle
A1
d1 n2 n3
d1 n2 n3
Y
Z
I
n1 n2 n3
n1 n2 d3
n1 n2 n3
n1 d2 n3
A2
n1 n2 n3
n1 d2 n3
X
n1 n2 d3
A3
Here,
n1 n2 d3 + n1 d2 n3 + d1 n2 n3 = d1 d2 d3 .
(1)
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This is a Heron triangle with sides
a = n1(d2n3 +n2d3 ), b = n2(d3n1 +n3d1 ), c = n3 (d1n2+n1 d2),
semiperimeter s = n1 n2d3 + n1d2 n3 + d1n2 n3 = d1d2 d3
and area △ = n1d1 n2d2 n3d3 .
A1
d1 n2 n3
d1 n2 n3
Y
Z
I
n1 n2 n3
n1 n2 d3
n1 n2 n3
n1 d2 n3
A2
n1 n2 n3
n1 d2 n3
X
n1 n2 d3
A3
A primitive Heron triangle Γ0 results by dividing by the
sides by g := gcd(a, b, c).
Indecomposable Hereon triangles
17
3. Decomposability of primitive Heron triangles
Theorem 1. A primitive Heron triangle can be decomposed
into two Pythagorean components in at most one way, i.e., it
can 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 Pythagorean
triangles. 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, is
partitioned 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 the
original triangle.
2Proof of (2). The triangle being isosceles and Heron, the perimeter and hence the base must be
even. 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 isosceles
triangle is
2mn(m2 − n2 )
,
m2 + n2
which clearly cannot be an integer. This shows that the only way of decomposing a primitive isosceles triangle is into two congruent Pythagorean triangles.
3
Proof of (3). Let (a, b, c; △) be a Heron triangle, not containing any right angle. Suppose the
heights on the sides b and c are integers. Clearly, b and c cannot be relatively prime, for otherwise,
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P. Yiu
the heights of the triangle on these sides are respectively ch and bh, for some integer h. This is
impossible since, the triangle not containing any right angle, the height on b must be less than c,
Suppose therefore gcd(b, c) = g > 1. We write b = b′ g and c = c′ g for relatively prime integers
′
b′ and c′ . If the height on c is h, then that on the side b is ch
= cb′h . If this is also an integer, then
b
h must be divisible by b′ . Replacing h by b′ h, we may now assume that the heights on b and c are
respectively c′ h and b′ h. The side c is divided into b′ k and ±(c − b′ k) 6= 0, where g 2 = h2 + k2 .
It follows that
a2
=
=
=
(b′ h)2 + (c′ g − b′ k)2
b′2 (h2 + k2 ) + c′2 g 2 − 2b′ c′ gk
g[g(b′2 + c′2 ) − 2b′ c′ k]
From this it follows that g divides a2 , and every prime divisor of g is a common divisor of a, b, c.
The Heron triangle cannot be primitive.
Indecomposable Hereon triangles
19
4. Triple of simplifying factors (for the similarity class of a
Heron triangle)
Unless explicitly stated otherwise, whenever the three indices i, j, k appear altogether in an expression or an equation,
they are taken as a permutation of the indices 1, 2, 3.
Note that from
t1 t2 + t2 t3 + t3 t1 = 1,
any one of ti , tj , tk can be expressed in terms of the remaining
two.
i
k
j
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P. Yiu
In the process of expressing ti =
n
tj = djj and tk = ndkk from
ni
di
in terms of
t1 t2 + t2 t3 + t3 t1 = 1,
we encounter certain cancellations. Namely,
dj dk − nj nk
1 − tj tk
=
tj + tk
nj dk + dj nk
is simplified by canceling the gcd
ti =
gi := gcd(dj dk − nj nk , nj dk + dj nk )
from the numerator and denominator. We call this gi the
simplifying factor of ti from tj and tk :
gini = dj dk − nj nk ,
gidi = dj nk + nj dk .
Likewise, there are simplifying factors gj and gk .
(g1, g2 , g3) is called the triple of simplifying factors for the
numbers (t1 , t2, t3 ), or of the similarity class of triangles they
define.
Indecomposable Hereon triangles
21
Example 1. For the (13, 14, 15; 84), we have t1 = 21 , t2 =
and t3 = 32 .
B
7
7
4
4
8
6
4
C
6
8
A
1 − t2 t3
7 · 3 − 4 · 2 13 1
=
=
= =⇒ g1 = 13,
t2 + t3
7 · 2 + 4 · 3 26 2
3·2−2·1 4
1 − t3 t1
=
= = 1 =⇒ g2 = 1,
t2 =
t3 + t1
3·1+2·2 7
1 − t1 t2
2 · 7 − 1 · 4 10 2
t3 =
=
=
= =⇒ g3 = 5.
t1 + t2
2 · 4 + 7 · 1 15 3
t1 =
4
7
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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
△
2
3
25 34 39 49 = 7 24 = 2 · 3 15 = 3 · 5 10 = 2 · 5 22 · 3 · 5 · 7 = 420
we have r =
t1 =
△
s
=
60
7,
r
5
= ,
s − a 14
t2 =
r
4
= ,
s−b 7
t3 =
r
6
= .
s−c 7
1 − t2 t3
7 · 7 − 4 · 6 25
5
=
=
=
=⇒ g1 = 5,
t2 + t3
4 · 7 + 6 · 7 70 14
1 − t3 t1
7 · 14 − 6 · 5
68
4
t2 =
=
=
= =⇒ g2 = 17,
t3 + t1
6 · 14 + 5 · 7 119 7
1 − t1 t2
14 · 7 − 5 · 4 78 6
t3 =
=
=
= =⇒ g3 = 13.
t1 + t2
5 · 7 + 14 · 4 91 7
The simplifying factors are (g1, g2 , g3) = (5, 17, 13).
t1 =
Indecomposable Hereon triangles
23
5. Gaussian integers
We shall associate with each positive rational number
n
t= ,
gcd(n, d) = 1,
d
a primitive, positive Gaussian integer
√
√
z(t) := d + n −1 ∈ Z[ −1].
√
Here, we say that a Gaussian integer x + y −1 is
• primitive if x and y are relatively prime, and
• positive if both x and y are positive.
√
The norm of the Gaussian integer z = x + y −1 is the
integer
N (z) := x2 + y 2 .
√
The argument of a Gaussian integer z = x + y −1 is the
unique real number φ = φ(z) ∈ [0, 2π) defined by
y
x
,
sin φ = p
.
cos φ = p
x2 + y 2
x2 + y 2
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P. Yiu
We also consider the complement of z = x +
√
√
√
z ∗ := y + x −1 = −1 · z.
√
z ∗ = y + x −1
√
z = x + y −1
0
√
z = x − y −1
φ(z) + φ(z ∗ ) =
π
.
2
−1y:
Indecomposable Hereon triangles
25
Some basic facts about the Gaussian integers
√
√
(1) The units of Z[ −1] are precisely ±1 and ± −1.
(2) Multiplicativity of norm: N (z1 z2 ) = N (z1 )N (z2).
(3) An odd (rational) prime number p ramifies
into two
√
nonassociate primes π(p) and π(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 number p is a sum
of two squares in a unique way if and only if p ≡ 1 mod 4.
(4) Let g > 1 be an odd number. There is a primitive Gaussian integer θ satisfying N (θ) = g if and only if each prime
divisor of g is congruent to 1 (mod 4).
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P. Yiu
6. Heron triangles and Gaussian integers
Corresponding to the rational numbers ti =
√
the Gaussian integer zi = di + −1ni .
ni
di ,
we consider
The relations
gi ni = dj dk − nj nk ,
can be rewritten as
gi zi =
gi di = dj nk + nj dk ,
√
−1 · zj zk = (zj zk )∗.
Indecomposable Hereon triangles
27
Lemma 2. N (zi ) = gj gk .
Proof. From gi zi = (zj zk )∗, we have
gi2 N (zi ) = N ((zj zk )∗) = N (zj zk ) = N (zj )N (zk ).
Similarly,
gj2 N (zj ) = N (zk )N (zi) and
gk2 N (zk ) = N (zi )N (zj ).
Multiplying these latter two, and simplifying, we obtain
gj2 gk2 = N (zi )2 =⇒ N (zi ) = gj gk .
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P. Yiu
Lemma 2. N (zi ) = gj gk .
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) Suppose g1 = g2 = 1. Then, N (z3 ) = 1, which is clearly impossible.
(3) is clear from the relation (??).
(4) Suppose gi is even. Then nj , dj , nk , dk are all odd. This means that gi , being a divisor of
N (zj ) = d2j + n2j ≡ 2 (mod 4), is not divisible by 4. Also, dj dk − nj nk and nj dk + dj nk are
both even, and
=
≡
(dj dk − nj nk ) + (nj dk + dj nk )
(dj + nj )(dk + nk ) − 2nj nk
2
(mod 4),
it follows that one of them is divisible by 4, and the other is 2 (mod 4). After cancelling the common
divisor 2, we see that exactly one of ni and di is odd. This means, by (c), that gj and gk cannot be
odd.
(5) If gi and nj admit a common prime divisor p, then p divides both nj and n2j + d2j , and hence
√
dj as well, contradicting the assumption that dj + 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) of gi , gj , gk , with positive exponents ri , rj , rk in their prime factorizations.
By the relation (??), the product zj zk is divisible by the rational prime power pri . This means that
the primitive Gaussian integers zj and zk should contain in their prime factorizations powers of the
distinct primes π(p) and π(p). The same reasoning also applies to each of the pairs (zk , zi ) and
(zi , zj ), so that zk and zi (respectively zi and zj ) each contains one of the non - associate Gaussian
primes π(p) and π(p) in their factorizations. But then this means that zj and zk are divisible by the
same Gaussian prime, a contradiction.
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P. Yiu
7. Construction of Heron triangles
with 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 = g13 (z1 z2 )∗
√
1 + 8√−1
4 + 7√−1
7 + 4√ −1
8 + −1
√
√
2 + 9 −1
6 + 7 −1
√
√
5 − 14 √−1 11 − 10√ −1
10 − 11 −1 14 − 5 −1
√
7 + 6 −1
9+
√
−1
√
√
14 + 5 √−1 10 + 11√ −1
11 + 10 −1 5 + 14 −1
There are four classes of rational triangles with triple of simplifying factors (17, 13, 5):
(t1 , t2 , t3 ) =
4 6 5
4 2 11
1 6 10
1 2 14
, ,
,
, ,
,
, ,
,
, ,
.
7 7 14
7 9 10
8 7 11
8 9 5
Indecomposable Hereon triangles
31
Primitive Heron triangles with triple of simplifying factors
(17, 13, 5):
4
7
6
7
5
14
(a, b, c; △)
(34, 39, 25; 420)
1
8
2
9
5
14
(68, 117, 175; 2520)
1
8
6
7
10
11
(68, 273, 275; 9240)
4
7
2
9
11
10
(238, 117, 275; 13860)
t1 t2 t3
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P. Yiu
Theorem 5. Let g1 , g2 , g3 be odd numbers satisfying the following 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 Heron
triangles with simplifying factors (g1, g2, g3).
Proof. Suppose (g1 , g2 , g3 ) satisfies these conditions. By (ii), there are primitive Gaussian
integers θi , i = 1, 2, 3, such that gi = N (θi ). Since gcd(g
√ 1 , g2 , g3 ) = 1, if a rational prime
p ≡ 1 (mod 4) divides gi and gj , then, in the ring Z[ −1], the prime factorizations of θi
and θj contain powers of the same Gaussian prime π or π.
Therefore, if g1 , g2 , g3 together contain λ rational prime divisors, then there are 2λ
choices of the triple of primitive Gaussian integers (θ1 , θ2 , θ3 ), corresponding to a choice
between the Gaussian primes π(p) and π(p) for each of these rational primes. Choose units
ǫ1 and ǫ2 such that z1 = ǫ1 θ2 θ3 and z2 = ǫ2 θ3 θ1 are positive.
Two positive Gaussian integers z1 and z2 define a positive Gaussian integer z3 via
g3 z3 = (z1 z2 )∗ if and only if
π
0 < φ(z1 ) + φ(z2 ) < .
(2)
2
Since φ(z1∗ ) + φ(z2∗ ) = π − (φ(z1 ) + φ(z2 )), it follows that exactly one of the two pairs
(z1 , z2 ) and (z1∗ , z2∗ ) 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 triangles
in terms of triple of simplifying factors
Proposition 6. A Heron triangle is Pythagorean if and only if
its triple of simplifying factors is of the form (1, 2, g), for an
odd number g whose prime divisors are all of the form 4m + 1.
Proof. If the Heron triangle contains a right
√ angle, we may take
π
t3 = tan 4 = 1 so that g1 g2 = N (1 + −1) = 2. From this
the numbers g1 and g2 must be 1 and 2 in some order.
Conversely,
3 ) = 2 =⇒
√ if g1 = 1 and g2 = 2, then N (z
π
z3 = 1 + −1, t3 = 1, and C = 2 arctan 1 = 2 .
Here is the main theorem on indecomposable Heron triangles.
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P. Yiu
Consider the Heron triangle Γ again.
A1
d1 n2 n3
d1 n2 n3
Y
Z
I
n1 n2 n3
n1 n2 d3
n1 n2 n3
n1 d2 n3
A2
n1 n2 n3
n1 d2 n3
X
n1 n2 d3
A3
a = n1(d2n3 + n2d3 ) = g1 n1d1 ,
b = n2(d3n1 + n3d1 ) = g2 n2d2 ,
c = n3(d1n2 + n1d2 ) = g3 n3d3 .
Proposition 7. gcd(a, b, c) = gcd(n1d1, n2d2 , n3d3).
Proof. For i = 1, 2, 3, ai = gini di .
Indecomposable Hereon triangles
35
Proposition 8. The Heron triangle Γ (assumed non-Pythagorean)
is indecomposable if and only if each of the simplifying factors
gi, i = 1, 2, 3, contains an odd prime divisor.
Proof. We first consider the triangle Γ:
A1
d1 n2 n3
d1 n2 n3
Y
I
Z
n1 d2 n3
A2
n1 n2 n3
n1 n2 d3
n1 n2 n3
n1 n2 n3
n1 d2 n3 X
n1 n2 d3
A3
Since Γ has area △ = n1 d1n2 d2n3 d3 , the height on the side
ai = gi ni di is given by
2nj dj nk dk
hi =
.
gi
Since the triangle does not contain a right angle, it is indecomposable if and only if none of the heights hi , i = 1, 2, 3, is an
integer. By Proposition 3(4), this is the case if and only if each
of g1 , g2, g3 contains an odd prime divisor.
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P. Yiu
Theorem 9. The primitive Heron triangle Γ0 with half-tangents
(t1, t2, t3 ) (assumed non-Pythagorean) is indecomposable if and
only if each of the simplifying factors gi , i = 1, 2, 3, contains
an odd prime divisor.
Proof. The sides (and hence also the heights) of Γ0 are
those of Γ, where
1
g
times
g := gcd(a1, a2 , a3) = gcd(n1d1, n2d2 , n3d3).
The heights of Γ0 are therefore
2nj dj nk dk
2
nj dj nk dk
= ·
.
h′i =
gi · g
gi gcd(n1d1, n2d2 , n3d3)
n d n d
j k k
Note that gcd(n1 dj 1 ,n
is an integer prime to gi . If h′i is
2 d2 ,n3 d3 )
not an integer, then gi must contain an odd prime divisor, by
Proposition 3(4) again.
Indecomposable Hereon triangles
37
9. Orthocentric Quadrangles and
triples of simplifying factors
A
Γ2
Γ3
H
Γ1
B
Triangle
Γ = ABC
C
Orthocenter Half tangents
Simplifying factors
H
t1 , t2, t3
g1, g2 , g3(assumed odd)
Γ1 = HBC
A
1 1−t3 1−t2
t1 , 1+t3 , 1+t2
2g1 , g3, g2
Γ2 = AHC
B
1−t3 1 1−t1
1+t3 , t2 , 1+t1
g3 , 2g2, g1
Γ3 = ABH
C
1−t2 1−t1 1
1+t2 , 1+t1 , t3
g2 , g1, 2g3
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P. Yiu
Proposition 10. The simplifying factors for the four (rational) 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 by
Γi, i = 1, 2, 3, the primitive Heron triangles in the similarity
classes of the remaining three rational triangles in the orthocentric 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 indecomposable (25, 34, 39; 420):
(a, b, c)
(25, 34, 39; 420)
(t1 , t2, t3 ) (g1, g2 , g3)
5 4 6
(5, 17, 13)
14 , 7 , 7
(700, 561, 169; 30030)
14 3
1
5 , 11 , 13
(10, 17, 13)
(855, 952, 169; 62244)
9 7 1
19 , 4 , 13
(5, 34, 13)
(285, 187, 364; 26334)
9
3 7
19 , 11 , 6
(5, 17, 26)
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P. Yiu
Summary: To construct indecomposable primitive Heron triangles, one begins with a triple of odd integers (g1, g2, g3 ), each
greater than 1, such that
(i) the prime divisors of gi, i = 1, 2, 3, are all congruent to
1 (mod 4),
(ii) gcd(g1 , g2, g3) = 1.
The primitive Heron triangle with N (zi ) = gj gk is decomposable.
Indecomposable Hereon triangles
41
(g1 , g2 , g3)
(5, 13, 17)
(d1 , n1 ) (d2 , n2 ) (d3 , n3 )
(a, b, c; △)
(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 orthocentric quadrangle of each of these triangles.
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P. Yiu
10. Examples of indecomposable Heron triangles
as lattice triangle
Theorem 12 (Y, M ONTHLY, 108 (2001) 261–263).
Every Heron triangle is a lattice triangle.
(15, 36)
25
39
(30, 16)
34
(0, 0)
The indecomposable Heron triangle
(25, 34, 39; 420) as a lattice triangle
Indecomposable Hereon triangles
43
(18, 24)
5
(21, 20)
30
29
(0, 0)
The smallest indecomposable Heron triangle
(5, 29, 30; 72) as a lattice triangle
44
P. Yiu
(21, 28)
15
(30, 16)
35
34
(0, 0)
The smallest acute indecomposable Heron triangle
(15, 34, 35; 252) as a lattice triangle
T HANK YOU