# Heronian triangle

In geometry, a Heronian triangle is a triangle that has side lengths and area that are all integers.[1][2] Heronian triangles are named after Hero of Alexandria. The term is sometimes applied more widely to triangles whose sides and area are all rational numbers.[3]

## Properties

Any right-angled triangle whose sidelengths are a Pythagorean triple is a Heronian triangle, as the side lengths of such a triangle are integers, and its area is also an integer, being half of the product of the two shorter sides of the triangle, at least one of which must be even.

A triangle with sidelengths c, e and b + d, and height a.

An example of a Heronian triangle which is not right-angled is the one with sidelengths 5, 5, and 6, whose area is 12. This triangle is obtained by joining two copies of the right-angled triangle with sides 3, 4, and 5 along the sides of length 4. This approach works in general, as illustrated in the picture to the right. One takes a Pythagorean triple (a, b, c), with c being largest, then another one (a, d, e), with e being largest, constructs the triangles with these sidelengths, and joins them together along the sides of length a, to obtain a triangle with integer side lengths c, e, and b + d, and with area

$A=\frac{1}{2}(b+d)a$ (one half times the base times the height).

If a is even then the area A is an integer. Less obviously, if a is odd then A is still an integer, as b and d must both be even, making b+d even too.

An interesting question to ask is whether all Heronian triangles can be obtained by joining together two right-angled triangles with integers sides as described above. The answer is no. For example a 5, 29, 30 Heronian triangle with area 72 cannot be constructed from two integer Pythagorean triangle since none of its altitudes are integers. Such Heronian triangles are known as in-decomposable.[4] However, if one allows Pythagorean triples with rational values, not necessarily integers, then the answer is affirmative,[5] because every altitude of a Heronian triangle is rational (since it equals twice the integer area divided by the integer base). So the Heronian triangle with sides 5, 29, 30 can be constructed from rational Pythagorean triangles with sides 7/5, 24/5, 5 and 143/5, 24/5, 29. Note that a Pythagorean triple with rational values is just a scaled version of a triple with integer values.

Other properties of Heronian triangles are given in Integer triangle#Heronian triangles.

## Theorem

Any Heronian triangle can be split into two right-angled triangles whose side lengths form Pythagorean triples with rational values.

Proof of the theorem

Consider again the illustration to the right, where it is known that c, e, b + d, and the triangle area A are integers. Assume that b + d is greater than or equal to c and e, so that the foot of the altitude perpendicular to this side falls within the side. To show that the triples (a, b, c) and (a, d, e) are rational Pythagorean triples, it suffices to show that a, b, and d are rational.

Since the triangle area is

$A=\frac{1}{2}(b+d)a\, ,$

one can solve for a to find

$a=\frac{2A}{b+d}\, ,$

which is rational, as both $A$ and $b+d$ are integers. It remains to show that b and d are rational.

From the Pythagorean theorem applied to the two right-angled triangles, one has

$a^2+b^2=c^2\, ,$

and

$a^2+d^2=e^2\, .$

One can subtract these two, to find

$b^2-d^2=c^2-e^2\, ,$
$\Rightarrow (b-d)(b+d)=c^2-e^2\, ,$
$\Rightarrow b-d=\frac{c^2-e^2}{b+d}\, .$

By assumption, c, e, and b + d are integers. So b − d is rational and hence

$b = \frac{(b+d)+(b-d)}{2}\, ,$
$d = \frac{(b+d)-(b-d)}{2}\, ,$

are both rational. Q.E.D.

## Exact formula for Heronian triangles

Every Heronian triangle has sides proportional to:[6]

$a=n(m^{2}+k^{2}) \,$
$b=m(n^{2}+k^{2}) \,$
$c=(m+n)(mn-k^{2}) \,$
$\text{Semiperimeter}=s=(a+b+c)/2=mn(m+n) \,$
$\text{Area}=mnk(m+n)(mn-k^{2}) \,$
$\text{Inradius}=k(mn-k^{2}) \,$
$s-a=n(mn-k^{2}) \,$
$s-b=m(mn-k^{2}) \,$
$s-c=(m+n)k^{2} \,$

for integers m, n and k where:

$\gcd{(m,n,k)}=1 \,$
$mn > k^2 \ge m^2n/(2m+n) \,$
$m \ge n \ge 1 \,$.

The proportionality factor is generally a rational  $\frac{p}{q}$  where  $q=\gcd{(a,b,c)}$  reduces the generated Heronian triangle to its primitive and  $p$  scales up this primitive to the required size. For example, taking m = 36, n = 4 and k = 3 produces a triangle with a = 5220, b = 900 and c = 5400, which is similar to the 5, 29, 30 Heronian triangle and the proportionality factor used has p = 1 and q = 180.

## Examples

The list of primitive integer Heronian triangles, sorted by area and, if this is the same, by perimeter, starts as in the following table. "Primitive" means that the greatest common divisor of the three side lengths equals 1.

Area Perimeter side length b+d side length e side length c
6 12 5 4 3
12 16 6 5 5
12 18 8 5 5
24 32 15 13 4
30 30 13 12 5
36 36 17 10 9
36 54 26 25 3
42 42 20 15 7
60 36 13 13 10
60 40 17 15 8
60 50 24 13 13
60 60 29 25 6
66 44 20 13 11
72 64 30 29 5
84 42 15 14 13
84 48 21 17 10
84 56 25 24 7
84 72 35 29 8
90 54 25 17 12
90 108 53 51 4
114 76 37 20 19
120 50 17 17 16
120 64 30 17 17
120 80 39 25 16
126 54 21 20 13
126 84 41 28 15
126 108 52 51 5
132 66 30 25 11
156 78 37 26 15
156 104 51 40 13
168 64 25 25 14
168 84 39 35 10
168 98 48 25 25
180 80 37 30 13
180 90 41 40 9
198 132 65 55 12
204 68 26 25 17
210 70 29 21 20
210 70 28 25 17
210 84 39 28 17
210 84 37 35 12
210 140 68 65 7
210 300 149 148 3
216 162 80 73 9
234 108 52 41 15
240 90 40 37 13
252 84 35 34 15
252 98 45 40 13
252 144 70 65 9
264 96 44 37 15
264 132 65 34 33
270 108 52 29 27
288 162 80 65 17
300 150 74 51 25
300 250 123 122 5
306 108 51 37 20
330 100 44 39 17
330 110 52 33 25
330 132 61 60 11
330 220 109 100 11
336 98 41 40 17
336 112 53 35 24
336 128 61 52 15
336 392 195 193 4
360 90 36 29 25
360 100 41 41 18
360 162 80 41 41
390 156 75 68 13
396 176 87 55 34
396 198 97 90 11
396 242 120 109 13

## Equable triangles

A shape is called equable if its area equals its perimeter. There are exactly five equable Heronian triangles: the ones with side lengths (5,12,13), (6,8,10), (6,25,29), (7,15,20), and (9,10,17).[7][8]

## Almost-equilateral Heronian triangles

Since the area of an equilateral triangle with rational sides is an irrational number, no equilateral triangle is Heronian. However, there is a unique sequence of Heronian triangles that are "almost equilateral" because the three sides are of the form n − 1, n, n + 1. The first few examples of these almost-equilateral triangles are listed in the following table (sequence A003500 in OEIS):

n − 1 n n + 1
3 4 5 6 1
13 14 15 84 4
51 52 53 1170 15
193 194 195 16296 56
723 724 725 226974 209
2701 2702 2703 3161340 780
10083 10084 10085 44031786 2911
37633 37634 37635 613283664 10864

Subsequent values of n can be found by multiplying the previous value by 4, then subtracting the value prior to that one (52 = 4 × 14 − 4, 194 = 4 × 52 − 14, etc.), thus:

$n_t = 4n_{t-1} - n_{t-2} \, ,$

where t denotes any row in the table. This is a Lucas sequence. Alternatively, the formula $(2 + \sqrt{3})^t + (2 - \sqrt{3})^t$ generates all n. Equivalently, let A = area and y = inradius, then,

$\big((n-1)^2+n^2+(n+1)^2\big)^2-2\big((n-1)^4+n^4+(n+1)^4\big) = (6n y)^2 = (4A)^2$

where {n, y} are solutions to n2 − 12y2 = 4. A small transformation n = 2x yields a conventional Pell equation x2 − 3y2 = 1, the solutions of which can then be derived from the regular continued fraction expansion for √3.[9]

The variable n is of the form $n=\sqrt{2 + 2 k}$, where k is 7, 97, 1351, 18817, …. The numbers in this sequence have the property that k consecutive integers have integral standard deviation.[10]