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In geometry, a Heronian triangle is a triangle whose sidelengths and area are all rational numbers. It is named after Hero of Alexandria. Any such rational triangle can be scaled up to a corresponding triangle with integer sides and area, and often the term Heronian triangle is used to refer to the latter.

Properties

Any triangle whose sidelengths are a Pythagorean triple is Heronian, as the sidelengths of such a triangle are integers, and its area (being a right-angled triangle) is just half of the product of the two sides at the right angle.
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 by the side 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 by the side of length a, to obtain a triangle with integer sidelengths c, e, and b + d, with rational area

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

An interesting question to ask is whether all Heronian triangles can be obtained by joining together two right-angled triangles described in this procedure. The answer is no. If one takes the Heronian triangle with sidelengths 0.5, 0.5, and 0.6, which is just the triangle described above shrunk 10 times, it clearly cannot be decomposed into two triangles with integer sidelengths. Nor for example can a 5, 29, 30 triangle with area 72, since none of its altitudes are integers. However, if one allows for Pythagorean triples with rational entries, not necessarily integers, then the answer is affirmative,[1] because every altitude of a Heronian triangle is rational (since it equals twice the rational area divided by the rational base). Note that a triple with rational entries is just a scaled version of a triple with integer entries.
Theorem

Given a Heronian triangle, one can split it into two right-angled triangles, whose sidelengths form Pythagorean triples with rational entries.

Proof of the theorem

Consider again the illustration to the right, where this time it is known that c, e, b + d, and the triangle area A are rational. We can assume that the notation was chosen so that the sidelength b + d is the largest, as then the perpendicular onto this side from the opposite vertex falls inside this segment. To show that the triples (a, b, c) and (a, d, e) are Pythagorean, one must prove 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 rational. Left is 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

\( b^2-d^2=c^2-e^2\, \)

or

\( (b-d)(b+d)=c^2-e^2\, \)

or

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

The right-hand side is rational, because by assumption, c, e, and b + d are rational. Then, b − d is rational. This, together with b + d being rational implies by adding these up that b is rational, and then d must be rational too. Q.E.D.
Exact formula for Heronian triangles

All Heronian triangles can be generated [2] as multiples of:

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

for integers m, n and k subject to the contraints:

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

See also Heronian triangles with one angle equal to twice another.
Examples

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

Area	Perimeter	length b+d	length e	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

Almost-equilateral Heronian triangles

A Heronian triangle is a triangle with rational sides, area and inradius. 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, expressed as integers, are of the form n − 1, n, n + 1. The first few examples of these almost-equilateral triangles are set forth in the following table.
Side length Area Inradius

Side length			Area	Inradius
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

37633 37634 37635 613283664 10864

Subsequent values of n can be found by multiplying the last known value by 4, then subtracting the next to the last one (52 = 4 × 14 − 4, 194 = 4 × 52 − 14, etc), as expressed in

\( n_t = 4n_{t-1} - n_{t-2} \,\! \)

where t denotes any row in the table.

This sequence (sequence A003500 in OEIS) can also be generated from the solutions to the Pell's equation x² − 3y² = 1, which can in turn be derived from the regular continued fraction expansion for √3.[3]
See also

Heronian tetrahedron
Robbins pentagon

External links

Weisstein, Eric W., "Heronian triangle" from MathWorld.
Online Encyclopedia of Integer Sequences Heronian
Wm. Fitch Cheney, Jr., Heronian Triangles Am. Math. Montly 36 (1) (1929) 22-28.
S. sh. Kozhegel'dinov On fundamental Heronian triangles Math. Notes 55 (2) (1994) 151-156.

References

^ Sierpiński, Wacław, Pythagorean Triangles, Dover Publ., 2003 (orig. 1962).
^ Carmichael, R. D., 1914, "Diophantine Analysis", pp.11-13; in R. D. Carmichael, 1959, The Theory of Numbers and Diophantine Analysis, Dover.
^ William H. Richardson (2007), Super-Heronian Triangles.

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