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 isosceles triangle 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 adjacent picture. 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.

Some Heronian triangles cannot be obtained by joining together two right-angled triangles with integer sides as described above. For example, a 5, 29, 30 Heronian triangle with area 72 cannot be constructed from two integer Pythagorean triangles since none of its altitudes are integers. Also no primitive Pythagorean triangle can be constructed from two smaller integer Pythagorean triangles.^[4]:p.17 Such Heronian triangles are known as indecomposable.^[4] However, if one allows Pythagorean triples with rational values, not necessarily integers, then a decomposition into right triangles with rational sides always exists,^[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.

Exact formula for Heronian triangles

The Indian mathematician Brahmagupta (598-668 A.D.) derived the parametric solution such that every Heronian triangle has sides proportional to:^[6][7]

$a=n(m^{2}+k^{2}) \,$

$b=m(n^{2}+k^{2}) \,$

$c=(m+n)(mn-k^{2}) \,$

${\text{Semiperimeter}}=s=(a+b+c)/2=mn(m+n)\,$

$\text{Area}=mnk(m+n)(mn-k^{2}) \,$

${\text{Inradius}}=k(mn-k^{{2}})\,$

$s-a=n(mn-k^{{2}})\,$

$s-b=m(mn-k^{{2}})\,$

$s-c=(m+n)k^{{2}}\,$

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   ^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.

The obstacle for a computational use of Brahmagupta’s parametric solution is the denominator q of the proportionality factor. q can only be determined by calculating the greatest common divisor of the three sides ( gcd(a, b, c) ) and introduces an element of unpredictability into the generation process.^[7] The easiest way of generating lists of Heronian triangles is to generate all integer triangles up to a maximum side length and test for an integral area.

Faster algorithms have been derived by Kurz (2008).

See also formulas for Heronian triangles with one angle equal to twice another, Heronian triangles with sides in arithmetic progression, and isosceles Heronian triangles.

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

Lists of primitive Heronian triangles whose sides do not exceed 6,000,000 can be found at "Lists of primitive Heronian triangles". Sascha Kurz, University of Bayreuth, Germany. Retrieved 29 March 2016.

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).^[8][9]

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. A method for generating all solutions to this problem based on continued fractions was described in 1864 by Edward Sang,^[10] and in 1880 Reinhold Hoppe gave a closed-form expression for the solutions.^[11] The first few examples of these almost-equilateral triangles are listed in the following table (sequence A003500 in the OEIS):

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

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,

${\big (}(n-1)^{2}+n^{2}+(n+1)^{2}{\big )}^{2}-2{\big (}(n-1)^{4}+n^{4}+(n+1)^{4}{\big )}=(6ny)^{2}=(4A)^{2}$

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.^[12]

The variable n is of the form $n={\sqrt {2+2k}}$, where k is 7, 97, 1351, 18817, …. The numbers in this sequence have the property that k consecutive integers have integral standard deviation.^[13]

See also

• Heronian tetrahedron
• Brahmagupta quadrilateral
• Robbins pentagon

External links

• Weisstein, Eric W. "Heronian triangle". MathWorld.

• Online Encyclopedia of Integer Sequences Heronian
• Wm. Fitch Cheney, Jr. (January 1929), "Heronian Triangles", Am. Math. Monthly, 36 (1): 22–28, JSTOR 2300173
• S. sh. Kozhegel'dinov (1994), "On fundamental Heronian triangles", Math. Notes, 55 (2): 151–6, doi:10.1007/BF02113294

References

1. Carlson, John R. (1970), "Determination of Heronian Triangles" (PDF), Fibonacci Quarterly, 8: 499–506
2. Beauregard, Raymond A.; Suryanarayan, E. R. (January 1998), "The Brahmagupta Triangles" (PDF), College Math Journal, 29 (1): 13–17, doi:10.2307/2687630
3. Weisstein, Eric W. "Heronian Triangle". MathWorld.
4. Yiu, Paul (2008), Heron triangles which cannot be decomposed into two integer right triangles (PDF), 41st Meeting of Florida Section of Mathematical Association of America
5. Sierpiński, Wacław (2003) [1962], Pythagorean Triangles, Dover Publications, Inc., ISBN 978-0-486-43278-6
6. Carmichael, R. D., 1914, "Diophantine Analysis", pp.11-13; in R. D. Carmichael, 1959, The Theory of Numbers and Diophantine Analysis, Dover Publications, Inc.
7. Kurz, Sascha (2008). "On the generation of Heronian triangles". Serdica Journal of Computing. 2 (2): 181–196. arXiv:1401.6150. MR 2473583..
8. Dickson, Leonard Eugene (2005), History of the Theory of Numbers, Volume Il: Diophantine Analysis, Dover Publications, p. 199, ISBN 9780486442334
9. Markowitz, L. (1981), "Area = Perimeter", The Mathematics Teacher, 74 (3): 222–3, Zbl 1982d.06561
10. Sang, Edward, "On the theory of commensurables", Transactions of the Royal Society of Edinburgh, 23: 721–760. See in particular p. 734.
11. Gould, H. W. (February 1973), "A triangle with integral sides and area" (PDF), Fibonacci Quarterly, 11 (1): 27–39.
12. Richardson, William H. (2007), Super-Heronian Triangles
13. Online Encyclopedia of Integer Sequences,  A011943.