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.[3] The term is sometimes applied more widely to triangles whose sides and area are all rational numbers,[4] since one can rescale the sides by a common multiple to obtain a triangle that is Heronian in the above sense.

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

${\displaystyle A={\frac {1}{2}}(b+d)a}$ (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.[5]^:p.17 Such Heronian triangles are known as indecomposable.[5] However, if one allows Pythagorean triples with rational values, not necessarily integers, then a decomposition into right triangles with rational sides always exists,[6] 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 as follows:

• The perimeter of a Heronian triangle is always an even number.[7] Thus every Heronian triangle has an odd number of sides of even length,[8]^:p.3 and every primitive Heronian triangle has exactly one even side.
• The semiperimeter s of a Heronian triangle with sides a, b and c can never be prime. This can be seen from the fact that s(s−a)(s−b)(s−c) has to be a perfect square and if s is a prime then one of the other terms must have s as a factor but this is impossible as these terms are all less than s.
• The area of a Heronian triangle is always divisible by 6.[7]
• All the altitudes of a Heronian triangle are rational.[9] This can be seen from the fact that the area of a triangle is half of one side times its altitude from that side, and a Heronian triangle has integer sides and area. Some Heronian triangles have three non-integer altitudes, for example the acute (15, 34, 35) with area 252 and the obtuse (5, 29, 30) with area 72. Any Heronian triangle with one or more non-integer altitudes can be scaled up by a factor equalling the least common multiple of the altitudes' denominators in order to obtain a similar Heronian triangle with three integer altitudes.
• Heronian triangles that have no integer altitude (indecomposable and non-Pythagorean) have sides that are all divisible by primes of the form 4k+1.[5] However decomposable Heronian triangles must have two sides that are the hypotenuse of Pythagorean triangles. Hence all Heronian triangles that are not Pythagorean have at least two sides that are divisible by primes of the form 4k+1. All that remains are Pythagorean triangles. Therefore, all Heronian triangles have at least one side that is divisible by primes of the form 4k+1. Finally if a Heronian triangle has only one side divisible by primes of the form 4k+1 it has to be Pythagorean with the side as the hypotenuse and the hypotenuse must be divisible by 5.
• All the interior perpendicular bisectors of a Heronian triangle are rational: For any triangle these are given by ${\displaystyle p_{a}={\tfrac {2aA}{a^{2}+b^{2}-c^{2}}},}$ ${\displaystyle p_{b}={\tfrac {2bA}{a^{2}+b^{2}-c^{2}}},}$ and ${\displaystyle p_{c}={\tfrac {2cA}{a^{2}-b^{2}+c^{2}}},}$ where the sides are a ≥ b ≥ c and the area is A;[10] in a Heronian triangle all of a, b, c, and A are integers.
• There are no equilateral Heronian triangles.[9]
• There are no Heronian triangles with a side length of either 1 or 2.[11]
• There exist an infinite number of primitive Heronian triangles with one side length equal to a provided that a > 2.[11]
• There are no Heronian triangles whose side lengths form a geometric progression.[12]
• If any two sides (but not three) of a Heronian triangle have a common factor, that factor must be the sum of two squares.[13]
• Every angle of a Heronian triangle has a rational sine. This follows from the area formula Area = (1/2)ab sin C, in which the area and the sides a and b are integers, and equivalently for the other angles.
• Every angle of a Heronian triangle has a rational cosine. This follows from the law of cosines , c² = a² + b² − 2ab cos C, in which the sides a, b, and c are integers, and equivalently for the other angles.
• Since all Heronian triangles have all angles' sines and cosines rational, this implies that each oblique angle of a Heron triangle has a rational tangent, cotangent, secant, and cosecant. Furthermore, half of each angle has a rational tangent because tan C/2 = sin C / (1 + cos C), and equivalently for other angles.
• There are no Heronian triangles whose three internal angles form an arithmetic progression. This is because all plane triangles with angles in an arithmetic progression must have one angle of 60°, which does not have a rational sine.[14]
• Any square inscribed in a Heronian triangle has rational sides: For a general triangle the inscribed square on side of length a has length ${\displaystyle {\tfrac {2Aa}{a^{2}+2A}}}$ where A is the triangle's area;[15] in a Heronian triangle, both A and a are integers.
• Every Heronian triangle has a rational inradius (radius of its inscribed circle): For a general triangle the inradius is the ratio of the area to half the perimeter, and both of these are rational in a Heronian triangle.
• Every Heronian triangle has a rational circumradius (the radius of its circumscribed circle): For a general triangle the circumradius equals one-fourth the product of the sides divided by the area; in a Heronian triangle the sides and area are integers.
• In a Heronian triangle the distance from the centroid to each side is rational, because for all triangles this distance is the ratio of twice the area to three times the side length.[16] This can be generalized by stating that all centers associated with Heronian triangles whose barycentric coordinates are rational ratios have a rational distance to each side. These centers include the circumcenter, orthocenter, nine-point center, symmedian point, Gergonne point and Nagel point.[17]
• All Heronian triangles can be placed on a lattice with each vertex at a lattice point.[18]

Exact formula for all Heronian triangles

The Indian mathematician Brahmagupta (598-668 A.D.) derived the parametric solution such that every Heronian triangle has sides proportional to:[19][20]

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

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

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

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

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

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

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

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

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

for integers m, n and k where:

${\displaystyle \gcd {(m,n,k)}=1}$

${\displaystyle mn>k^{2}\geq 1}$

${\displaystyle m\geq n\geq 1}$.

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.[20] 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).

There are infinitely many primitive and indecomposable non-Pythagorean Heronian triangles with integer values for the inradius and all three of the exradii, including the ones generated by[21]^{:Thm. 4}

${\displaystyle a=25n^{2}+5n-5=5(5n^{2}+n-1),}$

${\displaystyle b=25n^{3}-5n^{2}-7n+3=(5n+3)(5n^{2}-4n+1),}$

${\displaystyle c=25n^{3}+20n^{2}-2n-4=(5n-2)(5n^{2}+6n+2),}$

${\displaystyle A=(5n-2)(5n+3)(5n^{2}+n-1),}$

${\displaystyle r=5n-2,}$

${\displaystyle r_{a}=5n+3,}$

${\displaystyle r_{b}=5n^{2}+n-1,}$

${\displaystyle r_{c}=A=(5n-2)(5n+3)(5n^{2}+n-1).}$

There are infinitely many Heronian triangles that can be placed on a lattice such that not only are the vertices at lattice points, as holds for all Heronian triangles, but additionally the centers of the incircle and excircles are at lattice points.[21]^{:Thm. 5}

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

Second approach

A triangle with side lengths and interior angles labeled. Capital A, B and C are the angles, and lower-case a, b, and c are the sides opposite them.

The tangent of half of any interior angle of a Heronian triangle is necessarily rational; see properties above. These half angles are positive, and they sum to 90° (π/2 radians) because the interior angles (A, B, C) sum to 180° (π radians). We start by choosing r = tan(A/2) and s = tan(B/2) to be any positive rational numbers satisfying rs < 1. The limit of 1 ensures that angle A/2 + B/2 is less than 90° and thus the angle C/2 will be positive. The value t = tan(C/2) will also be a positive rational number because

${\displaystyle t=\tan(C/2)=\cot(90^{\circ }-C/2)=\cot(A/2+B/2)={\frac {1-\tan(A/2)\tan(B/2)}{\tan(A/2)+\tan(B/2)}}={\frac {1-rs}{r+s}}\,.}$

We can compute the sine of any angle using the formula ${\displaystyle \sin \theta ={\frac {2\tan(\theta /2)}{1+\tan ^{2}(\theta /2)}}}$. We use the Law of sines to conclude that the side lengths are proportional to the sines of the interior angles:

${\displaystyle (a,b,c)\propto \left({\frac {2r}{1+r^{2}}},{\frac {2s}{1+s^{2}}},{\frac {2t}{1+t^{2}}}\right).}$

The values a, b, and c are rational because the values of r, s, and t are rational. Integer values for the side lengths can be obtained by multiplying the side lengths by an integer that clears the denominators.

When it is also the case that r, s, or t equals 1 then the corresponding interior angle will be a right angle and the three sides will also define a Pythagorean triple.

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).[22][23]

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,[24] and in 1880 Reinhold Hoppe gave a closed-form expression for the solutions.[25] 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:

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

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

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

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.[26]

The variable n is of the form ${\displaystyle n={\sqrt {2+2k}}}$, where k is 7, 97, 1351, 18817, …. The numbers in this sequence have the property that k consecutive integers have integral standard deviation.[27]

See also

• Heronian tetrahedron
• Brahmagupta quadrilateral
• Robbins pentagon
• Integer triangle#Heronian triangles

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 Mathematics Journal, 29 (1): 13–17, doi:10.2307/2687630, JSTOR 2687630
3. Peddie, Jon. (2013). The history of visual magic in computers : how beautiful images are made in CAD, 3D, VR and AR. London: Springer. pp. 25–26. ISBN 978-1-4471-4932-3. OCLC 849634980.CS1 maint: date and year (link)
4. Weisstein, Eric W. "Heronian Triangle". MathWorld.
5. 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
6. Sierpiński, Wacław (2003) [1962], Pythagorean Triangles, Dover Publications, Inc., ISBN 978-0-486-43278-6
7. Friche, Jan (2 January 2002). "On Heron Simplices and Integer Embedding". Ernst-Moritz-Arndt Universät Greiswald Publication. arXiv:math/0112239. Cite journal requires |journal= (help)
8. Buchholz, R. H.; MacDougall, J. A. (2001). "Cyclic Polygons with Rational Sides and Area". CiteSeerX Penn State University: 3. CiteSeerX 10.1.1.169.6336. Cite journal requires |journal= (help)
9. Somos, M. (December 2014). "Rational triangles". Retrieved 2018-11-04.
10. Mitchell, Douglas W. (2013), "Perpendicular Bisectors of Triangle Sides", Forum Geometricorum 13, 53−59: Theorem 2.
11. Carlson, John R. (1970). "Determination of Heronian triangles" (PDF). San Diego State College. Cite journal requires |journal= (help)
12. Buchholz, R. H.; MacDougall, J. A. (1999). "Heron Quadrilaterals with sides in Arithmetic or Geometric progression". Bulletin of the Australian Mathematical Society. 59: 263–269. doi:10.1017/s0004972700032883.
13. Blichfeldt, H. F. (1896–1897). "On Triangles with Rational Sides and Having Rational Areas". Annals of Mathematics. 11 (1/6): 57–60. doi:10.2307/1967214. JSTOR 1967214.
14. Zelator, K., "Triangle Angles and Sides in Progression and the diophantine equation x²+3y²=z²", Cornell Univ. archive, 2008
15. Bailey, Herbert, and DeTemple, Duane, "Squares inscribed in angles and triangles", Mathematics Magazine 71(4), 1998, 278–284.
16. Clark Kimberling, "Trilinear distance inequalities for the symmedian point, the centroid, and other triangle centers", Forum Geometricorum, 10 (2010), 135−139. http://forumgeom.fau.edu/FG2010volume10/FG201015index.html
17. Clark Kimberling's Encyclopedia of Triangle Centers "Encyclopedia of Triangle Centers". Archived from the original on 2012-04-19. Retrieved 2012-06-17.
18. Yiu, P., "Heronian triangles are lattice triangles", American Mathematical Monthly 108 (2001), 261–263.
19. Carmichael, R. D., 1914, "Diophantine Analysis", pp.11-13; in R. D. Carmichael, 1959, The Theory of Numbers and Diophantine Analysis, Dover Publications, Inc.
20. Kurz, Sascha (2008). "On the generation of Heronian triangles". Serdica Journal of Computing. 2 (2): 181–196. arXiv:1401.6150. Bibcode:2014arXiv1401.6150K. MR 2473583..
21. Zhou, Li, "Primitive Heronian Triangles With Integer Inradius and Exradii", Forum Geometricorum 18, 2018, 71-77. http://forumgeom.fau.edu/FG2018volume18/FG201811.pdf
22. Dickson, Leonard Eugene (2005), History of the Theory of Numbers, Volume Il: Diophantine Analysis, Dover Publications, p. 199, ISBN 9780486442334
23. Markowitz, L. (1981), "Area = Perimeter", The Mathematics Teacher, 74 (3): 222–3
24. Sang, Edward, "On the theory of commensurables", Transactions of the Royal Society of Edinburgh, 23: 721–760, doi:10.1017/s0080456800020019. See in particular p. 734.
25. Gould, H. W. (February 1973), "A triangle with integral sides and area" (PDF), Fibonacci Quarterly, 11 (1): 27–39.
26. Richardson, William H. (2007), Super-Heronian Triangles
27. Online Encyclopedia of Integer Sequences, OEIS: A011943.

External links

• Weisstein, Eric W. "Heronian triangle". MathWorld.
• Online Encyclopedia of Integer Sequences Heronian
• Wm. Fitch Cheney, Jr. (January 1929), "Heronian Triangles", Amer. 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