# Heronian triangle

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In geometry, a Heronian triangle is a triangle that has side lengths and area that are all integers. 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, 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.

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.:p.17 Such Heronian triangles are known as indecomposable. However, if one allows Pythagorean triples with rational values, not necessarily integers, then a decomposition into right triangles with rational sides always exists, 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. Thus every Heronian triangle has an odd number of sides of even length,: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.
• All the altitudes of a Heronian triangle are rational. 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. 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 $p_{a}={\tfrac {2aA}{a^{2}+b^{2}-c^{2}}},$ $p_{b}={\tfrac {2bA}{a^{2}+b^{2}-c^{2}}},$ and $p_{c}={\tfrac {2cA}{a^{2}-b^{2}+c^{2}}},$ where the sides are abc and the area is A; in a Heronian triangle all of a, b, c, and A are integers.
• There are no equilateral Heronian triangles.
• There are no Heronian triangles with a side length of either 1 or 2.
• There exist an infinite number of primitive Heronian triangles with one side length equal to a provided that a > 2.
• There are no Heronian triangles whose side lengths form a geometric progression.
• If any two sides (but not three) of a Heronian triangle have a common factor, that factor must be the sum of two squares.
• 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.
• Any square inscribed in a Heronian triangle has rational sides: For a general triangle the inscribed square on side of length a has length ${\tfrac {2Aa}{a^{2}+2A}}$ where A is the triangle's area; 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. 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.
• All Heronian triangles can be placed on a lattice with each vertex at a lattice point.

## 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:

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

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

The proportionality factor is generally a rational  ​pq  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. 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:Thm. 4

$a=25n^{2}+5n-5=5(5n^{2}+n-1),$ $b=25n^{3}-5n^{2}-7n+3=(5n+3)(5n^{2}-4n+1),$ $c=25n^{3}+20n^{2}-2n-4=(5n-2)(5n^{2}+6n+2),$ $A=(5n-2)(5n+3)(5n^{2}+n-1),$ $r=5n-2,$ $r_{a}=5n+3,$ $r_{b}=5n^{2}+n-1,$ $r_{c}=A=(5n-2)(5n+3)(5n^{2}+n-1).$ 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.:Thm. 5

### 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

$t=\tan(C/2)=\cot(90^{\circ }-C/2)=\cot(A/2+B/2)={\frac {1-\tan(A/2)\tan(B/2)}{\tan(A/2)+\tan(B/2)}}={\frac {1-rs}{r+s}}\,.$ We can compute the sine of any angle using the formula $\sin \theta ={\frac {2\tan(\theta /2)}{1+\tan ^{2}(\theta /2)}}$ . We use the Law of sines to conclude that the side lengths are proportional to the sines of the interior angles:

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

## 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, and in 1880 Reinhold Hoppe gave a closed-form expression for the solutions. 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.

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.