Heronian triangle
[3] Heron's formula implies that the Heronian triangles are exactly the positive integer solutions of the Diophantine equation that is, the side lengths and area of any Heronian triangle satisfy the equation, and any positive integer solution of the equation describes a Heronian triangle.
produce a rational Heronian triangle such that its side lengths
A byproduct of the proof is that exactly one of the side lengths of a primitive Heronian triangle is an even integer.
Supposing that c is odd, the right-hand side of the Diophantine equation can be rewritten with
Starting with the Pythagorean triple 3, 4, 5 this gives two Heronian triangles with side lengths (5, 5, 6) and (5, 5, 8) and area 12.
Many quantities related to a Heronian triangle are rational numbers.
A parametric equation or parametrization of Heronian triangles consists of an expression of the side lengths and area of a triangle as functions—typically polynomial functions – of some parameters, such that the triangle is Heronian if and only if the parameters satisfy some constraints—typically, to be positive integers satisfying some inequalities.
In the 18th century, Leonhard Euler provided another parametrization and proved that it generates all Heronian triangles.
This provides a proof that Brahmagupta's and Euler's parametrizations generate all Heronian triangles.
The Indian mathematician Brahmagupta (598-668 A.D.) discovered the following parametric equations for generating Heronian triangles,[20] but did not prove that every similarity class of Heronian triangles can be obtained this way.
[citation needed] For three positive integers m, n and k that are setwise coprime (
The fact that the generated triangle is not primitive is an obstacle for using this parametrization for generating all Heronian triangles with size lengths less than a given bound (since the size of
(to guarantee positive side lengths): where s is the semiperimeter, A is the area, and r is the inradius.
Even when m, n, p, and q are pairwise coprime, the resulting Heronian triangle may not be primitive.
it can be seen that they are the half-angle tangents of the interior angles of a class of similar Heronian triangles.
To make this bijection explicit, one can choose, as a specific member of the similarity class, the triangle inscribed in a unit-diameter circle with side lengths equal to the sines of the opposite angles:[23] where
To obtain an (integral) Heronian triangle, the denominators of a, b, and c must be cleared.
Kurz (2008) has derived fast algorithms for generating Heronian triangles.
There are infinitely many primitive and indecomposable non-Pythagorean Heronian triangles with integer values for the inradius
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.
The list of primitive Heronian triangles whose sides do not exceed 6,000,000 has been computed by Kurz (2008).
As of February 2021, only two primitive Heronian triangles with perfect square sides are known: (1853², 4380², 4427², Area=32918611718880), published in 2013.
[26] 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),[27][28] though only four of them are primitive.
The first few examples of these almost-equilateral triangles are listed in the following table (sequence A102341 in the OEIS): 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,[29] and in 1880 Reinhold Hoppe gave a closed-form expression for the solutions.
generates all n for positive integers t. Equivalently, let A = area and y = inradius, then, 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 numbers in this sequence have the property that k consecutive integers have integral standard deviation.
