Editing Heronian triangle

{{Short description|Triangle whose side lengths and area are integers}}
In [[geometry]], a '''Heronian triangle''' (or '''Heron triangle''') is a [[triangle]] whose side lengths {{mvar|a}}, {{mvar|b}}, and {{mvar|c}} and [[area]] {{mvar|A}} are all positive [[integer]]s.<ref name=carlson>{{citation |first1=John R. |last1=Carlson |title=Determination of Heronian Triangles |journal=[[Fibonacci Quarterly]] |volume=8 |pages=499–506 |year=1970 |issue=5 |doi=10.1080/00150517.1970.12431055 |url=http://www.fq.math.ca/Scanned/8-5/carlson-a.pdf }}</ref><ref>{{citation |first1=Raymond A. |last1=Beauregard |first2=E. R. |last2=Suryanarayan |title=The Brahmagupta Triangles |journal=College Mathematics Journal |volume=29 |issue=1 |pages=13–17 |date=January 1998 |doi= 10.2307/2687630|url=http://www.maa.org/mathdl/CMJ/methodoflastresort.pdf |jstor=2687630 }}
</ref> Heronian triangles are named after [[Heron of Alexandria]], based on their relation to [[Heron's formula]] which Heron demonstrated with the example triangle of sides {{math|13, 14, 15}} and area {{math|84}}.<ref>{{cite journal |last=Sastry |first=K. R. S. |year=2001 |title=Heron triangles: A Gergonne-Cevian-and-median perspective |journal=[[Forum Geometricorum]] |volume=1 |number=2001 |pages=17–24 |url=https://forumgeom.fau.edu/FG2001volume1/FG200104.pdf }}</ref>

Heron's formula implies that the Heronian triangles are exactly the positive integer solutions of the [[Diophantine equation]]
:<math>16\,A^2=(a+b+c)(a+b-c)(b+c-a)(c+a-b);</math> 
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.<ref>The sides and area of any triangle satisfy the Diophantine equation obtained by squaring both sides of Heron's formula; see {{slink|Heron's formula|Proofs}}. Conversely, consider a solution of the equation where <math>(a, b, c, A)</math> are all positive integers. It corresponds to a valid triangle [[if and only if]] the [[triangle inequality]] is satisfied, that is, if the three integers <math>a+b-c,</math> <math>b+c-a,</math> and <math>c+a-b</math> are all positive. This is necessarily true in this case: if any of these sums were negative or zero, the other two would be positive and the right-hand side of the equation would thus be negative or zero and could not possibly equal the left-hand side <math>16\,A^2,</math> which is positive.</ref>

If the three side lengths are [[setwise coprime]] (meaning that the greatest common divisor of all three sides is 1), the Heronian triangle is called ''primitive''.

Triangles whose side lengths and areas are all [[rational number]]s (positive rational solutions of the above equation) are sometimes also called ''Heronian triangles'' or ''rational triangles'';<ref>{{MathWorld |title=Heronian Triangle |id=HeronianTriangle}}</ref> in this article, these more general triangles will be called ''rational Heronian triangles''. Every (integral) Heronian triangle is a rational Heronian triangle. Conversely, every rational Heronian triangle is [[similar (geometry)|geometrically similar]] to exactly one primitive Heronian triangle.

In any rational Heronian triangle, the three [[altitude (triangle)|altitude]]s, the [[circumradius]], the [[incircle and excircles of a triangle|inradius and exradii]], and the [[sine and cosine|sines and cosines]] of the three angles are also all rational numbers.

==Scaling to primitive triangles==
[[Scaling (geometry)|Scaling]] a triangle with a factor of {{mvar|s}} consists of multiplying its side lengths by {{mvar|s}}; this multiplies the area by <math>s^2</math> and produces a [[similarity (geometry)|similar]] triangle. Scaling a rational Heronian triangle by a [[rational number|rational]] factor produces another rational Heronian triangle.

Given a rational Heronian triangle of side lengths <math display=inline>\frac pd, \frac qd,\frac rd,</math> the scale factor <math display=inline>\frac d{\gcd(p,q,r)}</math> produces a rational Heronian triangle such that its side lengths <math display=inline>a, b,c</math> are setwise [[coprime number|coprime integer]]s. It is proved below that the area {{mvar|A}} is an integer, and thus the triangle is a Heronian triangle. Such a triangle is often called a ''primitive Heronian triangle.''

In summary, every similarity [[equivalence class|class]] of rational Heronian triangles contains exactly one primitive Heronian triangle. A byproduct of the proof is that exactly one of the side lengths of a primitive Heronian triangle is an even integer.

''Proof:'' One has to prove that, if the side lengths <math display=inline>a, b,c</math> of a rational Heronian triangle are coprime integers, then the area {{mvar|A}} is also an integer and exactly one of the side lengths is even.

The Diophantine equation given in the introduction shows immediately that <math>16A^2</math> is an integer. Its square root <math>4A</math> is also an integer, since the square root of an integer is either an integer or an [[irrational number]].

If exactly one of the side lengths is even, all the factors in the right-hand side of the equation are even, and, by dividing the equation by {{math|16}}, one gets that <math>A^2</math> and <math>A</math> are integers.

As the side lengths are supposed to be coprime, one is left with the case where one or three side lengths are odd. Supposing that {{mvar|c}} is odd, the right-hand side of the Diophantine equation can be rewritten
:<math>((a+b)^2-c^2)(c^2-(a-b)^2),</math>
with <math>a+b</math> and <math>a-b</math> even. As the square of an odd integer is [[modular arithmetic|congruent]] to <math>1</math> modulo {{math|4}}, the right-hand side of the equation must be congruent to <math>-1</math> modulo {{math|4}}. It is thus impossible, that one has a solution of the Diophantine equation, since <math>16A^2</math> must be the square of an integer, and the square of an integer is congruent to {{math|0}} or {{math|1}} modulo {{math|4}}.

==Examples==
Any [[Pythagorean triangle]] is a Heronian triangle. The side lengths of such a triangle are [[integer]]s, by definition. In any such triangle, one of the two shorter sides has even length, so the area (the product of these two sides, divided by two) is also an integer.

[[Image:Triangle-heronian.svg|thumb|right|A triangle with sidelengths {{mvar|c}}, {{mvar|e}} and {{math|''b'' + ''d''}}, and height {{mvar|a}}.]]
Examples of Heronian triangles that are not right-angled are the [[isosceles triangle]] obtained by joining a Pythagorean triangle and its mirror image along a side of the right angle. Starting with the [[Pythagorean triple]] {{math|3, 4, 5}} this gives two Heronian triangles with side lengths {{math|(5, 5, 6)}} and {{math|(5, 5, 8)}} and area {{math|12}}.

More generally, given two Pythagorean triples <math>(a,b,c)</math> and <math>(a,d,e)</math> with largest entries {{mvar|c}} and {{mvar|e}}, one can join the corresponding triangles along the sides of length {{mvar|a}} (see the figure) for getting a Heronian triangle with side lengths <math>c,e,b+d</math> and area <math display=inline>\tfrac12a(b+d)</math> (this is an integer, since the area of a Pythagorean triangle is an integer).

There are Heronian triangles that cannot be obtained by joining Pythagorean triangles. For example, the Heronian triangle of side lengths <math>5, 29, 30</math> and area 72, since none of its altitudes is an integer. Such Heronian triangles are known as {{Em|indecomposable}}.<ref name=Yiu>{{citation |first=Paul |last=Yiu |title=Heron triangles which cannot be decomposed into two integer right triangles |url=http://math.fau.edu/yiu/Southern080216.pdf |year=2008 |publisher=41st Meeting of Florida Section of Mathematical Association of America }}</ref> However, every Heronian triangle can be constructed from right triangles with [[rational number|rational]] side lengths, and is thus similar to a decomposable Heronian triangle. In fact, at least one of the altitudes of a triangle is inside the triangle, and divides it into two right triangles. These triangles have rational sides, since the cosine and the sine of the angles of a Heronian triangle are rational numbers, and, with notation of the figure, one has <math>a=c\sin \alpha</math> and <math>b=c\cos\alpha,</math> where <math>\alpha</math> is the left-most angle of the triangle.

==Rationality properties==

Many quantities related to a Heronian triangle are rational numbers. In particular:

*All the altitudes of a Heronian triangle are rational.<ref name=Somos>{{cite web|last=Somos|first=M.|author-link=Michael Somos|title=Rational triangles|url=http://grail.eecs.csuohio.edu/~somos/rattri.html|date=December 2014|access-date=2018-11-04|archive-date=2021-12-20|archive-url=https://web.archive.org/web/20211220133826/http://grail.eecs.csuohio.edu/~somos/rattri.html|url-status=dead}}</ref> 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 [[Similarity (geometry)|similar]] Heronian triangle with three integer altitudes.
*All the [[Bisection#Perpendicular bisectors|interior perpendicular bisectors]] of a Heronian triangle are rational: For any triangle these are given by <math>p_a=\tfrac{2aA}{a^2+b^2-c^2},</math> <math>p_b=\tfrac{2bA}{a^2+b^2-c^2},</math> and <math>p_c=\tfrac{2cA}{a^2-b^2+c^2},</math> where the sides are {{math|''a'' &ge; ''b'' &ge; ''c''}} and the area is {{mvar|A}};<ref>Mitchell, Douglas W. (2013), "Perpendicular Bisectors of Triangle Sides", ''[[Forum Geometricorum]]'' 13, 53−59: Theorem 2.</ref> in a Heronian triangle all of {{mvar|a}}, {{mvar|b}}, {{mvar|c}}, and {{mvar|A}} are integers.
*Every [[interior angle]] of a Heronian triangle has a rational sine. This follows from the area formula {{math|1=''Area'' = (1/2)''ab'' sin ''C''}}, in which the area and the sides {{mvar|a}} and {{mvar|b}} are integers, and equivalently for the other interior angles.
*Every interior angle of a Heronian triangle has a rational cosine. This follows from the [[law of cosines]], {{math|1=''c''{{sup|2}} = ''a''{{sup|2}} + ''b''{{sup|2}} − 2''ab'' cos ''C''}}, in which the sides {{mvar|a}}, {{mvar|b}}, and {{mvar|c}} are integers, and equivalently for the other interior angles.
*Because all Heronian triangles have all interior angles' sines and cosines rational, this implies that the tangent, cotangent, secant, and cosecant of each interior angle is either rational or infinite.
*Half of each interior angle has a rational tangent because {{math|1=tan ''C''/2 = sin ''C'' / (1 + cos ''C'')}}, and equivalently for other interior angles.  Knowledge of (at least two of) these half-angle tangent values is sufficient to reconstruct the side lengths of a primitive Heronian triangle ([[#Half-angle tangent parametrization|see below]]).
*For any triangle, the angle spanned by a side as viewed from the [[circumcenter|center]] of the [[circumcircle]] is twice the interior angle of the triangle vertex opposite the side.  Because the half-angle tangent for each interior angle of a Heronian triangle is rational, it follows that the quarter-angle tangent of each such central angle of a Heronian triangle is rational.  (Also, the quarter-angle tangents are rational for the central angles of a [[Brahmagupta quadrilateral]], but is an unsolved problem whether this is true for all [[Robbins pentagon]]s.)  [[Concyclic points#Integer area and side lengths|The reverse is true for all cyclic polygon]]s generally; if all such central angles have rational tangents for their quarter angles then the cyclic polygon can be scaled to simultaneously have integer area, sides, and diagonals (connecting any two vertices).
*There are no Heronian triangles whose three internal angles form an arithmetic progression. This is because all plane triangles with interior angles in an arithmetic progression must have one interior angle of 60°, which does not have a rational sine.<ref name=Zelator>[https://arxiv.org/ftp/arxiv/papers/0803/0803.3778.pdf Zelator, K., "Triangle Angles and Sides in Progression and the diophantine equation x{{sup|2}}+3y{{sup|2}}=z{{sup|2}}", ''Cornell Univ. archive'', 2008]</ref>
*Any square inscribed in a Heronian triangle has rational sides: For a general triangle the [[Triangle#Figures inscribed in a triangle|inscribed square]] on side of length {{mvar|a}} has length <math>\tfrac{2Aa}{a^2+2A}</math> where {{mvar|A}} is the triangle's area;<ref>Bailey, Herbert, and DeTemple, Duane, "Squares inscribed in angles and triangles", ''[[Mathematics Magazine]]'' 71(4), 1998, 278–284.</ref> in a Heronian triangle, both {{mvar|A}} and {{mvar|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 [[Circumscribed circle#Triangles|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.<ref>[[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</ref> This can be generalized by stating that all centers associated with Heronian triangles whose [[Barycentric coordinate system|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]].<ref name=ck>Clark Kimberling's [[Encyclopedia of Triangle Centers]] {{cite web|url=https://faculty.evansville.edu/ck6/encyclopedia/ETC.html |title=Encyclopedia of Triangle Centers |access-date=2012-06-17 }}</ref>
*Every Heronian triangle can be placed on a unit-sided [[square lattice]] with each vertex at a lattice point.<ref name=Yiu3>{{cite journal |last=Yiu |first=Paul |date= 2001|title= Heronian triangles are lattice triangles|journal= [[The American Mathematical Monthly]]|volume= 108|issue= 3|pages= 261–263|bibcode= |doi=  10.1080/00029890.2001.11919751}}</ref> As a corollary, every rational Heronian triangle can be placed into a two-dimensional [[Cartesian coordinate system]] with all rational-valued coordinates.

==Properties of side lengths==
Here are some properties of side lengths of Heronian triangles, whose side lengths are {{math|''a'', ''b'', ''c''}} and area is {{mvar|A}}.
*Every primitive Heronian triangle has one even and two odd sides (see {{slink||Scaling to primitive triangles}}). It follows that a Heronian triangle has either one or three sides of even length,<ref name=Buchholz1>{{cite journal |last1=Buchholz  |first1=Ralph H. |last2=MacDougall |first2=James A. |title=Cyclic Polygons with Rational Sides and Area |journal= [[Journal of Number Theory]] |date=2008 |volume= 128|issue= 1|pages= 17–48|bibcode= |doi=10.1016/j.jnt.2007.05.005|doi-access=free}}</ref>{{rp|p.3}} and that the perimeter of a primitive Heronian triangle is always an even number.<ref name=Friche>{{cite arXiv |title=On Heron Simplices and Integer Embedding |last=Fricke |first=Jan |date=2002-12-21 |eprint=math/0112239}}</ref> 
*There are no equilateral Heronian triangles, since a primitive Heronian triangle has one even side length and two odd side lengths.<ref name=Somos/>
*The area of a Heronian triangle is always divisible by 6.<ref>''Proof''. One can suppose that the Heronian triangle is primitive. The right-hand side of the Diophantine equation can be rewritten as <math>((a+b)^2-c^2)(c^2-(a-b)^2).</math>  If an odd length is chosen for {{mvar|c}}, all squares are odd, and therefore of the form <math>8k+1;</math> and the two differences are multiple of {{math|8{{math|}}}}. So <math>16A^2</math> is multiple of {{math|64}}, and {{mvar|A}} is even. For the divisibility by three, one chooses {{mvar|c}} as non-multiple of {{math|3}} (the triangle is supposed to be primitive). If one of <math>a+b</math> and <math>a-b</math> is not a multiple of {{mvar|3}}, the corresponding factor is a  nultiple of {{mvar|3}} (since the square of a non-multiple of {{mvar|3}} has the form <math>3k+1</math>), and this implies that {{mvar|3}} is a divisor of <math>16A^2.</math> Otherwise, {{mvar|3}} would divide both <math>a+b</math> and <math>a-b,</math>  and the right-hand side of the Diophantine would not be the square of <math>4A,</math> as being congruent to minus times a square  modulo {{math|3}}. So this last case is impossible.</ref><ref name=Friche/>
*There are no Heronian triangles with a side length of either 1 or 2.<ref>''Proof''. Supposing <math>a\ge b\ge c,</math> the [[triangle inequality]] implies <math>b\le a\le b+c.</math> If <math>c=1,</math> this implies that <math>a=b,</math> and the condition that there is exactly one even side length cannot be fulfilled. If <math>c=2,</math> one has two even side lengths if <math>a=b+1.</math> So, <math>a=b,</math> and the Diophantine equation becomes <math>16 A^2=16(a^2-1),</math> which is impossible for two positive integers.</ref><ref name=carlson/>
*There exist an infinite number of primitive Heronian triangles with one side length equal to a given {{mvar|a}}, provided that {{math|''a ''> 2}}.<ref name=carlson/>
*The semiperimeter {{mvar|s}} of a Heronian triangle cannot be prime (as <math>s(s-a)(s-b)(s-c)</math> is the square of the area, and the area is an integer, if {{mvar|s}} were prime, it would divide another factor; this is impossible as these factors are all less than {{mvar|s}}).
*In a Heronian triangles that has no integer altitude ([[#Properties of side lengths|indecomposable]] and non-Pythagorean), all side lengths have a prime factor of the form  {{math|4''k'' + 1}}.<ref name=Yiu/> In a primitive Pythagoran triangle, all [[prime factor]]s of the hypotenuse have the form {{math|4''k'' + 1}}. A decomposable Heronian triangle must have two sides that are the hypotenuse of a Pythagorean triangle, and thus two sides that have prime factors of the form {{math|4''k'' + 1}}. There may also be prime factors of the form {{math|4''k'' + 3}}, since the Pythagorean components of a decomposable Heronian triangle need not to be primitive, even if the Heronian triangle is primitive. In summary, all Heronian triangles have at least one side that is divisible by a prime of the form {{math|4''k'' + 1}}.
*There are no Heronian triangles whose side lengths form a [[geometric progression]].<ref name=Buchholz>{{Cite journal |title=Heron Quadrilaterals with sides in Arithmetic or Geometric progression |last1=Buchholz |first1=Ralph H. |last2=MacDougall |first2=James A. |journal=Bulletin of the Australian Mathematical Society |pages=263–269 |volume=59 |year=1999 |issue=2 |doi=10.1017/s0004972700032883|doi-access=free |hdl=1959.13/803798 |hdl-access=free }}</ref>
*If any two sides (but not three) of a Heronian triangle have a common factor, that factor must be the sum of two squares.<ref>{{Cite journal |title=On Triangles with Rational Sides and Having Rational Areas |last=Blichfeldt |first=H. F. |journal=Annals of Mathematics |volume=11 |issue=1/6 |year=1896–1897 |pages=57–60 |jstor=1967214 |doi=10.2307/1967214 }}</ref>

==Parametrizations==

A [[parametric equation]] or ''parametrization'' of Heronian triangles consists of an expression of the side lengths and area of a triangle as functions{{mdash}}typically [[polynomial function]]s{{mdash}}of some parameters, such that the triangle is Heronian if and only if the parameters satisfy some constraints{{mdash}}typically, to be positive integers satisfying some inequalities. It is also generally required that all Heronian triangles can be obtained up to a scaling for some values of the parameters, and that these values are unique, if an order on the sides of the triangle is specified.

The first such parametrization was discovered by [[Brahmagupta]] (598-668 A.D.), who did not prove that all Heronian triangles can be generated by the parametrization. In the 18th century, [[Leonhard Euler]] provided another parametrization and proved that it generates all Heronian triangles. These parametrizations are described in the next two subsections.

In the third subsection, a rational parametrization{{mdash}}that is a parametrization where the parameters are positive [[rational numbers]]{{mdash}}is naturally derived from properties of Heronian triangles. Both Brahmagupta's and Euler's parametrizations can be recovered from this rational parametrization by [[clearing denominators]]. This provides a proof that Brahmagupta's and Euler's parametrizations generate all Heronian triangles.

===Brahmagupta's parametric equation===
The Indian mathematician [[Brahmagupta]] (598-668 A.D.) discovered the following [[parametric equation]]s for generating Heronian triangles,<ref name="Kurz">{{cite journal
 | last = Kurz | first = Sascha
 | arxiv = 1401.6150
 | issue = 2
 | journal = Serdica Journal of Computing
 | mr = 2473583
 | pages = 181–196
 | title = On the generation of Heronian triangles
 | url = https://eudml.org/doc/11461
 | volume = 2
 | year = 2008
 | doi = 10.55630/sjc.2008.2.181-196
 | bibcode = 2014arXiv1401.6150K
 | s2cid = 16060132
 }}.</ref> but did not prove that every [[similarity (geometry)|similarity]] class of Heronian triangles can be obtained this way.{{citation needed|date=January 2023}}

For three positive integers {{mvar|m}}, {{mvar|n}} and {{mvar|k}} that are [[coprime integers#Coprimality in sets|setwise coprime]] (<math>\gcd(m,n,k)=1</math>) and satisfy <math>mn > k^2</math> (to guarantee positive side lengths) and {{nobr|<math>m \ge n</math>}} (for uniqueness):

:<math>\begin{align}
a &= n(m^2 + k^2), & s - a &= \tfrac12(b + c - a) = n(mn - k^2), \\
b &= m(n^2 + k^2), & s - b &= \tfrac12(c + a - b) = m(mn - k^2), \\
c &= (m + n)(mn - k^2), & s - c &= \tfrac12(a + b - c) = (m + n)k^2, \\
&& s &= \tfrac12(a + b + c) = mn(m + n), \\
A &= mnk(m+n)(mn-k^{2}), & r &= k(mn - k^2), \\
\end{align}</math>

where {{mvar|s}} is the semiperimeter, {{mvar|A}} is the area, and {{mvar|r}} is the inradius.

The resulting Heronian triangle is not always primitive, and a scaling may be needed for getting the corresponding primitive triangle. For example, taking {{math|1=''m'' = 36}}, {{math|1=''n'' = 4}} and {{math|1=''k'' = 3}} produces a triangle with {{math|1=''a'' = 5220}}, {{math|1=''b'' = 900}} and {{math|1=''c'' = 5400}}, which is similar to the {{math|(5, 29, 30)}} Heronian triangle with a proportionality factor of {{math|180}}.

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 <math>\gcd(a,b,c)</math> cannot be predicted.<ref name="Kurz" />

===Euler's parametric equation===
The following method of generating all Heronian triangles was discovered by [[Leonhard Euler]],{{sfn|Dickson|1920|p=193}} who was the first to provably parametrize all such triangles.

For four positive integers {{mvar|m}} coprime to {{mvar|n}} and {{mvar|p}} coprime to {{mvar|q}} {{nobr|(<math>\gcd{(m, n)} = \gcd{(p, q)} = 1</math>)}} satisfying <math>mp > nq</math> (to guarantee positive side lengths):

:<math>\begin{align}
a &= mn(p^2 + q^2),      & s - a &= mq(mp - nq), \\
b &= pq(m^2 + n^2),      & s - b &= np(mp - nq), \\
c &= (mq + np)(mp - nq), & s - c &= nq(mq + np), \\
&                        &     s &= mp(mq + np), \\
A &= mnpq(mq + np)(mp - nq), & r &= nq(mp - nq), \\
\end{align}</math>

where {{mvar|s}} is the semiperimeter, {{mvar|A}} is the area, and {{mvar|r}} is the inradius.

Even when {{mvar|m}}, {{mvar|n}}, {{mvar|p}}, and {{mvar|q}} are pairwise coprime, the resulting Heronian triangle may not be primitive. In particular, if {{mvar|m}}, {{mvar|n}}, {{mvar|p}}, and {{mvar|q}} are all odd, the three side lengths are even. It is also possible that {{mvar|a}}, {{mvar|b}}, and {{mvar|c}} have a common divisor other than {{math|2}}. For example, with {{math|1=''m'' = 2}}, {{math|1=''n'' = 1}}, {{math|1=''p'' = 7}}, and {{math|1=''q'' = 4}}, one gets {{math|1=(''a'', ''b'', ''c'') = (130, 140, 150)}}, where each side length is a multiple of {{math|10}}; the corresponding primitive triple is {{math|(13, 14, 15)}}, which can also be obtained by dividing the triple resulting from {{math|1=''m'' = 2, ''n'' = 1, ''p'' = 3, ''q'' = 2}} by two, then exchanging {{mvar|b}} and {{mvar|c}}.

===Half-angle tangent parametrization===
[[File:Triangle-tikz.svg|thumb|right|A triangle with side lengths and interior angles labeled as in the text]]
{{See also|Concyclic points#Integer area and side lengths}}

Let <math>a, b, c > 0</math> be the side lengths of any triangle, let <math>\alpha, \beta, \gamma</math> be the interior angles opposite these sides, and let <math display=inline>t = \tan\frac\alpha2,</math> <math display=inline>u = \tan\frac\beta2,</math> and <math display = inline>v = \tan\frac\gamma2</math> be the half-angle tangents. The values <math>t, u, v</math> are all positive and satisfy <math>tu + uv + vt = 1</math>; this "triple tangent identity" is the half-angle tangent version of the fundamental triangle identity written as <math display=inline>\frac\alpha 2 + \frac\beta 2 + \frac\gamma 2 = \frac\pi 2</math> radians (that is, 90°), as can be proved using the [[angle addition formulas|addition formula for tangents]]. By the [[law of sines|laws of sines]] and [[law of cosines|cosines]], all of the sines and the cosines of <math>\alpha, \beta, \gamma</math> are rational numbers if the triangle is a rational Heronian triangle and, because a [[Tangent half-angle formula|half-angle tangent is a rational function of the sine and cosine]], it follows that the half-angle tangents are also rational.

Conversely, if <math>t, u, v</math> are positive rational numbers such that <math>tu + uv + vt = 1,</math> it can be seen that they are the half-angle tangents of the interior angles of a class of similar Heronian triangles.<ref>{{cite journal |last=Cheney |first=William Fitch Jr. |year=1929 |title=Heronian Triangles |journal=American Mathematical Monthly |volume=36 |issue=1 |pages=22–28 |doi=10.1080/00029890.1929.11986902 |url=http://math.fau.edu/yiu/PSRM2015/yiu/New%20Folder%20(4)/Heron%20Triangles/Cheney.pdf}}</ref> The condition <math>tu + uv + vt = 1</math> can be rearranged to <math display=inline>v = \frac{1-tu}{t+u},</math> and the restriction <math>v > 0</math> requires <math>tu < 1.</math> Thus there is a [[bijection]] between the similarity classes of rational Heronian triangles and the pairs of positive rational numbers <math>(t, u)</math> whose product is less than {{math|1}}.

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:<ref>{{cite journal |last1=Kocik |first1=Jerzy |last2=Solecki |first2=Andrzej |date=2009 |title=Disentangling a triangle |journal=American Mathematical Monthly |volume=116 |number=3 |pages=228–237 |url=http://lagrange.math.siu.edu/Kocik/triangle/monthlyTriangle.pdf |doi=10.1080/00029890.2009.11920932|s2cid=28155804 }}</ref>
:<math>\begin{align}
a &= \sin\alpha = \frac{2t}{1+t^2}, & s - a = \frac{2u(1-tu)}{(1+t^2)(1+u^2)}, \\[5mu]
b &= \sin\beta = \frac{2u}{1+u^2},  & s - b = \frac{2t(1-tu)}{(1+t^2)(1+u^2)}, \\[5mu]
c &= \sin\gamma = \frac{2(t+u)(1-tu)}{(1+t^2)(1+u^2)},
                                        & s - c = \frac{2tu(t+u)}{(1+t^2)(1+u^2)}, \\[5mu]
&                                       &     s = \frac{2(t+u)}{(1+t^2)(1+u^2)}, \\
A &= \frac{4tu(t+u)(1-tu)}{(1+t^2)^2(1+u^2)^2}, & r = \frac{2tu(1-tu)}{(1+t^2)(1+u^2)},
\end{align}</math>
where <math>s = \tfrac12(a + b + c)</math> is the semiperimeter, <math>A = \tfrac12 ab \sin \gamma</math> is the area, <math>r = \sqrt{\tfrac{(s-a)(s-b)(s-c)}{s}}</math> is the inradius, and all these values are rational because <math>t</math> and <math>u</math> are rational.

To obtain an (integral) Heronian triangle, the denominators of {{mvar|a}}, {{mvar|b}}, and {{mvar|c}} [[clearing denominators|must be cleared]]. There are several ways to do this. If <math>t = m/n</math> and <math>u = p/q,</math> with <math>\gcd(m, n) = \gcd(p,q) = 1</math> ([[irreducible fraction]]s), and the triangle is scaled up by <math>\tfrac12(m^2 + n^2)(p^2 + q^2),</math> the result is Euler's parametrization. If <math>t = m/k</math> and <math>u = n/k</math> with <math>\gcd(m, n, k) = 1</math> (lowest common denominator), and the triangle is scaled up by <math>(k^2 + m^2)(k^2 + n^2)/2k,</math> the result is similar but not quite identical to Brahmagupta's parametrization. If, instead, this is <math>1/t</math> and <math>1/u</math> that are reduced to the lowest common denominator, that is, if <math>t = k/m</math> and <math>u = k/n</math> with <math>\gcd(m, n, k) = 1,</math> then one gets exactly Brahmagupta's parametrization by scaling up the triangle by <math>(k^2 + m^2)(k^2 + n^2)/2k.</math>

This proves that either parametrization generates all Heronian triangles.

The values of {{mvar|t}}, {{mvar|u}} and {{mvar|v}} that give the set of triangles that are geometrically similar to the triangle with side lengths {{mvar|a}}, {{mvar|b}}, and {{mvar|c}}, semiperimeter {{tmath|1=s = \tfrac12(a + b + c)}}, and area {{mvar|A}} are
<math display=block>(t, u, v) = \left( \frac{A}{s(s-a)}, \frac{A}{s(s-b)}, \frac{A}{s(s-c)} \right)\,.</math>

==Other results==

{{harvtxt|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]] <math>r</math> and all three of the [[inradius|exradii]] <math>(r_a, r_b, r_c)</math>, including the ones generated by<ref name=Yiu1>Zhou, Li, "Primitive Heronian Triangles With Integer Inradius and Exradii", ''[[Forum Geometricorum]]'' 18, 2018, 71-77. http://forumgeom.fau.edu/FG2018volume18/FG201811.pdf</ref>{{rp|Thm. 4}}

:<math>\begin{align}
a &= 5(5n^2 + n - 1),               & r_a &= 5n+3,     \\
b &= (5n + 3)(5n^2 - 4n + 1),       & r_b &= 5n^2+n-1, \\
c &= (5n - 2)(5n^2 + 6n + 2),       & r_c &= (5n - 2)(5n + 3)(5n^2 + n - 1), \\
&                                   & r &= 5n - 2, \\
A &= (5n - 2)(5n + 3)(5n^2 + n - 1) = r_c.
\end{align}</math>

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.<ref name=Yiu1/>{{rp|Thm. 5}}
 
See also {{slink|Integer triangle|Heronian triangles}} for parametrizations of some types of 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.
{| class="wikitable" style="text-align:right;"
|-
! Area
! Perimeter
! side length {{math|''b''+''d''}}
! side length {{mvar|e}}
! side length {{mvar|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
|}
The list of primitive Heronian triangles whose sides do not exceed 6,000,000 has been computed by {{harvtxt|Kurz|2008}}.

==Heronian triangles with perfect square sides==
As of February 2021, only two ''primitive'' Heronian triangles with perfect square sides are known:

(1853{{sup|2}}, 4380{{sup|2}}, 4427{{sup|2}}, Area={{val|32918611718880}}), published in 2013.<ref>{{cite journal
 | last1 = Stănică | first1 = Pantelimon
 | last2 = Sarkar | first2 = Santanu
 | last3 = Sen Gupta | first3 = Sourav
 | last4 = Maitra | first4 = Subhamoy
 | last5 = Kar | first5 = Nirupam
 | hdl = 10945/38838
 | journal = Integers
 | mr = 3083465
 | page = Paper No. A3, 17pp
 | title = Counting Heron triangles with constraints
 | volume = 13
 | year = 2013}}</ref>

(11789{{sup|2}}, 68104{{sup|2}}, 68595{{sup|2}}, Area={{val|284239560530875680}}), published in 2018.<ref>{{Cite web | url=https://listserv.nodak.edu/cgi-bin/wa.exe?A2=NMBRTHRY;e7c96423.1804&S=b | title=LISTSERV - NMBRTHRY Archives - LISTSERV.NODAK.EDU }}</ref>

Heronian triangles with perfect square sides are connected to the [[Perfect cuboid]] problem. The existence of a solution to the Perfect cuboid problem is equivalent to the existence of a solution to the Perfect square triangle problem:<ref>Florian Luca, Perfect Cuboids and Perfect Square Triangles (2000) Mathematics Magazine Vol. 73, No. 5 (Dec., 2000), pp. 400–401</ref> "Does there exist a triangle whose side lengths are perfect squares and whose angle bisectors are integers?".

==Equable triangles==
A shape is called [[equable shape|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),{{sfn|Dickson|1920|p=199}}<ref>{{citation
 | last = Markowitz | first = L.
 | issue = 3
 | journal = The Mathematics Teacher
 | pages = 222–3
 | title = Area = Perimeter
 | volume = 74
 | year = 1981
| doi = 10.5951/MT.74.3.0222
 }}</ref> though only four of them are primitive.

==Almost-equilateral Heronian triangles==

Since the area of an [[equilateral triangle]] with rational sides is an [[irrational number]], no equilateral triangle is Heronian. However, a sequence of isosceles Heronian triangles that are "almost equilateral" can be developed from the duplication of [[Pythagorean triple|right-angled triangles]], in which the hypotenuse is almost twice as long as one of the legs. The first few examples of these almost-equilateral triangles are listed in the following table {{OEIS|A102341}}:
{| class="wikitable" style="table-layout: fixed; width: 500px;"
! colspan="3" | Side length || rowspan="2" | Area
|-
! {{mvar|a}} || {{math|1=''b''=''a''}} || {{mvar|c}}
|- align="right"
| 5 || 5 || 6 || 12
|- align="right"
| 17 || 17 || 16 || 120
|- align="right"
| 65 || 65 || 66 || {{val|1848}}
|- align="right"
| 241 || 241 || 240 || {{val|25080}}
|- align="right"
| 901 || 901 || 902 || {{val|351780}}
|- align="right"
| {{val|3361}} || {{val|3361}} || {{val|3360}} || {{val|4890480}}
|- align="right"
| {{val|12545}} || {{val|12545}} || {{val|12546}} || {{val|68149872}}
|- align="right"
| {{val|46817}} || {{val|46817}} || {{val|46816}} || {{val|949077360}}
|}

There is a unique sequence of Heronian triangles that are "almost equilateral" because the three sides are of the form {{math|''n'' − 1}}, {{mvar|n}}, {{math|''n'' + 1}}. A method for generating all solutions to this problem based on [[continued fraction]]s was described in 1864 by [[Edward Sang]],<ref>{{citation|title=On the theory of commensurables|first=Edward|last=Sang|author-link=Edward Sang|journal=Transactions of the Royal Society of Edinburgh|year=1864|volume=23|issue=3|pages=721–760|doi=10.1017/s0080456800020019|s2cid=123752318 }}. See in particular [https://books.google.com/books?id=uXAxAQAAMAAJ&pg=PA734 p. 734].</ref> and in 1880 [[Reinhold Hoppe]] gave a [[closed-form expression]] for the solutions.<ref>{{citation|title=A triangle with integral sides and area|first=H. W.|last=Gould|journal=[[Fibonacci Quarterly]]|pages=27–39|date=February 1973|volume=11|issue=1|doi=10.1080/00150517.1973.12430863 |url=http://www.mathstat.dal.ca/FQ/Scanned/11-1/gould.pdf}}.</ref> The first few examples of these almost-equilateral triangles are listed in the following table {{OEIS|A003500}}:
{| class="wikitable" style="table-layout: fixed; width: 500px;"
! colspan="3" | Side length || rowspan="2" | Area || rowspan="2" | Inradius 
|-
! {{math|''n'' − 1}} || {{mvar|n}} || {{math|1=''n'' + 1}}
|- align="right"
| 1 || 2 || 3 || 0 || 0
|- align="right"
| 3 || 4 || 5 || 6 || 1
|- align="right"
| 13 || 14 || 15 || 84 || 4
|- align="right"
| 51 || 52 || 53 || {{val|1170}} || 15
|- align="right"
| 193 || 194 || 195 || {{val|16296}} || 56
|- align="right"
| 723 || 724 || 725 || {{val|226974}} || 209
|- align="right"
| {{val|2701}} || {{val|2702}} || {{val|2703}} || {{val|3161340}} || 780
|- align="right"
| {{val|10083}} || {{val|10084}} || {{val|10085}} || {{val|44031786}} || {{val|2911}}
|- align="right"
| {{val|37633}} || {{val|37634}} || {{val|37635}} || {{val|613283664}} || {{val|10864}}
|}

Subsequent values of {{mvar|n}} can be found by multiplying the previous value by 4, then subtracting the value prior to that one ({{nowrap|1=52 = 4 × 14 − 4}}, {{nowrap|1=194 = 4 × 52 − 14}}, etc.), thus:

:<math>n_t = 4n_{t-1} - n_{t-2} \, ,</math>

where {{mvar|t}} denotes any row in the table. This is a [[Lucas sequence]]. Alternatively, the formula <math>(2 + \sqrt{3})^t + (2 - \sqrt{3})^t</math> generates all {{mvar|n}} for positive integers {{mvar|t}}. Equivalently, let {{math|1=''A'' = area}} and {{math|1=''y'' = inradius}}, then,

:<math>\big((n-1)^2+n^2+(n+1)^2\big)^2-2\big((n-1)^4+n^4+(n+1)^4\big) = (6n y)^2 = (4A)^2</math>

where {{math|{{mset|''n'', ''y''}}}} are solutions to {{math|1=''n''{{sup|2}} − 12''y''{{sup|2}} = 4}}. A small transformation {{math|1=''n'' = 2''x''}} yields a conventional [[Pell equation]] {{math|1=''x''{{sup|2}} − 3''y''{{sup|2}} = 1}}, the solutions of which can then be derived from the [[continued fraction|regular continued fraction]] expansion for {{radic|3}}.<ref>{{citation |first1=William H. |last1=Richardson |title=Super-Heronian Triangles |date=2007 |url=http://www.math.wichita.edu/~richardson/heronian/heronian.html}}</ref>

The variable {{mvar|n}} is of the form <math>n=\sqrt{2 + 2 k}</math>, where {{mvar|k}} is 7, 97, 1351, 18817, .... The numbers in this sequence have the property that {{mvar|k}} consecutive integers have integral [[standard deviation]].<ref>Online Encyclopedia of Integer Sequences, {{OEIS2C|A011943}}.</ref>

==See also==
* [[Heronian tetrahedron]]
* [[Brahmagupta quadrilateral]]
* [[Brahmagupta triangle]]
* [[Robbins pentagon]]
* [[Integer triangle#Heronian triangles]]

==References==
{{reflist}}

== Further reading ==
* {{cite book |last=Carmichael |first=Robert Daniel |author-link= Robert Daniel Carmichael |year=1915 |title=Diophantine Analysis |publisher=Wiley |chapter=I. Introduction. Rational Triangles. Method of Infinite Descent |chapter-url=https://archive.org/details/diophantineanaly00carmuoft/page/1 |pages=1–23 }}
* {{cite book |last=Dickson |first=Leonard Eugene |author-link=Leonard Eugene Dickson |year=1920 |title=[[History of the Theory of Numbers]], Volume II: Diophantine Analysis |publisher=Carnegie Institution of Washington |pages=191–224 |chapter=V. Triangles, Quadrilaterals, and Tetrahedra with Rational Sides |chapter-url=https://archive.org/details/historyoftheoryo02dick_0/page/191 }}

==External links==
* {{mathworld|HeronianTriangle}}
* Online Encyclopedia of Integer Sequences [http://oeis.org/search?q=Heronian Heronian]
* {{citation |author=S. sh. Kozhegel'dinov |title=On fundamental Heronian triangles |journal=Math. Notes |volume=55 |issue=2 |pages=151–6 |year=1994 |doi=10.1007/BF02113294 |s2cid=115233024 }}

{{DEFAULTSORT:Heronian Triangle}}
[[Category:Arithmetic problems of plane geometry]]
[[Category:Types of triangles]]
[[Category:Articles containing proofs]]
[[Category:Eponymous geometric shapes]]