Integer triangles with given perimeter

Any triple of positive integers can serve as the side lengths of an integer triangle as long as it satisfies the triangle inequality: the longest side is shorter than the sum of the other two sides. Each such triple defines an integer triangle that is unique up to congruence. So the number of integer triangles (up to congruence) with perimeter p is the number of partitions of p into three positive parts that satisfy the triangle inequality. This is the integer closest to ${\displaystyle p^{2}/48}$ when p is even and to ${\displaystyle (p+3)^{2}/48}$ when p is odd.^[3][4] It also means that the number of integer triangles with even numbered perimeters ${\displaystyle p=2n}$ is the same as the number of integer triangles with odd numbered perimeters ${\displaystyle p=2n-3.}$ Thus there is no integer triangle with perimeter 1, 2 or 4, one with perimeter 3, 5, 6 or 8, and two with perimeter 7 or 10. The sequence of the number of integer triangles with perimeter p, starting at ${\displaystyle p=1,}$ is:

0, 0, 1, 0, 1, 1, 2, 1, 3, 2, 4, 3, 5, 4, 7, 5, 8, 7, 10, 8 ... (sequence A005044 in the OEIS)

This is called Alcuin's sequence.

Integer triangles with given largest side

The number of integer triangles (up to congruence) with given largest side c and integer triple ${\displaystyle (a,b,c)}$ is the number of integer triples such that ${\displaystyle a+b>c}$ and ${\displaystyle a\leq b\leq c.}$ This is the integer value ${\displaystyle \lceil {\tfrac {1}{2}}(c+1)\rceil \cdot \lfloor {\tfrac {1}{2}}(c+1)\rfloor .}$^[3] Alternatively, for c even it is the double triangular number ${\displaystyle {\tfrac {1}{2}}c{\bigl (}{\tfrac {1}{2}}c+1{\bigr )}}$ and for c odd it is the square ${\displaystyle {\tfrac {1}{4}}(c+1).}$ It also means that the number of integer triangles with greatest side c exceeds the number of integer triangles with greatest side c − 2 by c. The sequence of the number of non-congruent integer triangles with largest side c, starting at c = 1, is:

1, 2, 4, 6, 9, 12, 16, 20, 25, 30, 36, 42, 49, 56, 64, 72, 81, 90 ... (sequence A002620 in the OEIS)

The number of integer triangles (up to congruence) with given largest side c and integer triple (a, b, c) that lie on or within a semicircle of diameter c is the number of integer triples such that a + b > c , a² + b² ≤ c² and a ≤ b ≤ c. This is also the number of integer sided obtuse or right (non-acute) triangles with largest side c. The sequence starting at c = 1, is:

0, 0, 1, 1, 3, 4, 5, 7, 10, 13, 15, 17, 22, 25, 30, 33, 38, 42, 48 ... (sequence A236384 in the OEIS)

Consequently, the difference between the two above sequences gives the number of acute integer sided triangles (up to congruence) with given largest side c. The sequence starting at c = 1, is:

1, 2, 3, 5, 6, 8, 11, 13, 15, 17, 21, 25, 27, 31, 34, 39, 43, 48, 52 ... (sequence A247588 in the OEIS)

Area of an integer triangle

By Heron's formula, if T is the area of a triangle whose sides have lengths a, b, and c then

${\displaystyle 4T={\sqrt {(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}}.}$

Since all the terms under the radical on the right side of the formula are integers it follows that all integer triangles must have an integer value of 16T² and T² will be rational.

Angles of an integer triangle

By the law of cosines, every angle of an integer triangle has a rational cosine.

If the angles of any triangle form an arithmetic progression then one of its angles must be 60°.^[5] For integer triangles the remaining angles must also have rational cosines and a method of generating such triangles is given below. However, apart from the trivial case of an equilateral triangle, there are no integer triangles whose angles form either a geometric or harmonic progression. This is because such angles have to be rational angles of the form ${\displaystyle \pi p/q}$ with rational ${\displaystyle 0<p/q<1.}$ But all the angles of integer triangles must have rational cosines and this will occur only when ${\displaystyle p/q=1/3.}$^[6]: p.2 i.e. the integer triangle is equilateral.

The square of each internal angle bisector of an integer triangle is rational, because the general triangle formula for the internal angle bisector of angle A is ${\textstyle 2{\sqrt {bcs(s-a)}}{\big /}(b+c)}$ where s is the semiperimeter (and likewise for the other angles' bisectors).

Side split by an altitude

Any altitude dropped from a vertex onto an opposite side or its extension will split that side or its extension into rational lengths.

Medians

The square of twice any median of an integer triangle is an integer, because the general formula for the squared median m_a² to side a is ${\displaystyle {\tfrac {1}{4}}(2b^{2}+2c^{2}-a^{2})}$, giving (2m_a)² = 2b² + 2c² − a² (and likewise for the medians to the other sides).

Circumradius and inradius

Because the square of the area of an integer triangle is rational, the square of its circumradius is also rational, as is the square of the inradius.

The ratio of the inradius to the circumradius of an integer triangle is rational, equaling ${\displaystyle 4T^{2}/sabc}$ for semiperimeter s and area T.

The product of the inradius and the circumradius of an integer triangle is rational, equaling ${\displaystyle abc{\big /}2(a+b+c).}$

Thus the squared distance between the incenter and the circumcenter of an integer triangle, given by Euler's theorem as ${\displaystyle R^{2}-2Rr}$ is rational.

General formula

A Heronian triangle, also known as a Heron triangle or a Hero triangle, is a triangle with integer sides and integer area. Every Heronian triangle has sides proportional to^[8]

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

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

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

${\displaystyle {\text{Semiperimeter}}=mn(m+n)}$

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

for integers m, n and k subject to the constraints:

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

${\displaystyle mn>k^{2}\geq m^{2}n/(2m+n)}$

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

The proportionality factor is generally a rational ${\displaystyle p/q}$ where q = gcd(a,b,c) reduces the generated Heronian triangle to its primitive and ${\displaystyle p}$ scales up this primitive to the required size.

Pythagorean triangles

A Pythagorean triangle is right-angled and Heronian. Its three integer sides are known as a Pythagorean triple or Pythagorean triplet or Pythagorean triad.^[9] All Pythagorean triples ${\displaystyle (a,b,c)}$ with hypotenuse ${\displaystyle c}$ which are primitive (the sides having no common factor) can be generated by

${\displaystyle a=m^{2}-n^{2},\,}$

${\displaystyle b=2mn,\,}$

${\displaystyle c=m^{2}+n^{2},\,}$

${\displaystyle {\text{Semiperimeter}}=m(m+n)\,}$

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

where m and n are coprime integers and one of them is even with m > n.

Every even number greater than 2 can be the leg of a Pythagorean triangle (not necessarily primitive) because if the leg is given by ${\displaystyle a=2m}$ and we choose ${\displaystyle b=(a/2)^{2}-1=m^{2}-1}$ as the other leg then the hypotenuse is ${\displaystyle c=m^{2}+1}$.^[10] This is essentially the generation formula above with ${\displaystyle n}$ set to 1 and allowing ${\displaystyle m}$ to range from 2 to infinity.

Heronian triangles with sides in arithmetic progression

A triangle with integer sides and integer area has sides in arithmetic progression if and only if^[13] the sides are (b – d, b, b + d), where

${\displaystyle b=2(m^{2}+3n^{2})/g,}$

${\displaystyle d=(m^{2}-3n^{2})/g,}$

and where g is the greatest common divisor of ${\displaystyle m^{2}-3n^{2},}$ ${\displaystyle 2mn,}$ and ${\displaystyle m^{2}+3n^{2}.}$

Heronian triangles with one angle equal to twice another

All Heronian triangles with B = 2A are generated by^[14] either

${\displaystyle {\begin{aligned}a&={\tfrac {1}{4}}k^{2}(s^{2}+r^{2})^{2},\\[5mu]b&={\tfrac {1}{2}}k^{2}(s^{4}-r^{4}),\\[5mu]c&={\tfrac {1}{4}}k^{2}(3s^{4}-10s^{2}r^{2}+3r^{4}),\\[5mu]{\text{Area}}&={\tfrac {1}{2}}k^{2}csr(s^{2}-r^{2}),\end{aligned}}}$

with integers k, s, r such that ${\displaystyle s^{2}>3r^{2},}$ or

${\displaystyle {\begin{aligned}a&={\tfrac {1}{4}}q^{2}(u^{2}+v^{2})^{2},\\[5mu]b&=q^{2}uv(u^{2}+v^{2}),\\[5mu]c&={\tfrac {1}{4}}q^{2}(14u^{2}v^{2}-u^{4}-v^{4}),\\[5mu]{\text{Area}}&={\tfrac {1}{2}}q^{2}cuv(v^{2}-u^{2}),\end{aligned}}}$

with integers q, u, v such that ${\displaystyle v>u}$ and ${\displaystyle v^{2}<(7+4{\sqrt {3}})u^{2}.}$

No Heronian triangles with B = 2A are isosceles or right triangles because all resulting angle combinations generate angles with non-rational sines, giving a non-rational area or side.

Isosceles Heronian triangles

All isosceles Heronian triangles are decomposable. They are formed by joining two congruent Pythagorean triangles along either of their common legs such that the equal sides of the isosceles triangle are the hypotenuses of the Pythagorean triangles, and the base of the isosceles triangle is twice the other Pythagorean leg. Consequently, every Pythagorean triangle is the building block for two isosceles Heronian triangles since the join can be along either leg. All pairs of isosceles Heronian triangles are given by rational multiples of^[15]

${\displaystyle a=2(u^{2}-v^{2}),}$

${\displaystyle b=u^{2}+v^{2},}$

${\displaystyle c=u^{2}+v^{2},}$

and

${\displaystyle a=4uv,}$

${\displaystyle b=u^{2}+v^{2},}$

${\displaystyle c=u^{2}+v^{2},}$

for coprime integers u and v with u > v and u + v odd.

Heronian triangles whose perimeter is four times a prime

It has been shown that a Heronian triangle whose perimeter is four times a prime is uniquely associated with the prime and that the prime is congruent to ${\displaystyle 1}$ or ${\displaystyle 3}$ modulo ${\displaystyle 8}$.^[16][17] It is well known that such a prime ${\displaystyle p}$ can be uniquely partitioned into integers ${\displaystyle m}$ and ${\displaystyle n}$ such that ${\displaystyle p=m^{2}+2n^{2}}$ (see Euler's idoneal numbers). Furthermore, it has been shown that such Heronian triangles are primitive since the smallest side of the triangle has to be equal to the prime that is one quarter of its perimeter.

Consequently, all primitive Heronian triangles whose perimeter is four times a prime can be generated by

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

${\displaystyle b=m^{2}+4n^{2}}$

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

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

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

for integers ${\displaystyle m}$ and ${\displaystyle n}$ such that ${\displaystyle m^{2}+2n^{2}}$ is a prime.

Furthermore, the factorization of the area is ${\displaystyle 2mnp}$ where ${\displaystyle p=m^{2}+2n^{2}}$ is prime. However the area of a Heronian triangle is always divisible by ${\displaystyle 6}$. This gives the result that apart from when ${\displaystyle m=1}$ and ${\displaystyle n=1,}$ which gives ${\displaystyle p=3,}$ all other parings of ${\displaystyle m}$ and ${\displaystyle n}$ must have ${\displaystyle m}$ odd with only one of them divisible by ${\displaystyle 3}$.

Heronian triangles with rational angle bisectors

If in a Heronian triangle the angle bisector ${\displaystyle w_{a}}$ of the angle ${\displaystyle \alpha }$, the angle bisector ${\displaystyle w_{b}}$ of the angle ${\displaystyle \beta }$ and the angle bisector ${\displaystyle w_{c}}$ of the angle ${\displaystyle \gamma }$ have a rational relationship with the three sides then not only ${\displaystyle \alpha ,\beta ,\gamma }$ but also ${\displaystyle {\tfrac {1}{2}}\alpha }$, ${\displaystyle {\tfrac {1}{2}}\beta }$ and ${\displaystyle {\tfrac {1}{2}}\gamma }$ must be Heronian angles. Namely, if both angles ${\displaystyle {\tfrac {1}{2}}\alpha }$ and ${\displaystyle {\tfrac {1}{2}}\beta }$ are Heronian then ${\displaystyle {\tfrac {1}{2}}\gamma }$, the complement of ${\displaystyle {\tfrac {1}{2}}\alpha +{\tfrac {1}{2}}\beta }$, must also be a Heronian angle, so that all three angle-bisectors are rational. This is also evident if one multiplies:

${\displaystyle w_{a}={\frac {2{\sqrt {s(s-a)}}\cdot {\sqrt {bc}}}{b+c}}\quad w_{b}={\frac {2{\sqrt {s(s-b)}}\cdot {\sqrt {ac}}}{a+c}}\quad w_{c}={\frac {2{\sqrt {s(s-c)}}\cdot {\sqrt {ab}}}{a+b}}}$

together. Namely, through this one obtains:

${\displaystyle w_{a}\cdot w_{b}\cdot w_{c}={\frac {8s\cdot J\cdot a\cdot b\cdot c}{(a+b)(a+c)(b+c)}},}$

where ${\displaystyle s}$ denotes the semi-perimeter, and ${\displaystyle J}$ the area of the triangle.

All Heronian triangles with rational angle bisectors are generated by^[18]

${\displaystyle a=mn(p^{2}+q^{2})}$

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

${\displaystyle c=(mq+np)(mp-nq)}$

${\displaystyle {\text{Semiperimeter}}=s=(a+b+c)/2=mp(mq+np)}$

${\displaystyle s-a=mq(mp-nq)}$

${\displaystyle s-b=np(mp-nq)}$

${\displaystyle s-c=nq(mq+np)}$

${\displaystyle {\text{Area}}=J=mnpq(mq+np)(mp-nq)}$

where ${\displaystyle m,n,p,q}$ are such that

${\displaystyle m=t^{2}-u^{2}}$

${\displaystyle n=2tu}$

${\displaystyle p=v^{2}-w^{2}}$

${\displaystyle q=2vw}$

where ${\displaystyle t,u,v,w}$ are arbitrary integers such that

${\displaystyle t}$ and ${\displaystyle u}$ coprime,

${\displaystyle v}$ and ${\displaystyle w}$ coprime.

Heronian triangles with integer inradius and exradii

There are infinitely many decomposable, and infinitely many indecomposable, primitive Heronian (non-Pythagorean) triangles with integer radii for the incircle and each excircle.^{[19]: Thms. 3 and 4} A family of decomposible ones is given by

${\displaystyle a=4n^{2}}$

${\displaystyle b=(2n+1)(2n^{2}-2n+1)}$

${\displaystyle c=(2n-1)(2n^{2}+2n+1)}$

${\displaystyle r=2n-1}$

${\displaystyle r_{a}=2n+1}$

${\displaystyle r_{b}=2n^{2}}$

${\displaystyle r_{c}={\text{Area}}=2n^{2}(2n-1)(2n+1);}$

and a family of indecomposable ones is given by

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

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

${\displaystyle c=(5n-2)(5n^{2}+6n+2)}$

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

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

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

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

Heronian triangles as faces of a tetrahedron

There exist tetrahedra having integer-valued volume and Heron triangles as faces. One example has one edge of 896, the opposite edge of 190, and the other four edges of 1073; two faces have areas of 436800 and the other two have areas of 47120, while the volume is 62092800.^[9]: p.107

Heronian triangles in a 2D lattice

A 2D lattice is a regular array of isolated points where if any one point is chosen as the Cartesian origin (0, 0), then all the other points are at (x, y) where x and y range over all positive and negative integers. A lattice triangle is any triangle drawn within a 2D lattice such that all vertices lie on lattice points. By Pick's theorem a lattice triangle has a rational area that either is an integer or a half-integer (has a denominator of 2). If the lattice triangle has integer sides then it is Heronian with integer area.^[20]

Furthermore, it has been proved that all Heronian triangles can be drawn as lattice triangles.^[21][22] Consequently, an integer triangle is Heronian if and only if it can be drawn as a lattice triangle.

There are infinitely many primitive Heronian (non-Pythagorean) triangles which can be placed on an integer lattice with all vertices, the incenter, and all three excenters at lattice points. Two families of such triangles are the ones with parametrizations given above at #Heronian triangles with integer inradius and exradii.^{[19]: Thm. 5}

Integer triangles with a rational angle bisector

A triangle family with integer sides ${\displaystyle a,b,c}$ and with rational bisector ${\displaystyle d}$ of angle A is given by^[24]

${\displaystyle a=2(k^{2}-m^{2}),}$

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

${\displaystyle c=(k+m)^{2},}$

${\displaystyle d={\frac {2km(k^{2}-m^{2})}{k^{2}+m^{2}}},}$

with integers ${\displaystyle k>m>0}$.

Integer triangles with integer n-sectors of all angles

There exist infinitely many non-similar triangles in which the three sides and the bisectors of each of the three angles are integers.^[25]

There exist infinitely many non-similar triangles in which the three sides and the two trisectors of each of the three angles are integers.^[25]

However, for n > 3 there exist no triangles in which the three sides and the (n – 1) n-sectors of each of the three angles are integers.^[25]

Integer triangles with one angle with a given rational cosine

Some integer triangles with one angle at vertex A having given rational cosine h / k (h < 0 or > 0; k > 0) are given by^[26]

${\displaystyle a=p^{2}-2pqh+q^{2}k^{2},}$

${\displaystyle b=p^{2}-q^{2}k^{2},}$

${\displaystyle c=2qk(p-qh),}$

where p and q are any coprime positive integers such that p > qk.

Integer triangles with one angle equal to an arbitrary rational number times another angle

For positive coprime integers h and k, the triangle with the following sides has angles ${\displaystyle h\alpha }$, ${\displaystyle k\alpha }$, and ${\displaystyle \pi -(h+k)\alpha }$ and hence two angles in the ratio h : k, and its sides are integers:^[31]

${\displaystyle a=q^{h+k-1}{\frac {\sin h\alpha }{\sin \alpha }}=q^{k}\cdot \sum _{0\leq i\leq {\frac {h-1}{2}}}(-1)^{i}{\binom {h}{2i+1}}p^{h-2i-1}(q^{2}-p^{2})^{i},}$

${\displaystyle b=q^{h+k-1}{\frac {\sin k\alpha }{\sin \alpha }}=q^{h}\cdot \sum _{0\leq i\leq {\frac {k-1}{2}}}(-1)^{i}{\binom {k}{2i+1}}p^{k-2i-1}(q^{2}-p^{2})^{i},}$

${\displaystyle c=q^{h+k-1}{\frac {\sin(h+k)\alpha }{\sin \alpha }}=\sum _{0\leq i\leq {\frac {h+k-1}{2}}}(-1)^{i}{\binom {h+k}{2i+1}}p^{h+k-2i-1}(q^{2}-p^{2})^{i},}$

where ${\displaystyle \alpha =\cos ^{-1}\!{\frac {p}{q}}}$ and p and q are any coprime integers such that ${\displaystyle \cos {\frac {\pi }{h+k}}<{\frac {p}{q}}<1}$.

Integer triangles with three rational angles

The only integer triangle with three rational angles (rational numbers of degrees, or equivalently rational fractions of a full turn) is the equilateral triangle.^[2] This is because integer sides imply three rational cosines by the law of cosines, and by Niven's theorem a rational cosine coincides with a rational angle if and only if the cosine equals 0, ±1/2, or ±1. The only ones of these giving an angle strictly between 0° and 180° are the cosine value 1/2 with the angle 60°, the cosine value –1/2 with the angle 120°, and the cosine value 0 with the angle 90°. The only combination of three of these, allowing multiple use of any of them and summing to 180°, is three 60° angles.