Integer triangle#Properties of Heronian triangles

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 $p^2/48$ when p is even and to $(p\; +\; 3)^2\; /\; 48$ when p is odd.Tom Jenkyns and Eric Muller, Triangular Triples from Ceilings to Floors, American Mathematical Monthly 107:7 (August 2000) 634–639Ross Honsberger, Mathematical Gems III, pp. 39–37 It also means that the number of integer triangles with even numbered perimeters $p\; =\; 2n$ is the same as the number of integer triangles with odd numbered perimeters $p\; =\; 2n\; -\; 3.$ 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 $p\; =\; 1,$ is:

:0, 0, 1, 0, 1, 1, 2, 1, 3, 2, 4, 3, 5, 4, 7, 5, 8, 7, 10, 8 ... {{OEIS|A005044}}

This is called Alcuin's sequence.

The number of integer triangles (up to congruence) with given largest side c and integer triple $(a,\; b,\; c)$ is the number of integer triples such that $a\; +\; b\; >\; c$ and $a\; \backslash leq\; b\; \backslash leq\; c.$ This is the integer value $\backslash lceil\; \backslash tfrac12(c\; +\; 1)\backslash rceil\; \backslash cdot\; \backslash lfloor\; \backslash tfrac12(c\; +\; 1)\backslash rfloor.$ Alternatively, for c even it is the double triangular number $\backslash tfrac12c\; \backslash bigl(\; \backslash tfrac12c\; +\; 1\; \backslash bigr)$ and for c odd it is the square $\backslash tfrac14(c\; +\; 1).$ 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 ... {{OEIS|A002620}}

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 ... {{OEIS|A236384}}

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 ... {{OEIS|A247588}}

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

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

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

By the law of cosines, every angle of an integer triangle has a rational cosine. Every angle of an integer right triangle also has rational sine (see Pythagorean triple).

If the angles of any triangle form an arithmetic progression then one of its angles must be 60°. 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 $\backslash pi\; p/q$ with rational But all the angles of integer triangles must have rational cosines and this will occur only when $p/q\; =\; 1/3.${{cite arXiv |first=Jörg |last=Jahnel |title=When is the (Co)Sine of a Rational Angle equal to a rational number? |eprint=1006.2938 |year=2010 |class=math.HO }}{{rp|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 $2\backslash sqrt\{bcs(s-a)\}\backslash big/(b+c)$ where s is the semiperimeter (and likewise for the other angles' bisectors).

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

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 $\backslash tfrac14(2b^2+2c^2-a^2)$, giving (2m_a)² = 2b² + 2c² − a² (and likewise for the medians to the other sides).

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 $4T^2\; /\; sabc$ for semiperimeter s and area T.

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

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

{{Main article|Heronian triangle}}

A Heronian triangle, also known as a Heron triangle or a Hero triangle, is a triangle with integer sides and integer area.

All Heronian triangles can be placed on a lattice with each vertex at a lattice point.Yiu, P., "Heronian triangles are lattice triangles", American Mathematical Monthly 108 (2001), 261–263. Furthermore, if an integer triangle can be place on a lattice with each vertex at a lattice point it must be Heronian.

Every Heronian triangle has sides proportional toCarmichael, R. D. The Theory of Numbers and Diophantine Analysis. New York: Dover, 1952.

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

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

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

:$\backslash text\{Semiperimeter\}\; =\; mn(m+n)$

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

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

:$\backslash gcd\{(m,n,k)\}\; =\; 1$

:$mn\; >\; k^2\; \backslash ge\; m^2n/(2m+n)$

:$m\; \backslash ge\; n\; \backslash ge\; 1.$

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.

{{Main article|Pythagorean triple}}

A Pythagorean triangle is right-angled and Heronian. Its three integer sides are known as a Pythagorean triple or Pythagorean triplet or Pythagorean triad.Sierpiński, Wacław. Pythagorean Triangles, Dover Publications, 2003 (orig. 1962). All Pythagorean triples $(a,\; b,\; c)$ with hypotenuse $c$ which are primitive (the sides having no common factor) can be generated by

:$a\; =\; m^2\; -\; n^2,\; \backslash ,$

:$b\; =\; 2mn,\; \backslash ,$

:$c\; =\; m^2\; +\; n^2,\; \backslash ,$

:$\backslash text\{Semiperimeter\}\; =\; m(m+n)\; \backslash ,$

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

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 $a=2m$ and we choose $b=(a/2)^2-1=m^2-1$ as the other leg then the hypotenuse is $c=m^2+1$.{{Cite OEIS|1=A009111|2=List of ordered areas of Pythagorean triangles|accessdate=2017-03-03}} This is essentially the generation formula above with $n$ set to 1 and allowing $m$ to range from 2 to infinity.

There are no primitive Pythagorean triangles with integer altitude from the hypotenuse. This is because twice the area equals any base times the corresponding height: 2 times the area thus equals both ab and cd where d is the height from the hypotenuse c. The three side lengths of a primitive triangle are coprime, so $d\; =\; ab/c$ is in fully reduced form; since c cannot equal 1 for any primitive Pythagorean triangle, d cannot be an integer.

However, any Pythagorean triangle with legs x, y and hypotenuse z can generate a Pythagorean triangle with an integer altitude, by scaling up the sides by the length of the hypotenuse z. If d is the altitude, then the generated Pythagorean triangle with integer altitude is given byRichinick, Jennifer, "The upside-down Pythagorean Theorem", Mathematical Gazette 92, July 2008, 313–317.

:$(a,b,c,d)=(xz,\; yz,\; z^2,\; xy).\; \backslash ,$

Consequently, all Pythagorean triangles with legs a and b, hypotenuse c, and integer altitude d from the hypotenuse, with $\backslash gcd(a,\; b,\; c,\; d)\; =\; 1$, which necessarily satisfy both a² + b² = c² and $\backslash tfrac\{1\}\{a^\{2\}\}+\backslash tfrac\{1\}\{b^\{2\}\}=\backslash tfrac\{1\}\{d^\{2\}\}$, are generated byVoles, Roger, "Integer solutions of a⁻²+b⁻²=d⁻²", Mathematical Gazette 83, July 1999, 269–271.

:$a=(m^2-n^2)(m^2+n^2),\; \backslash ,$

:$b=2mn(m^2+n^2),\; \backslash ,$

:$c=(m^2+n^2)^2,\; \backslash ,$

:$d=2mn(m^2-n^2),\; \backslash ,$

:$\backslash text\{Semiperimeter\}=m(m+n)(m^2+n^2)\; \backslash ,$

:$\backslash text\{Area\}=mn(m^2-n^2)(m^2+n^2)^2\; \backslash ,$

for coprime integers m, n with m > n.

A triangle with integer sides and integer area has sides in arithmetic progression if and only if{{Cite journal |title=Heron Quadrilaterals with sides in Arithmetic or Geometric progression |last1=Buchholz |first1=R. H. |last2=MacDougall |first2=J. A. |journal=Bulletin of the Australian Mathematical Society |pages=263–269 |volume=59 |issue=2 |year=1999 |doi=10.1017/S0004972700032883 |doi-access=free |hdl=1959.13/803798 |hdl-access=free }} the sides are (b – d, b, b + d), where

:$b=2(m^2+3n^2)/g,$

:$d=(m^2-3n^2)/g,$

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

All Heronian triangles with B = 2A are generated byMitchell, Douglas W., "Heron triangles with ∠B=2∠A", Mathematical Gazette 91, July 2007, 326–328. either

:$\backslash begin\{align\}$

a &= \tfrac14 k^2(s^2+r^2)^2, \\[5mu]

b &= \tfrac12 k^2(s^4-r^4), \\[5mu]

c &= \tfrac14 k^2(3s^4-10s^2 r^2+3r^4), \\[5mu]

\text{Area} &= \tfrac12 k^2 csr(s^2-r^2),

\end{align}

with integers k, s, r such that $s^2\; >\; 3r^2,$ or

:$\backslash begin\{align\}$

a &= \tfrac14 q^2 (u^2 + v^2)^2, \\[5mu]

b &= q^{2}uv(u^2 + v^2), \\[5mu]

c &= \tfrac14 q^2(14u^2v^2 - u^4 - v^4), \\[5mu]

\text{Area} &= \tfrac12 q^2cuv(v^2 -u^2),

\end{align}

with integers {{math|q, u, v}} such that $v\; >\; u$ and

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.

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 the following side lengths:Sastry, K. R. S., [http://forumgeom.fau.edu/FG2005volume5/FG200515.pdf "Construction of Brahmagupta n-gons"] {{Webarchive|url=https://web.archive.org/web/20201205212607/http://forumgeom.fau.edu/FG2005volume5/FG200515.pdf |date=2020-12-05 }}, Forum Geometricorum 5 (2005): 119–126.

\begin{aligned}

a &= 2(u^2 - v^2), &\quad\quad a &= 4uv, \\

b &= u^2 + v^2, &\quad\quad b &= u^2 + v^2, \\

c &= u^2 + v^2, &\quad\quad c &= u^2 + v^2, \\

\text{Area} &= 2uv(u^2 - v^2),

\end{aligned}

for coprime integers of opposite parity $u$ and $v$, with $u\; >\; v$.

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 $1$ or $3$ modulo $8$.Yiu, P., [https://cms.math.ca/crux/v24/n3/page175-177.pdf "CRUX, Problem 2331, Proposed by Paul Yiu"] {{Webarchive|url=https://web.archive.org/web/20150905085516/http://cms.math.ca/crux/v24/n3/page175-177.pdf |date=2015-09-05 }}, Memorial University of Newfoundland (1998): 175-177Yui, P. and Taylor, J. S., [http://dongthapvietnam.informe.com/blog/uploads/2012/10/cruxv25_1999.pdf "CRUX, Problem 2331, Solution"] {{Webarchive|url=https://web.archive.org/web/20170216210921/http://dongthapvietnam.informe.com/blog/uploads/2012/10/cruxv25_1999.pdf |date=2017-02-16 }} Memorial University of Newfoundland (1999): 185-186 It is well known that such a prime $p$ can be uniquely partitioned into integers $m$ and $n$ such that $p=m^2+2n^2$ (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

:$a=m^2+2n^2$

:$b=m^2+4n^2$

:$c=2(m^2+n^2)$

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

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

for integers $m$ and $n$ such that $m^2+2n^2$ is a prime.

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

If in a Heronian triangle the angle bisector $w\_a$ of the angle $\backslash alpha$, the angle bisector $w\_b$ of the angle $\backslash beta$ and the angle bisector $w\_c$ of the angle $\backslash gamma$ have a rational relationship with the three sides then not only $\backslash alpha,\; \backslash beta,\; \backslash gamma$ but also $\backslash tfrac12\backslash alpha$, $\backslash tfrac12\backslash beta$ and $\backslash tfrac12\backslash gamma$ must be Heronian angles. Namely, if both angles $\backslash tfrac12\backslash alpha$ and $\backslash tfrac12\backslash beta$ are Heronian then $\backslash tfrac12\backslash gamma$, the complement of $\backslash tfrac12\backslash alpha\; +\; \backslash tfrac12\backslash beta$, must also be a Heronian angle, so that all three angle-bisectors are rational. This is also evident if one multiplies:

:$w\_a\; =\; \backslash frac\{2\backslash sqrt\{s(s-a)\}\; \backslash cdot\; \backslash sqrt\{bc\}\}\{b+c\}\; \backslash quad\; w\_b\; =\; \backslash frac\{2\backslash sqrt\{s(s-b)\}\; \backslash cdot\; \backslash sqrt\{ac\}\}\{a+c\}\; \backslash quad\; w\_c\; =\; \backslash frac\{2\backslash sqrt\{s(s-c)\}\; \backslash cdot\; \backslash sqrt\{ab\}\}\{a+b\}$

together. Namely, through this one obtains:

:$w\_a\; \backslash cdot\; w\_b\; \backslash cdot\; w\_c\; =\; \backslash frac\{8s\; \backslash cdot\; J\; \backslash cdot\; a\; \backslash cdot\; b\; \backslash cdot\; c\}\{(a+b)(a+c)(b+c)\},$

where $s$ denotes the semi-perimeter, and $J$ the area of the triangle.

All similarity classes of Heronian triangles with rational angle bisectors are generated byHermann Schubert, "Die Ganzzahligkeit in der Algebraischen Geometrie", Leipzig, 1905

:$a\; =\; mn(p^2+q^2)$

:$b\; =\; pq(m^2+n^2)$

:$c\; =\; (mq+np)(mp-nq)$

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

:$s-a=mq(mp-nq)$

:$s-b=np(mp-nq)$

:$s-c=nq(mq+np)$

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

where $m,\; n,\; p,\; q$ are such that

:$m\; =\; t^2-u^2$

:$n=2tu$

:$p\; =\; v^2-w^2$

:$q=2vw$

where $t,\; u,\; v,\; w$ are arbitrary integers such that

:$t$ and $u$ coprime,

:$v$ and $w$ coprime.

There are infinitely many decomposable, and infinitely many indecomposable, primitive Heronian (non-Pythagorean) triangles with integer radii for the incircle and each excircle.Li Zhou, "Primitive Heronian Triangles With Integer Inradius and Exradii", Forum Geometricorum 18, 2018, pp. 71–77.{{rp|Thms. 3 and 4}} A family of decomposible ones is given by

:$a=4n^2$

:$b=(2n+1)(2n^2\; -2n+1)$

:$c=(2n-1)(2n^2\; +2n+1)$

:$r=2n-1$

:$r\_a=2n+1$

:$r\_b=2n^2$

:$r\_c=\backslash text\{Area\}=2n^2(2n-1)(2n+1);$

and a family of indecomposable ones is given by

:$a=5(5n^2\; +n-1)$

:$b=(5n+3)(5n^2-4n+1)$

:$c=(5n-2)(5n^2\; +6n+2)$

:$r=5n-2$

:$r\_a=5n+3$

:$r\_b=5n^2+n-1$

:$r\_c=\backslash text\{Area\}=(5n-2)(5n+3)(5n^2\; +n-1).$

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.{{rp|p.107}}

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.{{cite journal

| last1 = Buchholz | first1 = Ralph H.

| last2 = MacDougall | first2 = James A.

| doi = 10.1016/j.jnt.2007.05.005

| issue = 1

| journal = Journal of Number Theory

| mr = 2382768

| pages = 17–48

| title = Cyclic polygons with rational sides and area

| url = https://scholar.archive.org/work/woebyikyrrb35dp6by6h7rwqsu

| volume = 128

| year = 2008}}

Furthermore, it has been proved that all Heronian triangles can be drawn as lattice triangles.P. Yiu, "Heronian triangles are lattice triangles", American Mathematical Monthly 108 (2001), 261–263.{{cite journal

| last1 = Marshall | first1 = Susan H. | author1-link = Susan H. Marshall

| last2 = Perlis | first2 = Alexander R.

| doi = 10.4169/amer.math.monthly.120.02.140

| issue = 2

| journal = American Mathematical Monthly

| jstor = 10.4169/amer.math.monthly.120.02.140

| mr = 3029939

| pages = 140–149

| title = Heronian tetrahedra are lattice tetrahedra

| url = https://www.math.lsu.edu/~aperlis/publications/latticetetrahedra/marshallperlis_latticetetrahedra_26Mar2012.pdf

| volume = 120

| year = 2013}} 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.{{rp|Thm. 5}}

{{Main article|Automedian triangle}}

An automedian triangle is one whose medians are in the same proportions (in the opposite order) as the sides. If x, y, and z are the three sides of a right triangle, sorted in increasing order by size, and if 2x < z, then z, x + y, and y − x are the three sides of an automedian triangle. For instance, the right triangle with side lengths 5, 12, and 13 can be used in this way to form the smallest non-trivial (i.e., non-equilateral) integer automedian triangle, with side lengths 13, 17, and 7.{{cite journal

| last = Parry | first = C. F.

| issue = 472

| journal = The Mathematical Gazette

| jstor = 3620241

| pages = 151–154

| title = Steiner–Lehmus and the automedian triangle

| volume = 75

| year = 1991| doi = 10.2307/3620241

| s2cid = 125374348

}}.

Consequently, using Euclid's formula, which generates primitive Pythagorean triangles, it is possible to generate primitive integer automedian triangles as

:$a=|m^2-2mn-n^2|$

:$b=m^2+2mn-n^2$

:$c=m^2+n^2$

with $m$ and $n$ coprime and $m+n$ odd, and

An important characteristic of the automedian triangle is that the squares of its sides form an arithmetic progression. Specifically, $c^2-a^2\; =\; b^2-c^2$ so $2c^2\; =\; a^2+b^2.$

A triangle family with integer sides $a,b,c$ and with rational bisector $d$ of angle A is given byZelator, Konstantine, Mathematical Spectrum 39(3), 2006/2007, 59−62.

:$a\; =\; 2(k^2-m^2),$

:$b\; =\; (k-m)^2,$

:$c\; =\; (k+m)^2,$

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

with integers $k>m>0$.

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

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

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.{{Cite web |url=http://forumgeom.fau.edu/FG2005volume5/FG200507.pdf |title=De Bruyn, Bart, "On a Problem Regarding the n-Sectors of a Triangle", Forum Geometricorum 5, 2005: pp. 47–52. |access-date=2012-05-04 |archive-date=2020-12-05 |archive-url=https://web.archive.org/web/20201205215007/http://forumgeom.fau.edu/FG2005volume5/FG200507.pdf |url-status=dead }}

Integer triangles with one angle at vertex A having given rational cosine h / k (h < 0 or > 0; k > 0) are given bySastry, K. R. S., "Integer-sided triangles containing a given rational cosine", Mathematical Gazette 68, December 1984, 289−290.

:$a\; =\; p^2-2pqh+q^2k^2,$

:$b\; =\; p^2-q^2k^2,$

:$c\; =\; 2qk(p-qh),$

where p and q are any coprime positive integers such that p > qk. All primitive solutions can be obtained by dividing a, b, and c by their greatest common divisor.

All integer triangles with a 60° angle have their angles in an arithmetic progression. All such triangles are proportional to:[https://arxiv.org/ftp/arxiv/papers/0803/0803.3778.pdf Zelator, K., "Triangle Angles and Sides in Progression and the diophantine equation x²+3y²=z²", Cornell Univ. archive, 2008]

:$a\; =\; 4mn,$

:$b\; =\; 3m^2+n^2,$

:$c\; =\; 2mn+|3m^2-n^2|$

with coprime integers m, n and 1 ≤ n ≤ m or 3m ≤ n. From here, all primitive solutions can be obtained by dividing a, b, and c by their greatest common divisor.

Integer triangles with a 60° angle can also be generated byGilder, J., Integer-sided triangles with an angle of 60°", Mathematical Gazette 66, December 1982, 261 266

:$a\; =\; m^2-mn+n^2,$

:$b\; =\; 2mn-n^2,$

:$c\; =\; m^2-n^2,$

with coprime integers m, n with 0 < n < m (the angle of 60° is opposite to the side of length a). From here, all primitive solutions can be obtained by dividing a, b, and c by their greatest common divisor (e.g. an equilateral triangle solution is obtained by taking {{nowrap|1=m = 2}} and {{nowrap|1=n = 1}}, but this produces a = b = c = 3, which is not a primitive solution). See also Burn, Bob, "Triangles with a 60° angle and sides of integer length", Mathematical Gazette 87, March 2003, 148–153.Read, Emrys, "On integer-sided triangles containing angles of 120° or 60°", Mathematical Gazette 90, July 2006, 299−305.

More precisely, If $m\; \backslash equiv\; -n\; \backslash !\backslash pmod\{3\}$, then $\backslash gcd(a,b,c)\; =\; 3$, otherwise $\backslash gcd(a,b,c)\; =\; 1$. Two different pairs $(m,\; n)$ and $(m,\; m-n)$ generate the same triple. Unfortunately the two pairs can both have a gcd of 3, so we can't avoid duplicates by simply skipping that case. Instead, duplicates can be avoided by $n$ going only till $m/2$. We still need to divide by 3 if the gcd is 3. The only solution for $n\; =\; m/2$ under the above constraints is $(3,3,3)\; \backslash equiv\; (1,1,1)$ for $m=2,\; n=1$. With this additional $n\; \backslash leq\; m/2$ constraint all triples can be generated uniquely.

An Eisenstein triple is a set of integers which are the lengths of the sides of a triangle where one of the angles is 60 degrees.

Integer triangles with a 120° angle can be generated bySelkirk, K., "Integer-sided triangles with an angle of 120°", Mathematical Gazette 67, December 1983, 251–255.

:$a\; =\; m^2\; +\; mn\; +\; n^2,$

:$b\; =\; 2mn+n^2,$

:$c\; =\; m^2\; -\; n^2,$

with coprime integers m, n with 0 < n < m (the angle of 120° is opposite to the side of length a). From here, all primitive solutions can be obtained by dividing a, b, and c by their greatest common divisor. The smallest solution, for m = 2 and n = 1, is the triangle with sides (3,5,7). See also.

More precisely, If $m\; \backslash equiv\; n\; \backslash !\backslash pmod\{3\}$, then $\backslash gcd(a,b,c)\; =\; 3$, otherwise $\backslash gcd(a,b,c)\; =\; 1$. Since the biggest side a can only be generated with a single $(m,\; n)$ pair, each primitive triple can be generated in precisely two ways: once directly with a gcd of 1, and once indirectly with a gcd of 3. Therefore, in order to generate all primitive triples uniquely, one can just add additional $m\; \backslash not\backslash equiv\; n\; \backslash !\backslash pmod\{3\}$ condition.{{citation needed|date=October 2020}}

For positive coprime integers h and k, the triangle with the following sides has angles $h\; \backslash alpha$, $k\; \backslash alpha$, and $\backslash pi\; -\; (h+k)\; \backslash alpha$ and hence two angles in the ratio h : k, and its sides are integers:Hirschhorn, Michael D., "Commensurable triangles", Mathematical Gazette 95, March 2011, pp. 61−63.

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

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

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

where $\backslash alpha\; =\; \backslash cos^\{-1\}\; \backslash !\backslash frac\{p\}\{q\}$ and p and q are any coprime integers such that .

With angle A opposite side $a$ and angle B opposite side $b$, some triangles with B = 2A are generated byDeshpande, M. N., "Some new triples of integers and associated triangles", Mathematical Gazette 86, November 2002, 464–466.

:$a\; =\; n^2,$

:$b\; =\; mn,$

:$c\; =\; m^2\; -\; n^2,$

with integers m, n such that 0 < n < m < 2n.

All triangles with B = 2A (whether integer or not) satisfyWillson, William Wynne, "A generalisation of the property of the 4, 5, 6 triangle", Mathematical Gazette 60, June 1976, 130–131. $a(a+c)\; =\; b^2.$

The equivalence class of similar triangles with $B\; =\; \backslash tfrac32\; A$ are generated by

:$a=mn^3,$

:$b=n^2(m^2-n^2),$

:$c=(m^2\; -\; n^2)^2\; -\; m^2\; n^2,$

with integers $m,\; n$ such that

All triangles with $B\; =\; \backslash tfrac32\; A$ (whether with integer sides or not) satisfy $(b^\{2\}-a^\{2\})(b^\{2\}-a^\{2\}+bc)\; =\; a^\{2\}c^\{2\}.$

We can generate the full equivalence class of similar triangles that satisfy B = 3A by using the formulas{{cite journal |last=Parris |first=Richard |journal=College Mathematics Journal |title=Commensurable Triangles |doi=10.1080/07468342.2007.11922259 |volume=38 |issue=5 |date=November 2007 |pages=345–355|s2cid=218549375 }}

:$a=n^\{3\},\; \backslash ,$

:$b=n(m^\{2\}-n^\{2\}),\; \backslash ,$

:$c=m(m^\{2\}-2n^\{2\}),\; \backslash ,$

where $m$ and $n$ are integers such that .

All triangles with B = 3A (whether with integer sides or not) satisfy $ac^2\; =\; (b-a)^\{2\}(b+a).$

The only integer triangle with three rational angles (rational numbers of degrees, or equivalently rational fractions of a full turn) is the equilateral triangle. 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.

Conditions are known in terms of elliptic curves for an integer triangle to have an integer ratio N of the circumradius to the inradius.{{Cite web |url=http://forumgeom.fau.edu/FG2010volume10/FG201017.pdf |title=MacLeod, Allan J., "Integer triangles with R/r = N", Forum Geometricorum 10, 2010: pp. 149−155. |access-date=2012-05-02 |archive-date=2022-01-20 |archive-url=https://web.archive.org/web/20220120114622/https://forumgeom.fau.edu/FG2010volume10/FG201017.pdf |url-status=dead }}[http://forumgeom.fau.edu/FG2012volume12/FG201203.pdf Goehl, John F. Jr., "More integer triangles with R/r = N", Forum Geometricorum 12, 2012: pp. 27−28] The smallest case, that of the equilateral triangle, has N = 2. In every known case, $N\; \backslash equiv\; 2\; \backslash !\backslash pmod\{8\}$ – that is, $N\; -\; 2$ is divisible by 8.

{{Main article|5-Con triangles}}

A 5-Con triangle pair is a pair of triangles that are similar but not congruent and that share three angles and two sidelengths. Primitive integer 5-Con triangles, in which the four distinct integer sides (two sides each appearing in both triangles, and one other side in each triangle) share no prime factor, have triples of sides

:$(x^3,\; x^2y,\; xy^2)$ and $(x^2y,\; xy^2,\; y^3)$

for positive coprime integers x and y. The smallest example is the pair (8, 12, 18), (12, 18, 27), generated by x = 2, y = 3.

• The only triangle with consecutive integers for sides and area has sides (3, 4, 5) and area 6.
• The only triangle with consecutive integers for an altitude and the sides has sides (13, 14, 15) and altitude from side 14 equal to 12.
• The (2, 3, 4) triangle and its multiples are the only triangles with integer sides in arithmetic progression and having the complementary exterior angle property.Barnard, T., and Silvester, J., "Circle theorems and a property of the (2,3,4) triangle", Mathematical Gazette 85, July 2001, 312−316.Lord, N., "A striking property of the (2,3,4) triangle", Mathematical Gazette 82, March 1998, 93−94.Mitchell, Douglas W., "The 2:3:4, 3:4:5, 4:5:6, and 3:5:7 triangles", Mathematical Gazette 92, July 2008. This property states that if angle C is obtuse and if a segment is dropped from B meeting perpendicularly AC extended at P, then ∠CAB=2∠CBP.
• The (3, 4, 5) triangle and its multiples are the only integer right triangles having sides in arithmetic progression.
• The (4, 5, 6) triangle and its multiples are the only triangles with one angle being twice another and having integer sides in arithmetic progression.
• The (3, 5, 7) triangle and its multiples are the only triangles with a 120° angle and having integer sides in arithmetic progression.
• The only integer triangle with area = semiperimeterMacHale, D., "That 3,4,5 triangle again", Mathematical Gazette 73, March 1989, 14−16. has sides (3, 4, 5).
• The only integer triangles with area = perimeter have sidesL. E. Dickson, History of the Theory of Numbers, vol.2, 181. (5, 12, 13), (6, 8, 10), (6, 25, 29), (7, 15, 20), and (9, 10, 17). Of these the first two, but not the last three, are right triangles.
• There exist integer triangles with three rational medians.{{rp|p. 64}} The smallest has sides (68, 85, 87). Others include (127, 131, 158), (113, 243, 290), (145, 207, 328) and (327, 386, 409).
• There are no isosceles Pythagorean triangles.
• The only primitive Pythagorean triangles for which the square of the perimeter equals an integer multiple of the area are (3, 4, 5) with perimeter 12 and area 6 and with the ratio of perimeter squared to area being 24; (5, 12, 13) with perimeter 30 and area 30 and with the ratio of perimeter squared to area being 30; and (9, 40, 41) with perimeter 90 and area 180 and with the ratio of perimeter squared to area being 45.[http://forumgeom.fau.edu/FG2009volume9/FG200927.pdf Goehl, John F. Jr., "Pythagorean triangles with square of perimeter equal to an integer multiple of area", Forum Geometricorum 9 (2009): 281–282.]
• There exists a unique (up to similitude) pair of a rational right triangle and a rational isosceles triangle which have the same perimeter and the same area. The unique pair consists of the (377, 135, 352) triangle and the (366, 366, 132) triangle.{{Cite journal|last1=Hirakawa|first1=Yoshinosuke|last2=Matsumura|first2=Hideki|date=2018|title=A unique pair of triangles|journal=Journal of Number Theory|volume=194|pages=297–302|doi=10.1016/j.jnt.2018.07.007|issn=0022-314X|arxiv=1809.09936|s2cid=119661968}} There is no pair of such triangles if the triangles are also required to be primitive integral triangles. The authors stress the striking fact that the second assertion can be proved by an elementary argumentation (they do so in their appendix A), whilst the first assertion needs modern highly non-trivial mathematics.

• Brahmagupta triangle, a Heronian triangle in which the side lengths are consecutive integers
• Robbins pentagon, a cyclic pentagon with integer sides and integer area
• Euler brick, a cuboid with integer edges and integer face diagonals
• {{section link|Tetrahedron|Integer tetrahedra}}

{{reflist|30em}}

{{Geometry}}

{{DEFAULTSORT:Integer Triangle}}

Category:Arithmetic problems of plane geometry

Category:Discrete geometry

Category:Squares in number theory

Category:Types of triangles