Aninteger triangle orintegral triangle is atriangle all of whose side lengths areintegers. Arational triangle is one whose side lengths arerational numbers; any rational triangle can berescaled by thelowest common denominator of the sides to obtain asimilar integer triangle, so there is a close relationship between integer triangles and rational triangles.

Sometimes other definitions of the termrational triangle are used: Carmichael (1914) and Dickson (1920) use the term to mean aHeronian triangle (a triangle with integral or rational side lengths and area);^[1] Conway and Guy (1996) define a rational triangle as one with rational sides and rationalangles measured in degrees—the only such triangles are rational-sidedequilateral triangles.^[2]

Any triple of positive integers can serve as the side lengths of an integer triangle as long as it satisfies thetriangle inequality: the longest side is shorter than the sum of the other two sides. Each such triple defines an integer triangle that is uniqueup tocongruence. So the number of integer triangles (up to congruence) withperimeterp is the number ofpartitions ofp into three positive parts that satisfy the triangle inequality. This is the integer closest to${\displaystyle p^2/48}$ whenp iseven and to${\displaystyle (p+3{)}^2/48}$ whenp isodd.^[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. Thesequence of the number of integer triangles with perimeterp, 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 ... (sequenceA005044 in theOEIS)

This is calledAlcuin's sequence.

The number of integer triangles (up to congruence) with given largest sidec and integer triple${\displaystyle (a,b,c)}$ is the number of integer triples such that${\displaystyle a+b>c}$ and${\displaystyle a\le b\le c.}$ This is the integer value${\displaystyle \lceil \frac{1}{2}(c+1)\rceil \cdot \lfloor \frac{1}{2}(c+1)\rfloor .}$^[3] Alternatively, forc even it is the doubletriangular number${\displaystyle \frac{1}{2}c(\frac{1}{2}c+1)}$ and forc odd it is thesquare${\displaystyle \frac{1}{4}(c+1).}$ It also means that the number of integer triangles with greatest sidec exceeds the number of integer triangles with greatest sidec − 2 byc. The sequence of the number of non-congruent integer triangles with largest sidec, starting atc = 1, is:

1, 2, 4, 6, 9, 12, 16, 20, 25, 30, 36, 42, 49, 56, 64, 72, 81, 90 ... (sequenceA002620 in theOEIS)

The number of integer triangles (up to congruence) with given largest sidec and integer triple (a, b, c) that lie on or within a semicircle of diameterc is the number of integer triples such thata + b > c , a² + b² ≤ c² anda ≤ b ≤ c. This is also the number of integer sidedobtuse orright (non-acute) triangles with largest sidec. The sequence starting atc = 1, is:

0, 0, 1, 1, 3, 4, 5, 7, 10, 13, 15, 17, 22, 25, 30, 33, 38, 42, 48 ... (sequenceA236384 in theOEIS)

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

1, 2, 3, 5, 6, 8, 11, 13, 15, 17, 21, 25, 27, 31, 34, 39, 43, 48, 52 ... (sequenceA247588 in theOEIS)

ByHeron's formula, ifT is thearea of a triangle whose sides have lengthsa,b, andc then

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

Since all the terms under theradical on the right side of the formula are integers it follows that all integer triangles must have16T² an integer andT² will be rational.

By thelaw of cosines, every angle of an integer triangle has a rationalcosine. Every angle of an integer right triangle also has rationalsine (seePythagorean triple).

If the angles of any triangle form anarithmetic 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 ageometric orharmonic progression. This is because such angles have to be rational angles of the form${\displaystyle \pi p/q}$ with rational 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 internalangle bisector of an integer triangle is rational, because the general triangle formula for the internal angle bisector of angleA is$2\sqrt{bcs(s-a)}/(b+c)$ wheres is thesemiperimeter (and likewise for the other angles' bisectors).

Anyaltitude 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 anymedian of an integer triangle is an integer, because the general formula for the squared medianm_a² to sidea is${\displaystyle \frac{1}{4}(2b^2+2c^2-a^2)}$, 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 itscircumradius is also rational, as is the square of theinradius.

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

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

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

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

All Heronian triangles can be placed on alattice with each vertex at a lattice point.^[7] 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 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 integersm,n andk subject to the constraints:

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

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

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

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

A Pythagorean triangle is right-angled and Heronian. Its three integer sides are known as aPythagorean triple orPythagorean triplet orPythagorean triad.^[9] All Pythagorean triples${\displaystyle (a,b,c)}$ withhypotenuse${\displaystyle c}$ which areprimitive (the sides having nocommon factor) can be generated by

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

${\displaystyle b=2mn,}$

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

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

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

wherem andn arecoprime integers and one of them is even withm > 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.

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 bothab andcd whered is the height from the hypotenusec. The three side lengths of a primitive triangle are coprime, so${\displaystyle d=ab/c}$ is in fully reduced form; sincec cannot equal 1 for any primitive Pythagorean triangle,d cannot be an integer.

However, any Pythagorean triangle with legsx, y and hypotenusez can generate a Pythagorean triangle with an integer altitude, by scaling up the sides by the length of the hypotenusez. Ifd is the altitude, then the generated Pythagorean triangle with integer altitude is given by^[11]

${\displaystyle (a,b,c,d)=(xz,yz,z^2,xy).}$

Consequently, all Pythagorean triangles with legsa andb, hypotenusec, and integer altituded from the hypotenuse, with${\displaystyle gcd(a,b,c,d)=1}$, which necessarily satisfy botha² + b² = c² and${\displaystyle \frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{d^2}}$, are generated by^[12][11]

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

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

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

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

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

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

for coprime integersm,n withm > n.

A triangle with integer sides and integer area has sides in arithmetic progressionif 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 whereg is thegreatest common divisor of${\displaystyle m^2-3n^2,}$${\displaystyle 2mn,}$ and${\displaystyle m^2+3n^2.}$

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

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

with integersq,u,v such that${\displaystyle v>u}$ and

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

Allisosceles 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:^[15]

for coprime integers of opposite parity${\displaystyle u}$ and${\displaystyle v}$, with${\displaystyle u>v}$.

It has been shown that a Heronian triangle whose perimeter is four times aprime is uniquely associated with the prime and that the prime iscongruent 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${\displaystyle p=m^2+2n^2}$ (seeEuler'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${\displaystyle 3}$.

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${\displaystyle \frac{1}{2}\alpha }$,${\displaystyle \frac{1}{2}\beta }$ and${\displaystyle \frac{1}{2}\gamma }$ must beHeronian angles. Namely, if both angles${\displaystyle \frac{1}{2}\alpha }$ and${\displaystyle \frac{1}{2}\beta }$ are Heronian then${\displaystyle \frac{1}{2}\gamma }$, the complement of${\displaystyle \frac{1}{2}\alpha +\frac{1}{2}\beta }$, must also be a Heronian angle, so that all three angle-bisectors are rational. This is also evident if one multiplies:

${\displaystyle w_a=\frac{2\sqrt{s(s-a)}\cdot \sqrt{bc}}{b+c}{w}_b=\frac{2\sqrt{s(s-b)}\cdot \sqrt{ac}}{a+c}{w}_c=\frac{2\sqrt{s(s-c)}\cdot \sqrt{ab}}{a+b}}$

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 similarity classes of 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.

There are infinitely many decomposable, and infinitely many indecomposable, primitive Heronian (non-Pythagorean) triangles with integer radii for theincircle and eachexcircle.^{[19]: Thms. 3 and 4} A family of decomposable 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

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

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

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

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

${\displaystyle r_a=5n+3}$

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

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

There existtetrahedra having integer-valuedvolume and Heron triangles asfaces. 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

A 2Dlattice is a regular array of isolated points where if any one point is chosen as theCartesian origin (0, 0), then all the other points are at (x, y) wherex andy 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. ByPick's theorem a lattice triangle has a rational area that either is an integer or ahalf-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 aninteger lattice with all vertices, theincenter, and all threeexcenters 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}

Anautomedian triangle is one whose medians are in the same proportions (in the opposite order) as the sides. Ifx,y, andz are the three sides of a right triangle, sorted in increasing order by size, and if 2x < z, thenz,x + y, andy − 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.^[23]

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

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

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

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

with${\displaystyle m}$ and${\displaystyle n}$ coprime and${\displaystyle m+n}$ odd, and  (if the quantity inside theabsolute value signs is negative) or  ${\displaystyle m>(2+\sqrt{3})n}$ (if that quantity is positive) to satisfy thetriangle inequality.

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

A triangle family with integer sides${\displaystyle a,b,c}$ and with rational bisector${\displaystyle d}$ of angleA 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${\displaystyle k>m>0}$.

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, forn > 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 at vertexA having given rational cosineh /k (h < 0 or > 0;k > 0) are given by^[26]

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

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

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

wherep andq are any coprime positive integers such thatp >qk. All primitive solutions can be obtained by dividinga,b, andc 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:^[5]

${\displaystyle a=4mn,}$

${\displaystyle b=3m^2+n^2,}$

${\displaystyle c=2mn+|3m^2-n^2|}$

with coprime integersm,n and 1 ≤ n ≤ m or 3m ≤ n. From here, all primitive solutions can be obtained by dividinga,b, andc by their greatest common divisor.

Integer triangles with a 60° angle can also be generated by^[27]

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

${\displaystyle b=2mn-n^2,}$

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

with coprime integersm,n with 0 < n < m (the angle of 60° is opposite to the side of lengtha). From here, all primitive solutions can be obtained by dividinga,b, andc by their greatest common divisor (e.g. an equilateral triangle solution is obtained by takingm = 2 andn = 1, but this producesa =b =c = 3, which is not a primitive solution). See also^[28][29]

More precisely, If${\displaystyle m\equiv -n(\mathrm{mod}3)}$, then${\displaystyle gcd(a,b,c)=3}$, otherwise${\displaystyle gcd(a,b,c)=1}$. Two different pairs${\displaystyle (m,n)}$ and${\displaystyle (m,m-n)}$ 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${\displaystyle n}$ going only till${\displaystyle m/2}$. We still need to divide by 3 if the gcd is 3. The only solution for${\displaystyle n=m/2}$ under the above constraints is${\displaystyle (3,3,3)\equiv (1,1,1)}$ for${\displaystyle m=2,n=1}$. With this additional${\displaystyle n\le m/2}$ constraint all triples can be generated uniquely.

AnEisenstein 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 by^[30]

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

${\displaystyle b=2mn+n^2,}$

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

with coprime integersm, n with 0 < n < m (the angle of 120° is opposite to the side of lengtha). From here, all primitive solutions can be obtained by dividinga,b, andc by their greatest common divisor. The smallest solution, form = 2 andn = 1, is the triangle with sides (3,5,7). See also.^[28][29]

More precisely, If${\displaystyle m\equiv n(\mathrm{mod}3)}$, then${\displaystyle gcd(a,b,c)=3}$, otherwise${\displaystyle gcd(a,b,c)=1}$. Since the biggest sidea can only be generated with a single${\displaystyle (m,n)}$ 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${\displaystyle m\not\equiv n(\mathrm{mod}3)}$ condition.^{[citation needed]}

For positive coprime integersh andk, 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 ratioh :k, and its sides are integers:^[31]

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

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

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

where${\displaystyle \alpha ={\cos }^{-1}\frac{p}{q}}$ andp andq are any coprime integers such that.

With angleA opposite side${\displaystyle a}$ and angleB opposite side${\displaystyle b}$, some triangles withB = 2A are generated by^[32]

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

${\displaystyle b=mn,}$

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

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

All triangles withB = 2A (whether integer or not) satisfy^[33]${\displaystyle a(a+c)=b^2.}$

Theequivalence class of similar triangles with${\displaystyle B=\frac{3}{2}A}$ are generated by^[32]

${\displaystyle a=m{n}^3,}$

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

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

with integers${\displaystyle m,n}$ such that, where${\displaystyle \phi }$ is thegolden ratio$\phi =\frac{1}{2}(1+\sqrt{5})\approx 1.61803$.

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

We can generate the full equivalence class of similar triangles that satisfyB = 3A by using the formulas^[34]

${\displaystyle a=n^3,}$

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

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

where${\displaystyle m}$ and${\displaystyle n}$ are integers such that.

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

The only integer triangle with three rational angles (rational numbers of degrees, or equivalently rational fractions of a full turn) is theequilateral triangle.^[2] This is because integer sides imply three rationalcosines by thelaw of cosines, and byNiven'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 ofelliptic curves for an integer triangle to have an integer ratioN of thecircumradius to theinradius.^[35][36] The smallest case, that of theequilateral triangle, hasN = 2. In every known case,${\displaystyle N\equiv 2(\mathrm{mod}8)}$ – that is,${\displaystyle N-2}$ is divisible by 8.

A 5-Con triangle pair is a pair of triangles that aresimilar but notcongruent 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

${\displaystyle (x^3,x^{2}y,x{y}^2)}$ and${\displaystyle (x^{2}y,x{y}^2,y^3)}$

for positive coprime integersx andy. The smallest example is the pair (8, 12, 18), (12, 18, 27), generated byx = 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.^[37][38][39] This property states that if angle C is obtuse and if a segment is dropped from B meeting perpendicularly ACextended 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.^[39]
• 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.^[39]
• The (3, 5, 7) triangle and its multiples are the only triangles with a 120° angle and having integer sides in arithmetic progression.^[39]
• The only integer triangle with area = semiperimeter^[40] has sides (3, 4, 5).
• The only integer triangles with area = perimeter have sides^[40][41] (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 rationalmedians.^{[9]: 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.^[15]
• 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.^[42]
• 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.^[43] There is no pair of such triangles if the triangles are also required to be primitive integral triangles.^[43] 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, aHeronian 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
• Tetrahedron § Integer tetrahedra

• Arithmetic problems of plane geometry
• Discrete geometry
• Squares in number theory
• Types of triangles

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