Brahmagupta triangle

Let the side lengths of a Brahmagupta triangle be $t\; -1$, $t$ and $t+1$ where $t$ is an integer greater than 1. Using Heron's formula, the area $A$ of the triangle can be shown to be

:$A=\backslash big(\backslash tfrac\{t\}\{2\}\backslash big)\backslash sqrt\{3\backslash big[\; \backslash big(\backslash tfrac\{t\}\{2\}\backslash big)^2\; -1\; \backslash big]\; \}$

Since $A$ has to be an integer, $t$ must be even and so it can be taken as $t=2x$ where $x$ is an integer. Thus,

:$A\; =\; x\backslash sqrt\{3(x^2-1)\; \}$

Since $\backslash sqrt\{3(x^2-1)\; \}$ has to be an integer, one must have $x^2-1\; =3y^2$ for some integer $y$. Hence, $x$ must satisfy the following Diophantine equation:

: $x^2-3y^2=1$.

This is an example of the so-called Pell's equation $x^2-Ny^2=1$ with $N=3$. The methods for solving the Pell's equation can be applied to find values of the integers $x$ and $y$.

File:BrahmaguptaTriangle.png

Obviously $x=2$, $y=1$ is a solution of the equation $x^2-3y^2=1$. Taking this as an initial solution $x\_1=2,\; y\_1=1$ the set of all solutions $\backslash \{(x\_n,\; y\_n)\backslash \}$ of the equation can be generated using the following recurrence relations

:$$

x_{n+1}=2x_n+3y_n, \quad y_{n+1}= x_n+2y_n \text{ for } n=1,2,\ldots

or by the following relations

:$$

\begin{align}

x_{n+1} & = 4x_{n}-x_{n-1}\text{ for }n=2,3,\ldots \text{ with } x_1=2, x_2=7\\

y_{n+1} & = 4y_{n}-y_{n-1}\text{ for }n=2,3,\ldots \text{ with } y_1=1, y_2=4.

\end{align}

They can also be generated using the following property:

:$$

x_n+\sqrt{3} y_n=(x_1+\sqrt{3}y_1)^n\text{ for } n=1,2, \ldots

The following are the first eight values of $x\_n$ and $y\_n$ and the corresponding Brahmagupta triangles:

::

The sequence $\backslash \{x\_n\backslash \}$ is entry {{OEIS link|A001075}} in the Online Encyclopedia of Integer Sequences (OEIS) and the sequence $\backslash \{y\_n\backslash \}$ is entry {{OEIS link|A001353}} in OEIS.

In a Brahmagupta triangle the side lengths form an integer arithmetic progression with a common difference 1. A generalized Brahmagupta triangle is a Heronian triangle in which the side lengths form an arithmetic progression of positive integers. Generalized Brahmagupta triangles can be easily constructed from Brahmagupta triangles. If $t-1,\; t,\; t+1$ are the side lengths of a Brahmagupta triangle then, for any positive integer $k$, the integers $k(t-1),\; kt,\; k(t+1)$ are the side lengths of a generalized Brahmagupta triangle which form an arithmetic progression with common difference $k$. There are generalized Brahmagupta triangles which are not generated this way. A primitive generalized Brahmagupta triangle is a generalized Brahmagupta triangle in which the side lengths have no common factor other than 1.{{cite journal |last1=James A. Macdougall |title=Heron Triangles With Sides in Arithmetic Progression |journal=Journal of Recreational Mathematics |date=January 2003 |volume=31 |pages=189–196}}

To find the side lengths of such triangles, let the side lengths be $t-d,\; t,\; t+d$ where $t,d$ are integers satisfying $1\backslash le\; d\backslash le\; t$. Using Heron's formula, the area $A$ of the triangle can be shown to be

:$A\; =\; \backslash big(\backslash tfrac\{b\}\{4\}\backslash big)\backslash sqrt\{3(t^2-4d^2)\}$.

For $A$ to be an integer, $t$ must be even and one may take $t=2x$ for some integer. This makes

:$A=x\backslash sqrt\{3(x^2-d^2)\}$.

Since, again, $A$ has to be an integer, $x^2-d^2$ has to be in the form $3y^2$ for some integer $y$. Thus, to find the side lengths of generalized Brahmagupta triangles, one has to find solutions to the following homogeneous quadratic Diophantine equation:

:$x^2-3y^2=d^2$.

It can be shown that all primitive solutions of this equation are given by

:$$

\begin{align}

d & = \vert m^2 - 3n^2\vert /g\\

x & = (m^2 + 3n^2)/g\\

y & = 2mn/g

\end{align}

where $m$ and $n$ are relatively prime positive integers and $g\; =\; \backslash text\{gcd\}(m^2\; -\; 3n^2,\; 2mn,\; m^2\; +\; 3n^2)$.

If we take $m=n=1$ we get the Brahmagupta triangle $(3,4,5)$. If we take $m=2,\; n=1$ we get the Brahmagupta triangle $(13,14,15)$. But if we take $m=1,\; n=2$ we get the generalized Brahmagupta triangle $(15,\; 26,\; 37)$ which cannot be reduced to a Brahmagupta triangle.

• Brahmagupta polynomials
• Brahmagupta quadrilateral

{{reflist}}

Category:Arithmetic problems of plane geometry

Category:Types of triangles

Category:Eponymous geometric shapes

Category:Elementary mathematics

Category:Elementary number theory

Category:Brahmagupta