Brahmagupta triangle explained

A Brahmagupta triangle is a triangle whose side lengths are consecutive positive integers and area is a positive integer.[1] [2] [3] The triangle whose side lengths are 3, 4, 5 is a Brahmagupta triangle and so also is the triangle whose side lengths are 13, 14, 15. The Brahmagupta triangle is a special case of the Heronian triangle which is a triangle whose side lengths and area are all positive integers but the side lengths need not necessarily be consecutive integers. A Brahmagupta triangle is called as such in honor of the Indian astronomer and mathematician Brahmagupta (c. 598 – c. 668 CE) who gave a list of the first eight such triangles without explaining the method by which he computed that list.[1] [4]

A Brahmagupta triangle is also called a Fleenor-Heronian triangle in honor of Charles R. Fleenor who discussed the concept in a paper published in 1996.[5] [6] [7] [8] Some of the other names by which Brahmagupta triangles are known are super-Heronian triangle[9] and almost-equilateral Heronian triangle.[10]

The problem of finding all Brahmagupta triangles is an old problem. A closed form solution of the problem was found by Reinhold Hoppe in 1880.[11]

Generating Brahmagupta triangles

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=(\tfrac{t}{2})\sqrt{3[(\tfrac{t}{2})2-1]}

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\sqrt{3(x2-1)}

Since

\sqrt{3(x2-1)}

has to be an integer, one must have

x2-1=3y2

for some integer

y

. Hence,

x

must satisfy the following Diophantine equation:

x2-3y2=1

.This is an example of the so-called Pell's equation

x2-Ny2=1

with

N=3

. The methods for solving the Pell's equation can be applied to find values of the integers

x

and

y

.

Obviously

x=2

,

y=1

is a solution of the equation

x2-3y2=1

. Taking this as an initial solution

x1=2,y1=1

the set of all solutions

\{(xn,yn)\}

of the equation can be generated using the following recurrence relations[1]

xn+1=2xn+3yn,yn+1=xn+2ynforn=1,2,\ldots

or by the following relations

\begin{align} xn+1&=4xn-xn-1forn=2,3,\ldotswithx1=2,x2=7\\ yn+1&=4yn-yn-1forn=2,3,\ldotswithy1=1,y2=4. \end{align}

They can also be generated using the following property:

xn+\sqrt{3}yn=(x1+\sqrt{3}y

nfor
1)

n=1,2,\ldots

The following are the first eight values of

xn

and

yn

and the corresponding Brahmagupta triangles:

n

1 2 3 4 5 6 7 8

xn

2 7 26 97 362 1351 5042 18817

yn

1 4 15 56 209 780 2911 10864
Brahmagupta
triangle
3,4,513,14,1551,52,53193,194,195723,724,7252701,2702,270310083,10084,1008537633,37634,337635
The sequence

\{xn\}

is entry in the Online Encyclopedia of Integer Sequences (OEIS) and the sequence

\{yn\}

is entry in OEIS.

Generalized Brahmagupta triangles

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.[12]

To find the side lengths of such triangles, let the side lengths be

t-d,t,t+d

where

b,d

are integers satisfying

1\led\let

. Using Heron's formula, the area

A

of the triangle can be shown to be

A=(\tfrac{b}{4})\sqrt{3(t2-4d2)}

.For

A

to be an integer,

t

must be even and one may take

t=2x

for some integer. This makes

A=x\sqrt{3(x2-d2)}

.Since, again,

A

has to be an integer,

x2-d2

has to be in the form

3y2

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:

x2-3y2=d2

.It can be shown that all primitive solutions of this equation are given by[12]

\begin{align} d&=\vertm2-3n2\vert/g\\ x&=(m2+3n2)/g\\ y&=2mn/g \end{align}

where

m

and

n

are relatively prime positive integers and

g=gcd(m2-3n2,2mn,m2+3n2)

.

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.

See also

Notes and References

  1. R. A. Beauregard and E. R. Suryanarayan . The Brahmagupta Triangles . The College Mathematics Journal . January 1998 . 29 . 1 . 13-17 . 6 June 2024.
  2. Web site: G. Jacob Martens . Rational right triangles and the Congruent Number Problem . arxiv.org . Cornell University . 6 June 2024.
  3. Herb Bailey and William Gosnell . Heronian Triangles with Sides in Arithmetic Progression: An Inradius Perspective . Mathematics Magazine . October 2012 . 85 . 4 . 290-294 . 10.4169/math.mag.85.4.290.
  4. Book: Venkatachaliyengar, K. . 1988 . Subbarayappa . B. V. . Scientific Heritage of India: Proceedings of a National Seminar, September 19-21, 1986, Bangalore . The Mythic Society, Bangalore . 36-48 . The Development of Mathematics in Ancient India: The Role of Brahmagupta .
  5. Charles R. Fleenor . Heronian Triangles with Consecutive Integer Sides . Journal of Recreational Mathematics . 1996 . 28 . 2 . 113-115.
  6. Web site: N. J. A. Sloane . A003500 . Online Encyclopedia of Integer Sequences . The OEIS Foundation Inc. . 6 June 2024.
  7. Web site: Definition:Fleenor-Heronian Triangle . Proof-Wiki . 6 June 2024.
  8. Vo Dong To . Finding all Fleenor-Heronian triangles . Journal of Recreational Mathematics . 2003 . 32 . 4 . 298-301.
  9. Web site: William H. Richardson . Super-Heronian Triangles . www.wichita.edu . Wichita State University . 7 June 2024.
  10. Roger B Nelsen . Almost Equilateral Heronian Triangles . Mathematics Magazine . 2020 . 93 . 5 . 378-379.
  11. H. W. Gould . A triangle with integral sides and area . Fibonacci Quarterly . 1973 . 11 . 27-39 . 7 June 2024.
  12. James A. Macdougall . Heron Triangles With Sides in Arithmetic Progression . Journal of Recreational Mathematics . January 2003 . 31 . 189-196.