Brahmagupta triangle

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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 t1, 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=(t2)3[(t2)21]

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=x3(x21)

Since 3(x21) has to be an integer, one must have x21=3y2 for some integer y. Hence, x must satisfy the following Diophantine equation:

x23y2=1.

This is an example of the so-called Pell's equation x2Ny2=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
A Brahmagupta triangle where xn and yn are integers satisfying the equation xn23yn2=1.

Obviously x=2, y=1 is a solution of the equation x23y2=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+2yn for n=1,2,

or by the following relations

xn+1=4xnxn1 for n=2,3, with x1=2,x2=7yn+1=4ynyn1 for n=2,3, with y1=1,y2=4.

They can also be generated using the following property:

xn+3yn=(x1+3y1)n for n=1,2,

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
3,4,5 13,14,15 51,52,53 193,194,195 723,724,725 2701,2702,2703 10083,10084,10085 37633,37634,37635

The sequence {xn} is entry A001075 in the Online Encyclopedia of Integer Sequences (OEIS) and the sequence {yn} is entry A001353 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 t1,t,t+1 are the side lengths of a Brahmagupta triangle then, for any positive integer k, the integers k(t1),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 td,t,t+d where b,d are integers satisfying 1dt. Using Heron's formula, the area A of the triangle can be shown to be

A=(b4)3(t24d2).

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

A=x3(x2d2).

Since, again, A has to be an integer, x2d2 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:

x23y2=d2.

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

d=|m23n2|/gx=(m2+3n2)/gy=2mn/g

where m and n are relatively prime positive integers and g=gcd(m23n2,2mn,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

References

  1. 1.0 1.1 1.2 R. A. Beauregard and E. R. Suryanarayan (January 1998). "The Brahmagupta Triangles" (PDF). The College Mathematics Journal. 29 (1): 13–17. doi:10.1080/07468342.1998.11973907. Retrieved 6 June 2024.
  2. G. Jacob Martens (2021). "Rational right triangles and the Congruent Number Problem". arXiv:2112.09553 [math.GM].
  3. Herb Bailey and William Gosnell (October 2012). "Heronian Triangles with Sides in Arithmetic Progression: An Inradius Perspective". Mathematics Magazine. 85 (4): 290–294. doi:10.4169/math.mag.85.4.290.
  4. Venkatachaliyengar, K. (1988). "The Development of Mathematics in Ancient India: The Role of Brahmagupta". In Subbarayappa, B. V. (ed.). Scientific Heritage of India: Proceedings of a National Seminar, September 19-21, 1986, Bangalore. The Mythic Society, Bangalore. pp. 36–48.
  5. Charles R. Fleenor (1996). "Heronian Triangles with Consecutive Integer Sides". Journal of Recreational Mathematics. 28 (2): 113–115.
  6. N. J. A. Sloane. "A003500". Online Encyclopedia of Integer Sequences. The OEIS Foundation Inc. Retrieved 6 June 2024.
  7. "Definition:Fleenor-Heronian Triangle". Proof-Wiki. Retrieved 6 June 2024.
  8. Vo Dong To (2003). "Finding all Fleenor-Heronian triangles". Journal of Recreational Mathematics. 32 (4): 298–301.
  9. William H. Richardson. "Super-Heronian Triangles". www.wichita.edu. Wichita State University. Retrieved 7 June 2024.
  10. Roger B Nelsen (2020). "Almost Equilateral Heronian Triangles". Mathematics Magazine. 93 (5): 378–379. doi:10.1080/0025570X.2020.1817708.
  11. H. W. Gould (1973). "A triangle with integral sides and area" (PDF). Fibonacci Quarterly. 11: 27–39. doi:10.1080/00150517.1973.12430863. Retrieved 7 June 2024.
  12. 12.0 12.1 James A. Macdougall (January 2003). "Heron Triangles With Sides in Arithmetic Progression". Journal of Recreational Mathematics. 31: 189–196.