History of Hindu Mathematics: A Source Book v2.c17.s18.t1: Right Triangles having a Given Side

History of Hindu Mathematics: A Source Book v2.c17.s18.t1

Language: English

Right Triangles having a Given Side

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In the *Śulba* we find an attempt to find rational right triangles having a given side, that is, rational solutions of x² + a² = z².

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² "He dvitribîjakasya kṣetrabhujârdhena cânyamutthâpya"—*GSS*, vii. 111½.

In particular, we find mention of two such right triangles having a common side *a*, *viz.*, (a, 3a/4, 5a/4) and (a, 5a/12, 13a/12).⁴ The principle underlying these solutions will be easily detected to be that of the reduction of the sides of any rational right triangle in the ratio of the given side to its corresponding side. This principle of finding rational right triangles having a given side has been followed explicitly by Mahâvîra (850).¹

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⁴ Datta, *Śulba*, p. 180.

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¹ *Vide infra*, p. 213

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It has been stated before that one rule of Brahmagupta² can be interpreted as giving rational solutions of x² + a² = z² as $a, ½((a²)/n-n), ½((a²)/n+n),$ where n is any rational number. In fact, he has used this solution in finding rational isosceles triangles having a given altitude.³ This solution has been expressly stated by Mahâvîra (850). He says:

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³ *Vide infra*, p. 223

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"The *saṅkramaṇa* between any optional divisor of the square of the given upright or the base and the (respective) quotient gives the diagonal and the base (or upright)."⁴

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⁴ *GSS*, vii. 97½.

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He has restated the solution thus:

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"The *saṅkramaṇa* between any (rational) divisor of the upright and the quotient gives the elements; or any (rational) divisor of half the side and the quotient are the elements."⁵

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⁵ *GSS*, vii. 95½.

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The right triangles formed according to the first half of this rule are:⁶ $a, ½((a²)/(p²)-p²), ½((a²)/(p²)+p²),$ and those according to the second half are :¹ $(a²)/(4q²)-q², a, (a²)/(4q²)+q².$

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² *Vide supra*, p. 206.

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⁶ The "elements" here are ½(a/p+p), ½(a/p-p), where p is an optional number.

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¹ The "elements" here are q, a/2q, where q is an optional number.

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Bhâskara II gives two solutions one of which is the same as that of Brahmagupta. He says :

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"The side is given : from that multiplied by twice an optional number and divided by the square of that optional number minus unity, is obtained the upright ; this again multiplied by the optional number and diminished by the given side becomes the hypotenuse. This triangle is a right-angled triangle.

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"Or the side is given : its square divided by an optional number is put down at two places ; the optional number is subtracted (at one place) and added (at another) and then halved ; these results are the upright and the hypotenuse. Similarly from the given upright can be obtained the side and the hypotenuse."²

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That is to say, the two solutions are $a, 2na/(n²-1), n(2na/(n²-1))-a,$ $and a, ½((a²)/n-n), ½((a²)/n+n).$

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Bhâskara II illustrates this by finding four right triangles having a side equal to 12, *viz.*, (12, 35, 37), (12, 16, 20), (12, 9, 15) and (12, 5, 13).³

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³ *L*, pp. 34f.

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The *rationale* of the first solution has been given by Sûryadâsa (1538) thus : Starting with the rational right triangle n²-1, 2n, n²+1, he observes that if x, y, z be the corresponding sides of another right triangle, then $x/(n² - 1) = y/2n = z/(n² + 1) = k.(say).$

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² *L*, p. 34.

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Hence x = k(n² - 1), y = 2nk, z = k(n² + 1).

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Therefore x + z = 2kn² = ny.

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If now we have x = a, then $k = a/(n² - 1).$

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Hence $y = 2na/(n² - 1),$ and $z = ny - a = n(2na/(n² - 1)) - a.$

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The second rule has been demonstrated by Sûryadâsa, Gaṇeśa and Raṅganâtha thus:

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Since x² + a² = z², we have a² = z² - x² = (z - x)(z + x).

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Assume z - x = n, where n is any rational number; then $z + x = (a²)/n.$ $∴ z = ½((a²)/n + n), x = ½((a²)/n - n).$

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Generalising the method of the *Âpastamba Śulba* the commentators obtained the solution¹ $a, ((m² + 2m)/(2m + 2))a, ((m² + 2m + 2)/(2m + 2))a.$

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¹ Datta, *Śulba*, p. 16.

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**Right Triangles having a Given Hypotenuse.** For finding all rational right triangles having a given hypotenuse (c), that is, for rational solutions of $x² + y² = c²,$ Mahâvîra gives three rules. The first rule is :

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“The square-root of half the sum and difference of the diagonal and the square of an optional number are they (the elements).”¹

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¹ *GSS*, vii. 95½.

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In other words, the required solution will be obtained from the “elements” √((c + p²)/2) and √((c - p²)/2), where p is any rational number. Hence the solution is $p², √(c² - p⁴), c.$

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The second rule is :

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“Or the square-root of the difference of the squares of the diagonal and of an optional number, and that optional number are the upright and the base.”²

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That is, the solution is $p, √(c² - p²), c.$

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These solutions are defective in the sense that √(c² - p⁴) or √(c² - p²) might not be rational unless p is suitably chosen. Mahâvîra’s third rule is of greater importance. He says :

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² *GSS*, vii. 97½.

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“Each of the various figures (rectangles) that can be formed from the elements are put down ; by its diagonal is divided the given diagonal. The perpendicular, base and the diagonal (of this figure) multiplied by this quotient (give rise to the corresponding sides of the figure having the given hypotenuse).”³

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³ *GSS*, vii. 122½.

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Thus having obtained the general solution of the rational right triangle, *viz.*, m² - n², 2mn, m² + n², Mahâvîra reduces it in the ratio c/(m² + n²), so that all rational right triangles having a given hypotenuse c will be given by $((m²-n²)/(m²+n²))c, (2mn/(m²+n²))c, c.$

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By way of illustration Mahâvîra finds four rectangles (39, 52), (25, 60), (33, 56) and (16, 63) having the same diagonal 65.¹

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¹ *GSS*, vii. 123-124½.

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This method was later on rediscovered in Europe by Leonardo Fibonacci of Pisa (1202) and Vieta. It has been pointed out before that the origin of the method can be traced to the *Sulba*.

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Bhâskara II (1150) says :

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"From the given hypotenuse multiplied by an optional number and doubled and then divided by the square of the optional number added to unity, is obtained the upright ; this is again multiplied by the optional number ; the difference between that (product) and the given hypotenuse is the side.
"Or divide twice the hypotenuse by the square of an optional number added to unity. The hypotenuse minus the quotient is the upright and the quotient multiplied by the optional number is the side."²

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Thus, according to the above, the sides of a right-angled triangle whose hypotenuse is c are : $2mc/(m²+1), m(2mc/(m²+1))-c, c;$ $or 2mc/(m²+1), c-2c/(m²+1), c.$

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² *L*, pp. 35, 36.

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By way of illustration Bhâskara II finds two right triangles (51, 68) and (40, 75) having the same hypotenuse 85.¹

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¹ *L*, pp. 35f.

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Sûryadâsa demonstrates the above substantially thus :

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If (x, y, z) be the sides of the right triangle, we have $x/(m² - 1) = y/2m = z/(m² + 1) = k (say),$ where m is any rational integer. Then $x = k(m² - 1), y = 2mk, z = k(m² + 1).$ Therefore $x + z = 2km² = my.$

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Since z is given to be equal to c, we have $k = c/(m² + 1).$ Hence $y = 2mc/(m² + 1),$ and $x = my - z = m(2mc/(m² + 1)) - c.$

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**Problems Involving Areas and Sides.** Mahâvîra proposes to find rational rectangles (or squares) in which the area will be *numerically* (*saṁkhyayâ*) equal to any multiple or submultiple of a side, diagonal or perimeter, or of any linear combination of two or more of them. Expressed symbolically, the problem is to solve $.x² + y² = z², / mx + ny + pz = rxy; (1)$ m, n, p, r being any rational numbers (r ≠ 0). For the solution of this problem he gives the following rule :

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"Divide the sides (or their sum) of any generated square or other figure as multiplied by their respective given multiples by the area of that figure taken into its given multiple. The sides of that figure multiplied by this quotient will be the sides of the (required) square or other figure."[^\mathcal]

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That is to say, starting with any rational solution of x′ 2 + y′ 2 = z′ 2,2 we shall have to calculate the value of mx′ + ny′ + pz′ = Q, say.3 Then the required solution of (1) will be obtained by reducing the values of x′, y′, z′ in the ratio of Q/rx′ y′. Thus .x &= x^\prime Q/rx^\prime y^\prime = Q/ry^\prime, / y &= y^\prime Q/rx^\prime y^\prime = Q/rx^\prime, / z &= z^\prime Q/rx^\prime y^\prime.}\tag

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Mahâvîra gives several illustrative examples some of which are very interesting:

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"In a rectangle the area is (numerically) equal to the perimeter; in another rectangle the area is (numerically) equal to the diagonal. What are the sides (in each of these cases)?"²

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Algebraically, we shall have to solve .x² + y² &= z², / 2(x + y) &= xy,}\tag{1.1} and .x² + y² &= z², / xy &= z.}\tag{1.2}

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Starting with the solution s² - t², 2st, s² + t² of (2) and putting m = n = 2, p = 0, r = 1 in (4), we get the solution of (1.¹) as $(2(s² - t²) + 4st)/2st, (2(s² - t²) + 4st)/(s² - t²),$ $(2(s² - t²) + 4st)/(2st(s² - t²))(s² + t²).$

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¹ *GSS*, vii. 112½.

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² *GSS*, vii. 115½.

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And putting m = n = 0, p = r = 1, in (4), we have the solution of (1.2) as $(s² + t²)/2st, (s² + t²)/(s² - t²), ((s² + t²)²)/(2st(s² - t²)).$

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Bhâskara II solves a problem similar to the second one above:

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Find a right triangle whose area equals the hypotenuse.¹

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¹ *BBi*, p. 56.

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He starts with the rational right triangle (3x, 4x, 5x); then by the condition, area = hypotenuse, finds the value x = 5/6. So that a right triangle of the required type is (5/2, 10/3, 25/6). He then observes: "In like manner, by virtue of various assumptions, other right triangles can also be found."² The general solution in this case is $(s² + t²)/st, (2(s² + t²))/(s² - t²), ((s² + t²)²)/(st(s² - t²)).$

Cites (3)

Gaṇitasārasaṅgraha — Mahāvīra (mathematician), Sanskrit
Līlāvaī — Kouhala, Māhārāṣṭrī Prakrit
Bījagaṇita — Bhāskara II, Sanskrit