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History of Hindu Mathematics: A Source Book v2.c17.s19.t1: Rational Trapeziums with Three Equal Sides

History of Hindu Mathematics: A Source Book v2.c17.s19.t1

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Rational Trapeziums with Three Equal Sides

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This problem is nearly the same as that of the rational isosceles trapezium with this difference that in this case one of the parallel sides also is equal to the slant sides. Brahmagupta states the following solution of the problem :

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“The square of the diagonal (of a *generated* rectangle) gives three equal sides; the fourth (is obtained) by subtracting the square of the upright from thrice the square of the side (of that rectangle). If greater, it is the base; if less, it is the face.”¹

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¹ *BrSpSi*, xii. 37.

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The rectangle generated from m, n is given by $m² - n², 2mn, m² + n².$

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If ABCD be a rational trapezium whose sides AB, BC, AD are equal, then $AB = BC = AD = (m² + n²)²,$ $CD = 3(2mn)² - (m² - n²)² = 14m²n² - m⁴ - n⁴,$ or CD = 3(m² - n²)² - (2mn)² = 3m⁴ + 3n⁴ - 10m²n².

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In particular, putting m = 2, n = 1, Pṛthûdakasvâmî deduces two rational trapeziums with three equal sides, *viz.*, (25, 25, 25, 39) and (25, 25, 25, 11).

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The first solution is also given by Mahâvîra who indicates the method for obtaining it. He says :

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"For a trapezium with three equal sides (proceed) as in the case of the isosceles trapezium with (the rectangle formed from) the quotient of the area of a generated rectangle divided by the square-root of its side multiplied by the difference of its elements and divisor; and that (formed) from the side and upright."

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That is to say, from any rectangle (m² - n², 2mn, m² + n²), calculate $(2mn(m² - n²))/(√(2mn(m - n))) = √(2mn(m + n)).$

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Then from √(2mn(m - n)), √(2mn(m + n)) form the rectangle $8m²n², 4mn(m² - n²), 4mn(m² + n²). 1$

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Again from 2mn, m² - n² form another rectangle $6m²n² - m⁴ - n⁴, 4mn(m² - n²), (m² + n²)² 2$

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By the juxtaposition of the rectangles (1) and (2) we get Brahmagupta's rational trapezium with three equal sides : CD = 8m²n² + (6m²n² - m⁴ - n⁴) = 14m²n² - m⁴ - n⁴, AB = 8m²n² - (6m²n² - m⁴ - n⁴) = (m² + n²)² = AD = BC, if m² + n² < 4mn.

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¹ *GSS*, vii. 101½.

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The segment (CH), altitude (AH), diagonals (AC, BD) and area of this trapezium are also rational, for CH = 6m²n² - m⁴ - n⁴, AH = 4mn(m² - n²), AC = BD = 4mn(m² + n²), area ABCD = 32m³n³(m² - n²).

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**Rational Inscribed Quadrilaterals.** Brahmagupta formulated a remarkable proposition: To find all quadrilaterals which will be inscribable within circles and whose sides, diagonals, perpendiculars, segments (of sides and diagonals by perpendiculars from vertices as also of diagonals by their intersection), areas, and also the diameters of the circumscribed circles will be expressible in integers. Such quadrilaterals are called Brahmagupta quadrilaterals.

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The solution given by Brahmagupta is as follows:

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"The uprights and bases of two right-angled triangles being reciprocally multiplied by the diagonals of the other will give the sides of a quadrilateral of unequal sides: (of these) the greatest is the base, the least is the face, and the other two sides are the two flanks."¹

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¹ *BrSpSi*, xii. 38.

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Taking Brahmagupta's integral solution, the sides of the two right triangles of reference are given by m² - n², 2mn, m² + n² ; p² - q², 2pq, p² + q² ;

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where *m*, *n*, *p*, *q* are integers. Then the sides of a Brahmagupta quadrilateral are $. (m² - n²)(p² + q²), & (p² - q²)(m² + n²), / 2mn(p² + q²), & 2pq(m² + n²). (A)$

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Mahâvîra says :

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“The base and the perpendicular (of the smaller and the larger derived rectangles of reference) multiplied reciprocally by the longer and the shorter diagonals and (each again) by the shorter diagonal will be the sides, the face and the base (of the required quadrilateral). The uprights and bases are reciprocally multiplied and then added together; again the product of the uprights is added to the product of the bases; these two sums multiplied by the shorter diagonal will be the diagonals. (These sums) when multiplied respectively by the base and perpendicular of the smaller figure of reference will be the altitudes; and they when multiplied respectively by the perpendicular and the base will be the segments of the base. Other segments will be the difference of these and the base. Half the product of the diagonals (of the required figure) will be the area.”¹

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¹ *GSS*, vii. 103½.

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If the rectangle generated from *m*, *n* be smaller than that from *p*, *q*, then, according to Mahâvîra, we obtain the rational inscribed quadrilateral of which the sides are $(m² - n²)(p² + q²)(m² + n²), (p² - q²)(m² + n²)²,$ $2mn(p² + q²)(m² + n²), 2pq(m² + n²)² ;$ whose diagonals are $2pq(m² - n²) + 2mn(p² - q²)(m² + n²),$ $(m² - n²)(p² - q²) + 4mnpq(m² + n²);$ whose altitudes are $2pq(m² - n²) + 2mn(p² - q²) 2mn,$ $(m² - n²)(p² - q²) + 4mnpq (m² - n²);$ whose segments are $2pq(m² - n²) + 2mn(p² - q²) (m² - n²),$ $(m² - n²)(p² - q²) + 4mnpq 2mn;$ and whose area is $½ 2pq(m² - n²) + 2mn(p² - q²) (m² - n²)(p² - q²)$ $+ 4mnpq (m² + n²)².$

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Śrîpati writes :

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“Of the two right triangles the sides and uprights are reciprocally multiplied by the hypotenuses; of the products the greatest is the base, the smallest is the face and the rest are the two flank sides of a quadrilateral with unequal sides.”∗1

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∗1 *SiŚe*, xiii. 42.

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Bhâskara II gives the rule :

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“The sides and uprights of two optional right triangles being multiplied by their reciprocal hypotenuses become the sides : in this way has been derived a quadrilateral of unequal sides. There the two diagonals can be obtained from those two triangles. The product of the uprights, added with the product of the sides, gives one diagonal; the sum of the reciprocal products of the uprights and sides is the other.”∗2

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∗2 *L*, p. 51.

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Bhâskara II³ illustrates by taking the right triangles (3, 4, 5) and (5, 12, 13) so that the resulting cyclic quadrilateral is (25, 39, 60, 52). The same example was given before by Mahâvîra¹ and Pṛthûdakasvâmî.² This cyclic quadrilateral also appears in the *Triśatikâ* of Śrîdhara³ and in the commentary of the *Āryabhaṭîya* by Bhâskara I (522). The diagonals of this quadrilateral are, states Bhâskara II, 56 (= 3.12 + 4.5) and 63 (= 4.12 + 3.5) (Fig. 7). He then observes:

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³ *L*, p. 52.

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¹ *GSS*, vii. 104½.

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² *BrSpSi*, xii. 38 (*Com.*).

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³ *Triś*, Ex. 80.

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"If the figure be formed by changing the arrangement of the face and flank then the second diagonal will be equal to the product of the hypotenuses of the two right triangles (of reference), *i.e.*, 65." (Fig. 8).

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Fig. 7

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Fig. 8

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By taking the right triangles (3, 4, 5) and (15, 8, 17) Bhâskara II gets another cyclic quadrilateral (68, 51, 40, 75), whose diagonals are (77, 85), altitude is 308/5, segments are 144/5 and 231/5, and area is 3234.⁴ (Fig. 9). With the sequence of the sides (68, 40, 51, 75) the diagonals are (77, 84) (Fig. 10), and with (68, 40, 75, 51) they are (84, 85). (Fig. 11).

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⁴ *L*, pp. 46ff.

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Fig. 9

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Fig. 10

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Fig. 11

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The deep significance of Brahmagupta's results has been demonstrated by Chasles¹ and Kummer.²

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¹ M. Chasles, *Aperçu historique sur l'origine et development des méthodes en géométrie*, Paris, 1875, pp. 436ff.

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² E. E. Kummer, "Über die Vierecke, deren Seiten und Diogonalen rational sind," *Journ. für Math.*, XXXVII, 1848, pp. 1-20.

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In fact, according to the sequence in which the quantities (A) are taken, there will be two varieties of Brahmagupta quadrilaterals having them as their sides : (1) one in which the two diagonals intersect at right angles and (2) the other in which the diagonals intersect obliquely. The arrangement (A) gives a quadrilateral of the first variety. For the oblique variety, the sides are in the following order : $.(p² - q²)(m² + n²), & (m² - n²)(p² + q²), / 2mn(p² + q²), & 2pq(m² + n²) ; (B)$ $or .(p² - q²)(m² + n²), & 2mn(p² + q²), / (m² - n²)(p² + q²), & 2pq(m² + n²). (C)$

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Bhâskara II points out that the diagonals of the Brahmagupta quadrilateral are in the (A) variety, $2pq(m² - n²) + 2mn(p² - q²), 4mnpq + (p² - q²)(m² - n²) ;$ in (B), $2pq(m² - n²) + 2mn(p² - q²), (p² + q²)(m² + n²) ;$ and in (C), $4mnpq + (p² - q²)(m² - n²), (p² + q²)(m² + n²).$ The diameter of the circumscribed circle in every case is (p² + q²)(m² + n²).

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Gaṇeśa (1545) shows that the quadrilateral is formed by the juxtaposition of four right triangles obtained by multiplying the sides of each of two rational right triangles by the upright and base of the other. He writes :

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“A quadrilateral is divided into four triangles by its intersecting diagonals. So by the juxtaposition of four triangles a quadrilateral will be formed. For that purpose the four triangles are assumed in this manner : Take two right triangles formed in the way indicated

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Compare also L. E. Dickson, “Rational Triangles and Quadrilaterals,” *Amer. Math. Mon.*, XXVIII, 1921, pp. 244-250.

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before. If the upright, base and hypotenuse of a rational right triangle be multiplied by any arbitrary rational number, there will be produced another right triangle with rational sides. Hence on multiplying the sides of each of the two right triangles by an optional number equal to the base of the other and again by an optional nnmber equal to the upright of the other, four right triangles will be obtained by the judicious juxtaposition of which the quadrilateral will be formed."

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He then shows with the help of specific examples (see Figs. 12, 13 & 14) that we can obtain in this way from the same set of two rational right triangles two varieties of rational convex quadrilaterals: One in which the diagonals intersect each other perpendicularly; and the other in which they do so obliquely.

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Fig. 12

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Fig. 13

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Fig. 14

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