History of Hindu Mathematics: A Source Book v2.c17.s18.t2: Problems Involving Sides but not Areas
History of Hindu Mathematics: A Source Book v2.c17.s18.t2
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Problems Involving Sides but not Areas
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Mahâvîra also obtained right triangles whose sides multiplied by arbitrary rational numbers have a given sum. Algebraically, the problems require the solution of $.x² + y² = z², / rx + sy + tz = A;$ where r, s, t, A are known rational numbers. His method of solution is the same as that described above. Starting with the general solution of $x'² + y'² = z'²$ we are asked to calculate the value $rx' + sy' + tz' = P, say.$ Then, says Mahâvîra, the required solution is $x = x'A/P, y = y'A/P, z = z'A/P.$
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One illustrative problem given by Mahâvîra is :
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"The perimeter of a rectangle is unity. Tell me quickly, after calculating, what are its base and upright."¹
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¹ *GSS*, vii. 118½.
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Starting with the rectangle m² - n², 2mn, m² + n², we have in this case P = 2(m² - n² + 2mn). Hence all rectangles having the same perimeter unity will be given by $(m² - n²)/(2(m² - n² + 2mn)), mn/(m² - n² + 2mn),$ m, n being any rational numbers.
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The isoperimetric right triangles will be given by $((m-n)/2m)p, np/(m+n), (m²+n²)/(2m(m+n))p ;$ where p is the given perimeter.
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Another example is :
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"(Find) a rectangle in which twice the diagonal, thrice the base, four times the upright and the perimeter together equal unity."²
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² *GSS*, vii. 119½.
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**Pairs of Rectangles.** Mahâvîra found "pairs of rectangles such that
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(*i*) their perimeters are equal but the area of one is double that of the other, or
(*ii*) their areas are equal but the perimeter of one is double that of the other, or
(*iii*) the perimeter of one is double that of the other and the area of the latter is double that of the former."
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These are particular cases of the following general problem contemplated in his rule:
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To find (*x*, *y*) and (*u*, *v*) representing the base and upright respectively of two rectangles which are related, such that $.2m(x + y) &= 2n(u + v), / pxy &= quv; (A)$ where *m*, *n*, *p*, *q* are known integers.
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His rule for the solution of this general problem is :
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"Divide the greater multiples of the area and the perimeter by the (respective) smaller ones. The square of the product of these ratios multiplied by an optional number is the upright of one rectangle. That diminished by unity will be its base, when the areas are equal. Otherwise, multiply the bigger ratio of the areas by that optional number and subtract unity; three times the upright diminished by this (difference) will be the base. The upright and base of the other rectangle should be obtained from its area and perimeter (thus determined) with the help of the rule, 'From the square of half the perimeter, etc.,' described before."¹
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¹ *GSS*, vii. 131½-133. The reference in the concluding line is to rule vii. 129½.
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In other words, to solve (A), assume $y = s (ratio of perimeters)(ratio of areas)², (1)$ and x = y - 1, if p = q, \hfill (2) or x = 3 [y - s (ratio of areas) - 1], if p ≠ q, \hfill (2') where s is an arbitrary number, and the ratios are to be so presented as always to remain greater than or equal to unity.
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Let m n, q p. Then we shall have to assume .y &= s (m² q²)/(n² p²), / x &= 3 (s (m² q²)/(n² p²) - s q/p + 1).} (3)
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Substituting these values in (A), we get $u + v = m/n (4 s (m² q²)/(n² p²) - 3 s q/p + 3), (4)$ $u v = 3 s (m² q)/(n² p) (s (m² q²)/(n² p²) - s q/p + 1).$
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Then $(u - v)² = (m²)/(n²) (4 s (m² q²)/(n² p²) - (9 s q)/(2 p) + 3)² + (3 s q)/(4 p) ((s q)/p - 4).$
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Now, if the arbitrary multiplier s be chosen such that $(s q)/p = 4, (5)$ we have $u - v = m/n (4 s (m² q²)/(n² p²) - (9 s q)/(2 p) + 3). (6)$
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From (4) and (6) we get .u &= m/n (4 s (m² q²)/(n² p²) - (15 s q)/(4 p) + 3), / v &= (3 s m q)/(4 n p).} (7)
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Substituting the value of s from (5) in (3) and (7) we have finally the solution of (A), when m n, q p, as
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y &= 4(m²q)/(n²p), & v &= 3m/n, \\ x &= 3(4(m²q)/(n²p) - 3), & u &= 4m/n(4(m²q)/(n²p) - 3). \end{aligned}} (I)
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Mahâvîra has observed that “when the areas are equal” we are to assume¹ y &= s(m²)/(n²), / x &= s(m²)/(n²) - 1.
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¹ Bibhutibhusan Datta has shown that this restriction is *not necessary*. In fact, starting with the assumption .y &= s(m²q²)/(n²p²), / x &= s(m²q²)/(n²p²) - 1;} m \geqslant n, q \geqslant p, and proceeding in the same way as above, he has obtained another solution of (A) in the form .y &= 2(m²q)/(n²p), & v &= m/n, / x &= 2(m²q)/(n²p) - 1, & u &= m/q(4(m²q)/(n²p) - 2).} (II) Datta finds two general solutions of (A), *viz.* .y &= (rm²q²)/(n²p²) + t, / x &= (rm²q²)/(tn²p²)((rm²q²)/(n²p²) - rq/p + t), / v &= rmq/np, / u &= m/nt((rm²q²)/(n²p²) + t)((rm²q²)/(n²p²) - rq/p + t);} (III)
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**Isosceles Triangles with Integral sides.** Brahmagupta says :
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“The sum of the squares of two unequal numbers is the side ; their product multiplied by two is the altitude, and twice the difference of the squares of those two unequal numbers is the base of an isosceles triangle.”
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*BrSpSi*, xii. 33.
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Mahâvîra gives the following rule for obtaining an isosceles triangle from a single generated rectangle:
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“In the isosceles triangle (required), the two diagonals (of a generated rectangle) are the two sides, twice its side is the base, the upright is the altitude, and the area (of the generated rectangle) is the area.”
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A rectangle generated from the numbers *m* and *n* has its sides equal to m² - n² and 2mn and its diagonal equal to m² + n². *Cf.* pp. 208-9.
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*GSS*, vii. 108½.
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Thus if *m*, *n* be two integers such that *m* ≠ *n*, the sides of all rational isosceles triangles with integral sides are :
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$(i) m² + n², m² + n², 2(m² - n²) ;$ or (ii) m² + n², m² + n², 4mn.$
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and
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.y &= (rm²q²)/(n²p²) - t, & v &= m/n(rq/p - t), / x &= (rm²q²)/(n²p²)(rq/tp - 1), & u &= rmq/np((rm²q²)/(tn²p²) - 1)} (IV)
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where m n, q p and *r*, *t* are any two integers.
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See Datta, “On Mahâvîra’s solution of rational triangles and quadrilaterals,” *BCMS*, XX, 1928-9, pp. 267-294; particularly p. 285.
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The altitude of the former is 2mn and of the latter m² - n² and the area in either case is the same, that is, 2mn(m² - n²).
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**Juxtaposition of Right Triangles.** It will be noticed that the device employed by Brahmagupta and Mahâvîra to find the above solutions is to juxtapose two rational right triangles—equal in this case—so as to have a common leg. It is indeed a very powerful device. For, every rational triangle or quadrilateral may be formed by the juxtaposition of two or four rational right triangles. So, in order to construct such rational figures, it suffices to know only the complete solution of x² + y² = z² in integers. The beginning of this principle is found as early as the *Baudhâyana Śulba*¹ (800 B.C.) wherein is described the formation of a kind of brick, called *ubhayî* (born of two), by the juxtaposition of the eighths of two suitable rectangular bricks of the same breadth (and thickness) but of different lengths.
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¹ *BŚl*, iii. 122; Compare Datta, *Śulba*, p. 45, where necessary figures are given.
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**Isosceles Triangles with a Given Altitude.** Brahmagupta gives a rule to find all rational isosceles triangles having the same altitude. He says :
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“The (given) altitude is the producer (*karaṇî*). Its square divided by an optional number is increased and diminished by that optional number. The smaller is the base and half the greater is the side.”²
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That is to say, the sides and bases of rational isosceles triangles having the same altitude a are respectively, $½((a²)/m + m), ½((a²)/m + m) and ((a² - m)/m),$ where m is any rational number.
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² *BrSpSi*, xviii. 37.
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In particular, let the given altitude be 8. Then taking m = 4 Pṛthûdakasvâmî (860) obtains the rational isosceles triangle (10, 10, 12).
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**Pairs of Rational Isosceles Triangles.** Mahâvîra gives the following rule for finding two isosceles triangles whose perimeters, as also their areas, are related in given proportions :
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“Multiply the square of the ratio-numbers of the perimeters by the ratio-numbers of the areas mutually and then divide the larger product by the smaller. Multiply the quotient by 6 and 2 (severally) and then diminish the smaller by unity : again (find severally) the difference between the results, and twice the smaller one : these are the two sets of elements for the figures to be generated. From them the sides, etc., can be obtained in the way described before.”¹
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¹ *GSS*, vii. 137.
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If (s₁, s₂) and (Δ₁, Δ₂) denote the perimeters and areas of two rational isosceles triangles, such that $s₁ : s₂ = m : n, Δ₁ : Δ₂ = p : q, (1)$ where the ratio-numbers m, n, p, q are known integers, then the triangles will be obtained, says Mahâvîra, from the rectangles *generated from* $(6(n²p)/(m²q), 2(n²p)/(m²q) - 1) and (4(n²p)/(m²q) + 1, 4(n²p)/(m²q) - 2),$ where n²p > m²q, when the dimensions of the first are multiplied by m and those of the second by n.
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The dimensions of the isosceles triangle formed from the first set of *bîja* are : $side = m (6(n²p)/(m²q))² + (2(n²p)/(m²q) - 1)²,$
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$base = 24m (n²p)/(m²q)(2 (n²p)/(m²q) - 1),$ $altitude = m(6 (n²p)/(m²q))² - (2 (n²p)/(m²q) - 1)²;$ and from the second set $side = n(4 (n²p)/(m²q) + 1)² + (4 (n²p)/(m²q) - 2)²,$ $base = 4n(4 (n²p)/(m²q) + 1)(4 (n²p)/(m²q) - 2),$ $altitude = n(4 (n²p)/(m²q) + 1)² - (4 (n²p)/(m²q) - 2)².$
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It can be easily verified that the perimeters and areas of the isosceles triangles thus obtained satisfy the conditions (1).
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In particular, putting m = n = p = q = 1, we have two isosceles triangles of sides, bases and altitudes (29, 40, 21) and (37, 24, 35) which have equal perimeters (98) and equal areas (420). This particular case was treated by Frans van Schooten the Younger (1657), J. H. Rahn (1697) and others.¹
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¹ Dickson, *Numbers*, II, p. 201.
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It is evident that multiplying the above values by m⁴q² we get pairs of isosceles triangles whose dimensions are integral.
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**Rational Scalene Triangles.** Brahmagupta says:
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“The square of an optional number is divided twice by two arbitrary numbers; the moieties of the sums of the quotients and (respective) optional numbers are the sides of a scalene triangle; the sum of the moieties of the differences is the base.”²
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² *BrSpSi*, xii. 34.
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That is to say, the sides of a rational scalene triangle are $½((m²)/p+p),½((m²)/q+q),½((m²)/p-p)+½((m²)/q-q)$ where m, p, q are any rational numbers. The altitude (m), area and segments of the base of this triangle are all rational.
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Mahāvîra gives the rule :
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“Half the base of a derived rectangle is divided by any optional number. With this divisor and the quotient is obtained another rectangle. The sum of the uprights (of these two rectangles) will be the base of the scalene triangle, the two diagonals, its sides and the base (of either rectangle) its altitude.”
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If m, n be any two rational numbers, the rational rectangle (AB′ BH)
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Fig. 3
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Fig. 4
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formed from them is $m²-n², 2mn, m²+n².$
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If p, q be any two rational factors of mn, that is, if mn=pq, the second rectangle (AC′ CH) is $p²-q², 2pq, p²+q².$
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¹ *GSS*, vii. 110½.
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Now, juxtaposing these two rectangles so that they do not overlap (Fig. 3), the sides of the rational scalene triangle are obtained as $p² + q², m² + n², (p² - q²) + (m² - n²),$ where mn = pq. Evidently the two rectangles can be juxtaposed so as to overlap (Fig. 4). So the general solution will be $p² + q², m² + n², (p² - q²) ± (m² - n²).$ The altitude of the rational scalene triangle thus obtained is 2mn or 2pq, its area pq(p² - q²) ± mn(m² - n²) and the segments of the base are p² - q² and m² - n².
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In particular, putting m = 12, p = 6, q = 8 in Brahmagupta's general solution, Pṛthûdakasvâmî derives a scalene triangle of which the sides (13, 15), base (14), altitude (12), area (84) and the segments of the base (5, 9) are all integral numbers.
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In order to get the above solutions of the rational scalene triangle the method employed was, it will be noticed, the juxtaposition of two rational right triangles so as to have a common leg. In Europe, it is found to have been employed first by Bachet (1621). The credit for the discovery of this method of finding rational scalene triangles should rightly go to Brahmagupta (628), but not to Bachet as is supposed by Dickson.¹
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¹ Dickson, *Numbers*, II, p. 192.
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**Triangles having a Given Area.** Mahâvîra proposes to find all triangles having the same given area A. His rules are :
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“Divide the square of four times the given area by three; The quotient is the square of the square of a side of the equilateral triangle.”\∗2
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\∗2 *GSS*, vii. 154½.
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"Divide the given area by an optional number; the square-root of the sum of the squares of the quotient and the optional number is a side of the isosceles triangle formed. Twice the optional number is its base and the area divided by the optional number is the altitude."¹
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¹ *GSS*, vii. 156½.
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"The cube of the square-root of the sum of eight times the given area and the square of an optional number is divided by the product of the optional number and that square-root; the quotient is diminished by half the optional number which is the base (of the required triangle). The *saṅkramaṇa* between this remainder and the quotient of the square of the optional number divided by twice that square-root will give the two sides."²
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The last rule has been re-stated differently.³
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³ *GSS*, vii. 160½-161½.
Linked texts (3)
- Gaṇitasārasaṅgraha — Mahāvīra (mathematician), Sanskrit
- Baudhāyanaśulbasūtra — Baudhāyana, Sanskrit
- Brāhmasphuṭasiddhānta — Brahmagupta, Sanskrit