Chapter 8

Use ?ryabhata's cube root algorithm to find the cube root of \(13,312,053\).

This is the method presented in the text for finding a circle whose area is equal to a given square: In square \(A B C D\), let \(M\) be the intersection of the diagonals (see Fig. 8.5). Draw the circle with \(M\) as center and \(M A\) as radius; let \(M E\) be the radius of the circle perpendicular to the side \(A D\) and cutting \(A D\) in \(G\). Let \(G N=\frac{1}{3} G E\). Then \(M N\) is the radius of the desired circle. Show that if \(A B=s\) and \(M N=r\), then \(\frac{r}{s}=\frac{2+\sqrt{2}}{6}\). Show that this implies a value for \(\pi\) equal to \(3.088311755\).

The Sulbas?tra method of "squaring a circle" of diameter \(d\) takes the side of the desired square to be \(\frac{7}{8}+\frac{1}{8 \times 29}-\) \(\frac{1}{8 \times 29 \times 6}+\frac{1}{8 \times 29 \times 6 \times 8}\) times \(d\). Show that this is equivalent to using a value for \(\pi\) equal to \(3.088326491\).

Brahmagupta asserts that if \(A B C D\) is a quadrilateral inscribed in a circle, with side lengths \(a, b, c, d\) (in cyclic order) (see Fig. 8.8), then the lengths of the diagonals \(A C\) and \(B D\) are given by $$ A C=\sqrt{\frac{(a c+b d)(a d+b c)}{a b+c d}} $$ and similarly $$ B D=\sqrt{\frac{(a c+b d)(a b+c d)}{a d+b c}} $$ Prove this result as follows: a. Let \(\angle A B C=\theta\). Then \(\angle A D C=\pi-\theta\). Let \(x=A C\). Use the law of cosines on each of triangles \(A B C\) and \(A D C\) to express \(x^{2}\) two different ways. Then, since \(\cos (\pi-\theta)=-\cos \theta\), use these two formulas for \(x^{2}\) to determine \(\cos \theta\) as a function of \(a, b, c\), and \(d\). b. Replace \(\cos \theta\) in your expression for \(x^{2}\) in terms of \(a\) and \(b\) by the value for the cosine determined in part \(\mathrm{a}\). c. Show that \(c d\left(a^{2}+b^{2}\right)+a b\left(c^{2}+d^{2}\right)=(a c+b d)(a d+b c)\). d. Simplify the expression for \(x^{2}\) found in part \(\mathrm{b}\) by using the algebraic identity found in part c. By then taking square roots, you should get the desired expression for \(x=A C\). (Of course, a similar argument will then give you the expression for \(y=B D\).)

Brahmagupta asserts that if \(A B C D\) is a quadrilateral inscribed in a circle, as in Exercise 7 , then if \(s=\frac{1}{2}(a+\) \(b+c+d\) ), the area of the quadrilateral is given by \(S=\) \(\sqrt{(s-a)(s-b)(s-c)(s-d)}\) (Fig. 8.12). Prove this result as follows: Area of a quadrilateral inscribed in a circle a. In triangle \(A B C\), drop a perpendicular from \(B\) to point \(E\) on \(A C\). Use the law of cosines applied to that triangle to show that \(b^{2}-a^{2}=x(x-2 A E)\). b. Let \(M\) be the midpoint of \(A C\), so \(x=2 A M\). Use the result of part a to show that \(E M=\left(b^{2}-a^{2}\right) / 2 x\). c. In triangle \(A D C\), drop a perpendicular from \(D\) to point \(F\) on \(A C\). Use arguments similar to those in parts a and \(\mathrm{b}\) to show that \(F M=\left(d^{2}-c^{2}\right) / 2 x\). d. Denote the area of quadrilateral \(A B C D\) by \(P\). Show that \(P=\frac{1}{2} x(B E+D F)\) and therefore that \(P^{2}=\frac{1}{4} x^{2}(B E+\) \(D F)^{2} .\) e. Extend \(B E\) to \(K\) such that \(\angle B K D\) is a right angle, and complete the right triangle \(B K D\). Then \(B E+D F=\) \(B K\). Substitute this value in your expression from part d; then use the Pythagorean Theorem to conclude that \(P^{2}=\) \(\frac{1}{4} x^{2}\left(y^{2}-E F^{2}\right)\) f. Since \(E F=E M+F M\), conclude that \(E F=\left[\left(b^{2}+\right.\right.\) \(\left.\left.d^{2}\right)-\left(a^{2}+c^{2}\right)\right] / 2 x\). Substitute this value into the expression for \(P^{2}\) found in part e, along with the values for \(x^{2}\) and \(y^{2}\) found in Exercise 7. Conclude that $$ \begin{aligned} P^{2} &=\frac{1}{4}(a c+b d)^{2}-\frac{1}{16}\left[\left(b^{2}+d^{2}\right)-\left(a^{2}+c^{2}\right)\right]^{2} \\\ &=\frac{1}{16}\left(4(a c+b d)^{2}-\left[\left(b^{2}+d^{2}\right)-\left(a^{2}+c^{2}\right)\right]^{2}\right) \end{aligned} $$ g. Since \(s=\frac{1}{2}(a+b+c+d)\), show that \(s-a=\) \(\frac{1}{2}(b+c+d-a), s-b=\frac{1}{2}(a+c+d-b), s-c=\) \(\frac{1}{2}(a+b+d-c)\), and \(s-d=\frac{1}{2}(a+b+c-d)\) h. To prove the theorem, it is necessary to show that the final expression for \(P^{2}\) given in part \(\mathrm{f}\) is equal to the product of the four expressions in part \(\mathrm{g}\). It is clear that the denominators are both equal to 16 . To prove that the numerators are equal involves a lot of algebraic manipulation. Work carefully and show that the two numerators are in fact equal.

Solve the following problem from the Bakhsh?li manuscript: One person goes 5 yojanas a day. When he has proceeded for seven days, the second person, whose speed is 9 yojanas a day, departs. In how many days will the second person overtake the first?

Solve the following problem from Mah?vira: There are 4 pipes leading into a well. Among these, each fills the well (in order) in \(1 / 2,1 / 3,1 / 4\), and \(1 / 5\) of a day. In how much of a day will all of them together fill the well and each of them to what extent?

Another problem from Mah?v?ra: Of a collection of mango fruits, the king took \(1 / 6\); the queen took \(1 / 5\) of the remainder, and three chief princes took \(1 / 4,1 / 3,1 / 2\) of what remained at each step; and the youngest child took the remaining three mangoes. O you, who are clever in working miscellaneous problems on fractions, give out the measure of that collection of mangoes.

Another problem from Mah?vira: One-third of a herd of elephants and three times the square root of the remaining part of the herd were seen on a mountain slope; and in a lake was seen a male elephant along with three female elephants constituting the ultimate remainder. How many were the elephants here?

Another problem from Mah?v?ra: If 3 peacocks cost 2 coins, 4 pigeons cost 3 coins, 5 swans cost 4 coins, and 6 s?rasa birds cost 5 coins, and if you buy 72 birds for 56 coins, how many of each type of bird do you have?

This problem is from Brahmagupta's work on congruences. Given that the sun makes 30 revolutions through the ecliptic in 10,960 days, how many days have elapsed (since the sun was at a given starting point) if the sun has made an integral number of revolutions plus \(8080 / 10,960\) of a revolution, that is, "when the remainder of solar revolutions is 8080 ." If \(y\) is the number of days sought and \(x\) is the number of revolutions, then, because 30 revolutions take 10,960 days, \(x\) revolutions take \((1096 / 3) x\) days. Therefore, \(y=(x+808 / 1096)(1096 / 3)\), or \(1096 x+808=3 y\). Thus, solve \(N \equiv 808(\) mod 1096) and \(N \equiv 0(\bmod 3)\).

Solve the congruence \(N \equiv 23(\bmod 137), N \equiv 0(\bmod 60)\) using Brahmagupta's procedure.

Solve \(1096 x+1=3 y\) using Brahmagupta's method. Given a solution to this equation (with "additive" 1 ), it is easy to find solutions to equations with other additives by simply multiplying. For example, solve \(1096 x+10=3 y\).

Prove that Brahmagupta's procedure does give a solution to the simultaneous congruences. Begin by noting that the Euclidean algorithm allows one to express the greatest common divisor of two positive integers as a linear combination of these integers. Note further that a condition for the solution procedure to exist is that this greatest common divisor must divide the "additive." Brahmagupta does not mention this, but Bh?skara and others do.

Solve the problem \(N \equiv 5(\bmod 6) \equiv 4(\bmod 5) \equiv\) \(3(\bmod 4)) \equiv 2(\bmod 3))\) by the Indian procedure and by the Chinese procedure. Compare the methods.

Solve the indeterminate equation \(17 n-1=75 m\) by both the Indian and Chinese methods explicitly using the Euclidean algorithm. Compare the solutions.

Prove that \(D\left(u_{0} v_{1}+u_{1} v_{0}\right)^{2}+c_{0} c_{1}=\left(D u_{0} u_{1}+v_{0} v_{1}\right)^{2}\) given that \(D u_{0}^{2}+c_{0}=v_{0}^{2}\) and \(D u_{1}^{2}+c_{1}=v_{1}^{2}\)

Solve \(83 x^{2}+1=y^{2}\) by Brahmagupta's method. Begin by noting that \((1,9)\) is a solution for subtractive 2 .

Show that if \((u, v)\) is a solution to \(D x^{2}-4=y^{2}\), then \(\left(u_{1}, v_{1}\right)=\left(\frac{1}{2} u v\left(v^{2}+1\right)\left(v^{2}+3\right),\left(v^{2}+2\right)\left[\frac{1}{2}\left(v^{2}+1\right)\left(v^{2}+\right)\right.\right.\) 3) \(-1]\) ) is a solution to \(D x^{2}+1=y^{2}\) and that both \(u_{1}\) and \(v_{1}\) are integers regardless of the parity of \(u\) or \(v\).

Show that if \((u, v)\) is a solution to \(D x^{2}+2=y^{2}\), then \(\left(u_{1}, v_{1}\right)=\left(u v, v^{2}-1\right)\) is a solution to \(D x^{2}+1=y^{2} .\) Deduce a similar rule if \((u, v)\) is a solution to \(D x^{2}-2=y^{2}\).

Solve \(61 x^{2}+1=y^{2}\) by Bh?skara's process. The solution is \(x=226,153,980, y=1,766,319,049\).

A combinatorics problem from Bh?skara: In a pleasant, spacious, and elegant edifice, with eight doors, constructed by a skillful architect as a palace for the lord of the land, tell me the combinations of apertures taken one, two, three, and so on, at a time.

Show that Bh?skara's algebraic formula for the Sine can be rewritten as an approximation formula for the modern sine in the form $$ \sin x \approx \frac{16 x(\pi-x)}{5 \pi^{2}-4 x(\pi-x)} $$ where \(x\) is given in radians. Graph this function on a graphing calculator from 0 to \(\pi\) and compare it with the graph of \(\sin x\) on that interval.

Use both the interpolation scheme of Brahmagupta and the algebraic formula of Bh?skara I to approximate \(\sin \left(16^{\circ}\right)\). Compare the two values to each other and to the exact value. What are the respective errors?

Why would the Indians have thought it better to use an algebraic approximation to the sine function rather than calculate values using geometric methods and methods of interpolation?