More Classroom Activities Based on Ancient Indian Rope Geometry - Transforming a square into a circle

Cynthia J. Huffman and Scott V. Thuong (Pittsburg State University)

This applet demonstrates the procedure found in Section 2.9 of the Śulba-sūtra of Baudhāyana (BSS 2.9) for transforming a square into a circle of equal area. The construction actually produces a circle (shown in blue) with area approximately equal to that of the square. The circle with area exactly equal to that of the square is shown in pink. The actual construction of the circle with area exactly equal to that of the square is impossible using only straightedge and compass, a consequence of the transcendental nature of \(\pi.\) Click “Go” to advance through the construction and “Reset” when the construction is completed.

Finally, we now turn our attention to the method that Baudhāyana gave in BSS 2.9 to construct a circle with area approximately equal to that of a given square. The translated original source is as follows [Sen and Bag, p. 4]:

If it is desired to transform a square into a circle, (a cord of length) half the diagonal (of the square) is stretched from the center to the east (a part of it lying outside the eastern side of the square); with one-third (of the part lying outside) added to the remainder (of the half diagonal), the (required) circle is drawn.

This method only gives an approximate answer. In the applet, consider square \(ABCD,\) the dimensions of which can be adjusted by sliding point \(A.\) First a rope is extended from the center of the square, \(O,\) to the corner \(B.\) The length of this rope is of course half the diagonal of the square. Keeping one end of the rope attached to \(O,\) pulling the other end southward traces out arc \(BEA.\) Let \(E\) be the point of the arc that points toward the east, and let \(G\) be the intersection with the side \(AB\) of the square.

As suggested by the translation, next, point \(P\) is marked so that \(GP\) is one-third of \(EG:\) \[GP=\frac{1}{3}EG.\]
In the next step, we define point \(F\) to be a point on segment \(EP.\) Slide point \(F\) to observe how the area of the blue circle with radius \(OF\) changes. When point \(F\) reaches point \(P,\) the area of the blue circle will have area only approximately equal to that of \(ABCD.\)

Point \(Z\) is marked so that a circle of radius \(OZ\) has area exactly equal to that of \(ABCD.\) The reader may observe that the blue approximate circle is just slightly bigger than the pink exact circle. We now calculate just how good of an approximation Baudhāyana gave.

Let \(L\) denote half the side length of \(ABCD.\) Thus the length of \(OG\) is equal to \(L.\) The length of \(BO\) (and hence that of \(EO\)) is \({\sqrt{2}}L.\) Hence the radius that Baudhāyana gave is \[OP=OG+GP=OG+\frac{1}{3}EG=L+\frac{1}{3}\left({\sqrt{2}}L-L\right)=\frac{2+\sqrt{2}}{3}L.\]

To find the exact radius, note that the area of \(ABCD\) is \((2L)^2=4L^2.\) Solving \({\pi}r^2=4L^2\) yields an exact radius of \[r=\frac{2}{\sqrt{\pi}}L.\]

Comparing the coefficients on \(L,\) we see that \[\frac{2+\sqrt{2}}{3}=1.138071\dots\quad{\rm{while}}\quad\frac{2}{\sqrt{\pi}}=1.128379\dots.\]

This amounts to a relative error of less than 1%. Note that we can also infer an approximation of \(\pi\) from this calculation. We solve for \(\pi\) in \[\frac{2+\sqrt{2}}{3}\approx\frac{2}{\sqrt{\pi}}\] to obtain \[\pi\approx\left(\frac{6}{2+\sqrt{2}}\right)^2.\]

Baudhāyana derived a rational estimation of \(\sqrt{2}\) as \[\frac{577}{408}=1.414215686\dots\] in Section 2.12 of BSS. (See "Ancient Indian Rope Geometry in the Classroom" [Huffman and Thuong] for an exposition.) Thus we obtain the rational approximation of \(\pi:\) \[\pi\approx\left(\frac{6}{2+\frac{577}{408}}\right)^2=\left(\frac{2448}{1393}\right)^2=3.0883\dots.\]