**An ancient secret? Squaring a circle with peg and rope.**

**(Inspired by Archimedes' essay, ***Measurement of a circle*)

*Measurement of a circle*)

NOTE The problem of squaring the circle might be one of the oldest. It dates back to at least 414 BCE.

*From Archimedes' triangulation of a circle, we now proceed to a square! From an idea of Abraham bar Ḥiyya ha-Nasi, imagine cutting a slice of onion from the top to the middle and all the layers flop onto the table. You get Archimedes' triangle in another form. The height is 1 and the base is the circumference of the circle.*

*Now cut down the onion from top to bottom and rearrange to make the a rectangle below. This rectangle has the same area as the circle. As the height remains 1, the base, half the circumference, gives the numerical area of the circle. (If you know the formula for the area of a unit circle, you will understand why... It's as easy as pi.)*

*Enough about onions. and cutting circles. Let's head to the beach and imagine the following. *

**Step 1 ** A rope of Unit length (1) is pegged at an end and rotated around O to draw a pink circle.

**Step 2** A red rope is laid on half the circumference and cut to present a red (lined) semi-circle.

**Step 3** The red rope the length of the semi-circle is straightened and extended left from O to S.

**NOTE** The distance of the line SO equals both the area of the circle and the required square. So we need to determine the square root of SO and this will be the side of our desired square.

**Step 4** Half the rope length of SA then draws the Green dashed circle around C.

**Step 5** A perpendicular (black dashed) line is drawn from O, meeting the Green circle at P.

**Step 6** The line from O to P is a side our our desired square.

**Step 7** The line from O to P draws the required circles to form the blue square OPMN ** which has the same area as the initial pink unit circle**.

It's that simple! The more detailed diagram that follows is discussed in the notes below and further below..

Today the area of a circle is given by the formula A = π r ²** **

When r = 1 the area of a circle is π.

So we need to create a square with area π.

So the sides of the square must be the square root of π.

And that is what is created with the line segment OP, via Euclid.

Thus OP is the mean proportional or geometric mean between π and one.

π : OP :: OP : 1

The product of the extremes (outer terms) equals the product of the mean (inner) terms.

So π × 1 = OP × OP meaning π = OP² and that √π = OP

The string/rope on the semi-circle gets around the problem of a compass and straight edge alone not being able to construct transcendental numbers such as π.

*The voodoo magic done here was to work out how to convert the circle into a rectangle. Bar Hiyya gave us the triangle, yet nobody it seems, took the next step of converting the triangle into a rectangle. This is a critical step, because Euclid gave a proof and construction for converting any rectangle into a square. This appears in Book II Proposition 14. Yet the sneaky part was to create the rectangle with a unit side! That meant I could use Descartes' technique (adopted from Euclid) and find the square root of the length of string on the semi-circle to get OP.*

**NOTES:**

**1**. The 10⁻¹⁵ or one quadrillionth rounding error in area calculation I got when I created these images at geogebra.org is due to a 15 decimal place restriction. Given we accept limits and the reals, the concept behind the construction is exact. No it can't be drawn to perfection with actual rope and pegs, yet neither can any Euclidean diagram!

**2**. The method of constructing the first side OP of the square OPMN is in Descartes' 1637 La Géométrie, via Euclid's Elements, (300 BCE) Book VI Prop. 13, “To two given lines to find a mean proportional.” Thus SO is to OP as OP is to 1 and we write this as SO : OP :: OP : 1 . (The yellow triangles SOP and POA are similar, so their side lengths are proportional.) In the proportion the product of the extreme (outer) terms = the product of the mean (inner) terms, SO × 1 = OP × OP. Thus SO = OP² and √SO = OP. Thus OP is the side of the square with the same area as the area of the given circle.

**3**. Squaring the circle to perfection cannot be done with compass and straight edge. Yet before the compass, in both India and Egypt, circles were drawn via 'peg and rope/cord' methods. As the construction shown does NOT appear in the Indian Śulbasūtras (Rules of the cord, 600 BCE) which circled the square, it may be assumed the general method of finding a mean proportional had not emerged, either in India, (which made use of the Pythagorean theorem before Pythagoras) or in Greece. Reversing the Indian method of circling the square is entirely different to this construction. Similarly, the Indian method of converting a rectangle into a square is different to Euclid's.

**4**. Irrational numbers can be constructed with compass and straight edge, such as √2, which is the hypotenuse of a square with unit sides. As π is transcendental, unable to be constructed with compass and straight edge alone, we use rope to straighten the semi-circle, π.

**ENDNOTE**

Did the ancient Egyptians know this method of squaring a circle? Egyptian rope stretchers created right angles for the corners of their pyramids via the Pythagorean theorem. Had they drawn a few more circles, they might have stumbled across this method. Perhaps they did, yet it appears to not have been recorded in extant historical records.

*BTW The headline doesn't claim I am the first to square the circle as that would instantly classify me as a crackpot and stop mathematicians visiting this page. And I really can't believe nobody else has managed to solve this 2500 year-old problem with such a simple solution. The key was to use string rather than a compass to construct the circle. *

My thanks to mathematics professor, Dr Avinash Sathaye for gifting me the book, Geometry in Ancient and Medieval India. Without having been aware of the Śulbasūtras (Rules of the cord, 600 BCE) I would not have thought of this solution.

**UPDATES as at 3.11.2016**

- Mathematics professors, Dr Steven Strogatz tweeted about this yesterday and Dr Keith Devlin 'liked' it. https://twitter.com/stevenstrogatz/status/707712621097709569
- The American Mathematical Society also retweeted it.
- An editor of a math journal has also agreed to publish an article on this solution later this year.