Here is a possibly new twist on the old problem to construct, by as few simple steps as possible, a square nearly equal in area to a given circle. The length of side of a square equal in area to a circle diameter d is (Ök)d, and Ök = 0.8862269....

The Rhind Papyrus c.1650 BC, approximates the circle area by a square of side (8/9)d = 0.8888888...d, giving a ratio of 1.003, correct to 3 parts in 1000; very good for such a simple construction.

Another way I have found of approximating the equal area square is to draw a line from a corner A of the circumsquare, taken as origin, to a point with coordinates (7, 33). This crosses the circle at a point C on the required square.

[**Proof**: Draw BCD parallel to the side AE. Take the radius of the circle to be 1.

Then AE = EO = OC = 1, and CD = sin q, where q = angle COE.

The slope of the line AC is CB/BA = (BD − CD)/(EO − DO) =
(1 − sin q)/(1 − cos q) = m.

Taking CD = Ök we find q = 62.402886°
and m = 0.2119672, which can be approximated by 7/33 = 0.2121212.

Working back the other way, starting with m = 7/33 we get, (1 + m²)q² − 2(1 − m)q + (1 − 2m) = 0,

where q = sin q; a quadratic equation giving q = CD = 0.886168.

The ratio of this to Ök is 0.9999335, i.e. correct to 7 parts in 100,000.]

This was originally published in *The Games and Puzzles Journal* vol.2, #18, March 2001, p.325.
where it is acknowledged to have been inspired by a less accurate construction
in *Ein Dutzend mathematischer Betrachtungen* by C. R. R. von Schinnern (1826).