Squaring the circle was one of the famous Ancient Greek mathematical problems. Although studied intensively for millennia by many brilliant scholars, no solution was ever found. The problem requires the construction of a square having area equal to that of a given circle. This must be done in a finite number of steps, using only ruler and compass.
Taking unit radius for the circle, the area is π, so the square must have a side length of √π. If we could construct a line segment of length π, we could also draw one of length √π. However, the only constructable numbers are those arising from a unit length by addition, subtraction, multiplication and division, together with the extraction of square roots.
In 1882, Ferdinand Lindemann proved that π is transcendental, that is, not a solution of any simple polynomial equation with integer coefficients. Therefore, π is not constructable with ruler and compass: the circle cannot be squared within the rules of classical geometry.
Bending the Rules
However, there are many simple ways to construct a line of length π, given one of unit length. Perhaps the simplest is to take the unit circle and roll it along a line for precisely half a revolution. Since the circumference is 2π, the point of contact traces a line segment of length π.
The next challenge is to construct a segment of length √π. This is a standard problem in Euclidean geometry. We can use a result known as the intersecting chords theorem. If two chords of a circle intersect, dividing each into two segments, then the product of the two segments of one chord equals the corresponding product for the other chord.
The figure below (left) shows the idea, and the result is easily demonstrated by using the properties of similar triangles. If the chord ab is a diameter, and chord cc is perpendicular to it, then cc is bisected (figure, right). If b = 1, then c = √a.
We use the properties of intersecting chords to find the square root of a given length. The line segment covered by the rolling unit circle has length π. We extend this by one unit and draw the circle of which it is a diameter. The diameter AB has two segments AC and CB, of length π and 1, so their product is π. We draw a chord through C perpendicular to AB. This has equal segments, whose product must be π. Therefore, the segment CD has length √π (Figure below).
Using CD as a side of a square, the area must be π, as required. Thus, by stretching the rules of the game, we have squared the circle.
Quad Erat Demonstrandum — but not by the methods permitted within Euclidean geometry.
Last week’s post: Bloom’s attempt to Square the Circle.
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