## 05 January 2014

### How to draw a circle with given radius

One of the greatest achievements of the human mind is The Elements, compiled by the Greek mathematician Euclid in the 3th century bC. Euclidean geometry allows but two 'instruments': a straightedge and a pair of compasses. With the former, one constructs a straight line through any two points, with the latter: a circle having any point as its centre and passing through any other point. Both 'instruments' are platonically perfect: infinitely large and infinitely precise.

The first problem treated in the first of the thirteen books is the following: how to draw a circle with a given point as centre and a given segment as radius. Any child can come up with the following solution: place the points of the compasses on the endpoints of the segment, then —leaving the opening of the compasses unchanged— draw a circle with the given point as centre. While this does solve the question, it uses the compasses in a way not allowed by Euclid. (Actually, the Greeks had no compasses at all; they drew circles with a piece of string.)

Here comes a second try, demanding considerably more creativity. The points A,B,C are given (A the intended centre, BC the intended radius), and D,E,F,G are obtained in that order.

The construction involves three uses of the straightedge and equally many of the compasses, before the point G is obtained, giving rise to the yellow circle solving the problem. It is a Euclidean construction all right, but nonetheless useless for Euclid's purpose. Indeed, to prove that this construction is valid requires considerably more than what is available at that point in the Elements. Actually, this being the first real problem, nothing is available besides the very definitions. The triangles AFG and CFB are congruent, but how do we know that?

Now that we are somewhat aware how Euclid's axiomatic building of geometry is intricate (and impressive), let's look up Book I, Proposition 2.

The straightedge is used twice, and the compasses four times, yielding D,E and finally F. This time, the proof demands nothing besides the definition of 'circle':

AF=DF-DA=DE-DC=CE=CB.

Q.E.D.