Neusis constructions (4): the Delian problem

The Delian problem is that of constructing ; since the polynomial equation is irreducible over the equation can’t be solved by ruler and compass.  The problem is also known as “doubling the cube”; and part of its mythology is that the oracle at Delphi told the people of Delos that to be rid of a plague they needed to double in size a cubical altar.

The length  can be constructed by a neusis, and the first such construction seemed to have been made by Nicomedes, about whom nothing is known.  Here is the simplest version of his construction.

Start with an equilateral triangle of side length 1, with base ; extend the base to , and construct the perpendicular to .  From , draw a line which crosses at and at and for which :

Then length .  To see this, create a line parallel to the base through which meets at .  Let and .  Drop the perpendicular from to meet at :

By chasing angles note that is half of an equilateral triangle, so that .  Triangles and are similar, so that .  Applying Pythagoras’ theorem to we have

or after expanding and simplifying:

.

From the similar triangles equation, we have so that

and after multiplying across by :

and this equation can be factored as

.

This construction can be generalized.  Start, instead of with an equilateral triangle, with a isosceles triangle of base and side lengths 1. Extend by another length of 1 to , and draw a line .  Then construct the neusis as above with :

Then .

The proof mirrors the proof above; starting with a line from parallel to and meeting at ; by similar triangles and we have .  The rest is the same as above.

For more information, and greater discussion, see the excellent book Geometric Constructions by George Martin.