Neusis constructions (4): the Delian problem

The Delian problem is that of constructing \sqrt[3]{2}; since the polynomial equation x^3-2 is irreducible over \mathbb{Q} 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 \sqrt[3]{2} 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 ABC of side length 1, with base AB; extend the base to D, and construct the perpendicular BE to BC.  From C, draw a line which crosses BE at P and BD at Q and for which PQ=1:

Rendered by

Then length CP=\sqrt[3]{2}.  To see this, create a line parallel to the base AB through C which meets BE at F.  Let CP=x and BQ=y.  Drop the perpendicular from C to meet AB at M:

Rendered by

By chasing angles note that CBF is half of an equilateral triangle, so that CF=2.  Triangles CPF and PQB are similar, so that y/1=2/x.  Applying Pythagoras’ theorem to QMC we have


or after expanding and simplifying:


From the similar triangles equation, we have y=2/x so that


and after multiplying across by x^2:


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 k/4 and side lengths 1. Extend CA by another length of 1 to G, and draw a line GBE.  Then construct the neusis as above with PQ=1:

Rendered by

Then BQ=k^3.

The proof mirrors the proof above; starting with a line from C parallel to AB and meeting BE at F; by similar triangles ABG and CFG we have CE=k/2.  The rest is the same as above.

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

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