Friday, 18 December 2009
The Delian Problem
One of the great geometry problems of antiquity is doubling of the cube (using ruler and compass). This is equivalent to extracting the cube root of two. While doing so is not possible I noticed the following (non ruler and compass construction). If one has an L shaped junction of two hallways, one 1 meter wide and the other W meters wide, then the longest board (red) that passes the bend involves extracting a cube root:
L = W / sinϑ + 1 / cosϑ
so that the derivative
L' = -W cosϑ / (sinϑ)2 + sinϑ/ (cosϑ)2 = ( -W (cosϑ)3 + (cosϑ)3)/ (sinϑ cosϑ)2
which, noting that for 0 < ϑ < Π has 0 < sin ϑ , cos ϑ < 1, is zero when
W1/3 = tanϑ = opposite /adjacent
at the point where the board touches the two walls and the corner.
L = W / sinϑ + 1 / cosϑ
so that the derivative
L' = -W cosϑ / (sinϑ)2 + sinϑ/ (cosϑ)2 = ( -W (cosϑ)3 + (cosϑ)3)/ (sinϑ cosϑ)2
which, noting that for 0 < ϑ < Π has 0 < sin ϑ , cos ϑ < 1, is zero when
W1/3 = tanϑ = opposite /adjacent
at the point where the board touches the two walls and the corner.
Posted by at 10:15 PM in Uncategorised
