To construct a regular decagon by cutting off the corners of a regular pentagon,
the question is what is x? If the decagon is regular then we have
is the proportion of the golden mean.Transpose the 2x to the other side of the equation
Rationalize the denominator.
which reduces to
While this format doesn't have rationalized denominators, it is a constructible number. One way to construct the regular decagon by cutting off the corners of a regular pentagon is as follows.
In the figure, ABC is a square. E is the midpoint of BC. As a result, we can use the Pythagorean theorem
to find that
F is where the unit circle centered at A meets AE. G is the foot of F in AB.
By a simple ratio,
we get that
We now return to our earlier figure
to see that
If we let I be the midpoint of HF then
which is the amount we have to take off the corners of the pentagon to get a regular decagon.
If you compare the construction of the regular octagon from the square and the regular dodecagon from the regular hexagon, this one is much more complicated. It is not clear if there even is a method for verifying this construction geometrically as there is for the octagon and dodecagon. It also casts doubt on our quest to find a generic method which will work for all regular polygons. However, there is a generic method which will work for all regular polygons which, while it isn't as interesting as our methods for the octagon and dodecagon, is simpler in the case of the decagon.