## An Ancient Problem

I ran across this problem on page 90 in Mathematics in Historical Context by Jeff Suzuki: Find the length of the side of a equilateral pentagon inscribed in a square with one vertex coinciding with a corner of the square.  Here is my solution.

Inscribed Pentagon-Quartic Equation

I basically just pushed Pythagorean relations around the square to get a polynomial equation of degree four.  My TI-84 Plus found an approximate zero.  Abu Kamil found the exact answer in radical form.  I wondered how he did that.

I had originally thought this might be good group problem for our Southern Oregon Math League high competition but it is too hard and doesn’t seem to lend itself to a group process.  I however did pass the problem on to Dr. Curtis Feist.  Here is his solution (my handwriting).

Now I saw how an exact solution could be found.  I could relate Dr. Feist’s quadratic equation is related to my quartic equation this way.

Inscribed Pentagon-Quartic Equation Factors

His quadratic equation was of course a factor of my fourth degree equation.  And with a little more trouble I found that my equation could be expressed as $(x-2)^4-8x^2$.  In the spirit of Annalisa Crannell (see The Shad-Fack Transom) I wonder if this beautiful compact form reflects something fundamental about the geometry.  The  fourth power term looks daunting.