In his book, The Irrationals, Julian Havil presents a method used by classical Arab scholars for finding the diameter of a circle inscribing a regular pentagon - and then asks the reader to prove the method is correct.

Can someone outline the general method for this (Harvil uses a particular size of regular pentagon) and a proof.

I have been filling pages of a notebook for a day and a half now and cannot get an answer with less than two unknowns.

  • $\begingroup$ It would help to have a diagram and a description of the method. $\endgroup$ – Mark Bennet Jul 31 '13 at 16:25
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    $\begingroup$ The book (near the beginning of chapter 2) quotes The Pentagon and the Decagon by Abu Kamil, then transcribes his (hard to understand) instructions into modern notation, ending up with $$\sqrt{2a^2+\frac25\sqrt{\left(2a^2\right)^2}+\left(a^2\right)^2} = \frac15a\sqrt{50+10\sqrt5}$$ the right side of which is called the “standard result” for writing the diameter of the circumcircle of a pentagon with side $a$. But this method in the book is not a proof, so I guess the OP is simply looking for a proof for this formula. $\endgroup$ – MvG Jul 31 '13 at 16:45
  • $\begingroup$ Yes, precisely so. The author invites the reader to show that the left hand side does "give the standard result" - obviously I can do that, but then invites the reader "to prove it" - hence I am hoping for a proof, as I cannot work one out myself $\endgroup$ – adrianmcmenamin Jul 31 '13 at 17:07

From this answer (or perhaps this one if you prefer a more geometric approach) you get

$$\cos(36°) = \frac{1+\sqrt5}4$$

and using the equality $\sin^2 t + \cos^2 t = 1$ you get

$$\sin(36°) = \frac{\sqrt{10-2\sqrt5}}4$$

Now consider the right triangle formed by the center of your circumcircle, one corner of the pentagon, and a midpoint of one of the adjacent sides.

Illustration of right triangle

Its hypothenuse will be the radius of the circle (i.e. half the diameter you want to obtain), and its one leg will be half the edge. The angle opposite that leg is a tenth of a full circle, i.e. $36°$. So now you have

$$\frac{a/2}{r} = \frac{a}{d} = \frac{\sqrt{10-2\sqrt5}}4$$

Solving this for $d$ you get

$$d = \frac{4}{\sqrt{10-2\sqrt5}}a = a\sqrt{\frac25\sqrt5+2} = \frac a5\sqrt{10\sqrt5+50}$$

which is the “standard result” mentioned in the book.


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