A [lune](http://en.wikipedia.org/wiki/Lune_%28mathematics%29) and a [lens](http://en.wikipedia.org/wiki/Lens_%28geometry%29) are both planar figures delineated by two circular arcs; the difference is that a lune has a concave and a convex arc (it is one circle minus another) whereas a lens has two convex arcs (it is the intersection of two circles).


It is well-known that there are exactly five lunes that can be squared; that is, there are five essentially different lunes where you can construct both the lune and a square of the same area using compass and straightedge in finitely many steps. ([Proof](http://www.mathpages.com/home/kmath171/kmath171.htm).) Are there any similar results for lenses?


(I was inspired to ask this question by [this Google+ post](https://plus.google.com/u/0/117663015413546257905/posts/TQaYtXscbR6) and [this page of fascinating background information, linked already above](http://www.mathpages.com/home/kmath171/kmath171.htm).)



Which lenses can be squared?
I suspect a complete proof is not so easy. There is a field called "the constructible numbers," sometimes denoted $E$ for Euclid, which is the smallest subfield of the reals closed under square root of positive numbers. A length $x$ is constructible if $x \in E.$ An angle $\alpha$ is constructible if $\cos \alpha \in E,$ or $\sin \alpha \in E,$ or $\tan \alpha \in E,$ these conditions being equivalent. The constructible angles that are rational multiples of $\pi$ are known precisely. However, we get a mess as soon as we consider, say, $\arctan 2,$ which is a transcendental multiple of $\pi.$


Using roughly the symbols on your link, we want to show that, for **acute** constructible angles $\alpha, \beta$ and **positive** constructible lengths x,y, that
$$ \color{magenta}{ x \alpha + y \beta \; \notin \; E.} $$
Now, if $x, y \in \mathbb Z,$ this follows by Hermite-Lindemann, page 131, Theorem 9.11 (b) in Niven, *Irrational Numbers.* Diving by a different integer, it is also true if $x, y \in \mathbb Q.$ Finally, multiplying by a single irrational constructible number, it is also true if the ratio $x/ y \in \mathbb Q.$


Anyway, a proof of the proposition in color would finish this. Probably true, but...


Now that I compare the two, evidently Cebotarev and Dorodnow proved (I think) that the expression in color is only in the field $E$ if equal to $0,$ where they were concentrating on $x,y$ of opposite signs. So it would seem they had enough technology to finish this. I just don't know how it was done. 


It seems [Postnikov gave a summary of Cebotarev and Dorodnov](http://en.wikipedia.org/wiki/Nikolai_Chebotaryov) in his book on Galois theory, an excerpt was translated in the MAA Monthly, and many of those columns [collected in a book](http://oskicat.berkeley.edu/record=b15911600~S1). see also [Quadrature of the Lune](https://mathoverflow.net/questions/92326/quadrature-of-the-lune)


Hmmm...Postnikov [evidently assumed commensurable angles](http://books.google.com/books?id=I5uYV1BIT6wC&pg=PA182&lpg=PA182&dq=The%20Problem%20of%20Squarable%20Lunes%20%20Postnikov&source=bl&ots=sFxQZkZVbR&sig=Sef3QVGoM0yUbkWEL1F_Ri4zgGU&hl=en&sa=X&ei=0hepUorTBJHzoASyzoGoDA&ved=0CEsQ6AEwAw#v=onepage&q=The%20Problem%20of%20Squarable%20Lunes%20%20Postnikov&f=false) which means he does not show how to deal with the problem above. At this point, and with my experience of Nestorovich and Mordukhai-Boltovskoi in about the same time, I have got to wonder if Cebotarev and his student really gave a complete proof of the five lunes, or made some "commensuability" assumption. As Samuel L. Jackson said in the Long Kiss Goodnight, "When you make an assumption, you make an ass out of you and umption." <http://www.imdb.com/title/tt0116908/> and 


Mitch Henessey: ...everyone knows, when you make an assumption, you make an ass out of "u" and "umption". 


<http://www.imdb.com/title/tt0116908/trivia?tab=qt&ref_=tt_trv_qu>


NOTE: **it is supposed to be** When you assume, you make an "ass" out of "u" and "me."


