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A lune and a lens 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.) Are there any similar results for lenses?

(I was inspired to ask this question by this Google+ post and this page of fascinating background information, linked already above.)

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    $\begingroup$ At first glance, the answer is 'none', because the magical cancellation that occurs for lunes can't occur for lenses. It shouldn't be too hard to stretch this out into a formal proof, using the incommensurability of $\theta$ and $\sin(\theta)$. $\endgroup$ – Steven Stadnicki Dec 11 '13 at 22:32
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    $\begingroup$ @StevenStadnicki, proof may be difficult. I have been unable to show that, if $\theta_1, \theta_2$ are acute constructible angles in radians, and $c_1, c_2$ are positive constructible lengths, then $$ c_1 \theta_1 + c_2 \theta_2 $$ is not a constructible length. True if $c_1,c_2$ are rational, or their ratio is rational. $\endgroup$ – Will Jagy Dec 11 '13 at 23:55
  • $\begingroup$ @WillJagy That's a good point. I was thinking that if $\sin(\theta)$ is algebraic then $\theta$ is an algebraic multiple of $\pi$, but that implication only goes the other direction. $\endgroup$ – Steven Stadnicki Dec 12 '13 at 0:09
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    $\begingroup$ @StevenStadnicki, the problem is completely finished, both for lunes and lenses, see mathoverflow.net/questions/151684/… $\endgroup$ – Will Jagy Dec 13 '13 at 2:23
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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 in his book on Galois theory, an excerpt was translated in the MAA Monthly, and many of those columns collected in a book. see also Quadrature of the Lune

Hmmm...Postnikov evidently assumed commensurable angles 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."

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  • $\begingroup$ In the article by Lenstra and Stevenhagen on Chebotarev's density theorem they mention this result of Chebotarev on constructing lunes. See pages 7 and 8 of math.leidenuniv.nl/~hwl/papers/cheb.pdf $\endgroup$ – KConrad Dec 12 '13 at 7:07
  • $\begingroup$ @KConrad, thanks. As the page from Postnikov seemed to indicate, Chebotarev and Dorodnov did assume commensurability, pages 7,8 in the pdf. So, a proof that there are no bizarre examples is still not done. This was exactly the state of the problem of squaring the circle in the hyperbolic plane when I looked into that problem. But that one had an easy finish with Gelfond-Schneider. This problem seems harder. $\endgroup$ – Will Jagy Dec 12 '13 at 7:19

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