Timeline for Which lenses can be squared?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 13, 2013 at 2:23 | comment | added | Will Jagy | @StevenStadnicki, the problem is completely finished, both for lunes and lenses, see mathoverflow.net/questions/151684/… | |
| Dec 12, 2013 at 15:18 | vote | accept | Erik P. | ||
| Dec 12, 2013 at 0:19 | answer | added | Will Jagy | timeline score: 5 | |
| Dec 12, 2013 at 0:09 | comment | added | Steven Stadnicki | @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. | |
| Dec 11, 2013 at 23:55 | comment | added | Will Jagy | @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. | |
| Dec 11, 2013 at 22:32 | comment | added | Steven Stadnicki | 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 $\sin(\theta)$. | |
| Dec 11, 2013 at 22:06 | review | First posts | |||
| Dec 11, 2013 at 22:16 | |||||
| Dec 11, 2013 at 21:48 | history | asked | Erik P. | CC BY-SA 3.0 |