Timeline for Thales' semicircle theorem in higher dimensions
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 2, 2015 at 19:57 | comment | added | The Masked Avenger | @Wolfgang <a href='mathoverflow.net/q/201374/6094' >Good Question!</a> | |
| Apr 1, 2015 at 10:51 | comment | added | Wolfgang | So imagine we want to replace the hemisphere by a (the?) surface such that the solid angle is the same for each point on it. What would be the equation of the curve, say in the x-z-plane, whose rotation around the z-axis yields that surface? | |
| Mar 29, 2015 at 1:16 | history | edited | Douglas Zare | CC BY-SA 3.0 |
Mainly fixed typos.
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| Mar 28, 2015 at 23:43 | comment | added | Joseph O'Rourke | This is wonderful confirmation of the start @$ 90^\circ, \Omega=0.59\pi \approx 1.85$ to @$0^\circ, \Omega = \pi$. Great that you could compute the solid angle explicitly! | |
| Mar 28, 2015 at 21:17 | history | answered | Douglas Zare | CC BY-SA 3.0 |