Timeline for Thales' semicircle theorem in higher dimensions
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 28, 2015 at 20:49 | comment | added | Joseph O'Rourke | The images show that as $\phi \to 0$, the boundary of the spherical polygon approaches the two semicircular arcs, surrounding a quarter of the sphere, just as Will describes. Thanks to you both for resolving this. | |
| Mar 28, 2015 at 20:46 | vote | accept | Joseph O'Rourke | ||
| Mar 28, 2015 at 18:16 | comment | added | The Masked Avenger | I don't think I should challenge your reasoning as much as I should challenge my intuition. My main stumbling block is that the diameter that is perpendicular to the radial line from the viewer appears to stay the same length. The pictures Joseph provided do seem to confirm the no answer. | |
| Mar 28, 2015 at 4:52 | comment | added | Will Sawin | @TheMaskedAvenger Can you explain the flaw (or the most dubious-seeming step) in my reasoning? | |
| Mar 28, 2015 at 4:15 | comment | added | The Masked Avenger | Thank you for lending credance to my earlier thoughts. When I use the flap model, I find the quadrant being partly covered as the flap goes toward the center of the steradial sphere. However, the base of the cone is two such flaps, and the flap moving away produces a smaller steradial projection that goes to zero. I am hoping Joseph will make a picture of the flap model to confirm. I am uncertain about the steradial variation: I am certain that the area does not approach pi steradians. | |
| Mar 28, 2015 at 2:43 | history | answered | Will Sawin | CC BY-SA 3.0 |