Timeline for Hiding $k$ disks inside a larger disk
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 13, 2014 at 4:43 | comment | added | Jan Kyncl | And for four discs, there are a continuum many configurations where the integral is $\pi +4$. | |
| Jun 12, 2014 at 4:30 | comment | added | Jan Kyncl | Now for two such touching discs, the integral is $\pi + 2$, and for three discs like on the picture the integral is $\pi + 3$. Asymptotically, and approximately, it seems you want to pack the discs into another disc of smallest diameter (and thus perimeter). | |
| Jun 12, 2014 at 1:50 | vote | accept | Joseph O'Rourke | ||
| Jun 11, 2014 at 10:50 | comment | added | Joseph O'Rourke | Nice! So for one disk of diameter $1$, the integral is $\pi$. | |
| Jun 11, 2014 at 5:16 | history | answered | Jan Kyncl | CC BY-SA 3.0 |