Timeline for Packing an upwards equilateral triangle efficiently by downwards equilateral triangles
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
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| Mar 31 at 17:55 | comment | added | Zach Hunter | One can also make this little trick work 'away from the boundary'. The idea is that if we do not have polynomially large perimeter, we must have some large triangle $T^*$ with side-length $>\epsilon^c$ (for some small choice of $c>0$). But then, if we flip $T^*$ triangle along its vertical base, we can look at the $(100 \epsilon^{1-c})$-fraction of right-most points in this flipped area. And the trick works here too, which allows you to find values of $x$ away from the right-most boundary of the host triangle (since $T^*$ should appear in lots of places). | |
| Mar 31 at 16:52 | comment | added | Zach Hunter | For $\sup_x K(x)$, couldn't you look at the set of points in the right-most $(100\epsilon)$-fraction of our big triangle? At most 10 percent (say) of these points are uncovered. So by averaging, there is some $x$ in this very-right region with at least 90 percent of points in the vertical slice covered. However, every right-pointing triangle contributes intervals of length at most $O(\epsilon)$ inside this region. So we need $\Omega(1/\epsilon)$ intervals. | |
| Mar 31 at 14:57 | vote | accept | Terry Tao | ||
| Apr 4 at 22:00 | |||||
| Mar 31 at 14:57 | comment | added | Terry Tao | Thanks! For my application, it is actually the lower bound on $\sup_x K(x)$ which is important, rather than the perimeter, and your argument actually gives a polynomial bound in that case, so I have all that I need for my application, thanks! The general case (where one seeks a polynomial lower bound on the perimeter) still seems intellectually interesting, though. | |
| Mar 31 at 4:16 | history | answered | fedja | CC BY-SA 4.0 |