I don’t see any “gaps” in Michael’s proof. So that brings up another question: what is the minimum area that cannot be covered?
On May 6, 2018, at 11:17 AM, Cris Moore <moore@santafe.edu> wrote:
I see. Indeed for open disks my claim about two copies of the close-packing in the plane covering everything was also false, by the same argument.
- Cris
On May 5, 2018, at 9:42 PM, Tom Karzes <karzes@sonic.net> wrote:
The argument seems sound to me. The only missing detail is the case where D2 contains D1. In this case, the boundaries either intersect at a single tangent point, or not at all. The latter case is problematic, but easily avoided by choosing D1 to be a maximal size disk. (In fact, any properly contained disks could be removed from a solution, since their contributions would be entirely redundant.)
Tom
Michael Kleber writes:
Seems like this is impossible.
If you take a collection C of disjoint open disks, and let -C be the set of antipodal disks to C, then C union -C would be a set of disks that cover every point on the sphere at least once, but no point more than twice. But you can't have such a collection: any open disk D1 must overlap with some other open disk D2 (otherwise the points on the boundary of D1 are uncovered), but then the two points on the intersection of boundary(D1) and boundary(D2) are uncovered. That is, those two points have zero coverage from D1 and D2, and there's no way a D3 can possibly cover an on-both-boundaries point without triple-covering something in D1 intersect D2.
--Michael
On Sat, May 5, 2018 at 12:58 PM, Cris Moore <moore@santafe.edu> wrote:
It is tempting that there is a solution when the disks/caps are small—so that the curvature of the sphere is unimportant—since two copies of the close-packing of disks can cover the flat plane (I think).
On May 5, 2018, at 10:03 AM, James Propp <jamespropp@gmail.com> wrote:
My bogus solution was to insribe a regular tetrahedron in the sphere and then take four open spherical caps (Dan calls them disks), as large as possible, centered at the four points.
This doesn’t quite work: look at the midpoints of edges of the inscribed tetrahedron, projected outward onto the sphere.
I like this puzzle!
Jim
On Friday, May 4, 2018, James Propp <jamespropp@gmail.com> wrote:
Dan said “disjoint”.
I think I see an answer, but my solution seems so easy that I fear that I have misunderstood something.
But I fear to say more, less my solution turn out to be valid and I deprive others of the chance to solve the puzzle. I’ll consult with Dan and make sure my solution is bogus before posting it! :-)
Jim
> On Friday, May 4, 2018, Allan Wechsler <acwacw@gmail.com> wrote: > > Either p or -p, but not both? Because you can certainly cover the whole > sphere with such disks. > > On Fri, May 4, 2018 at 5:17 PM, Dan Asimov <dasimov@earthlink.net> wrote: > >> Define an open disk on the unit sphere S^2 centered at >> the point c in S^2 to be a set of the form >> >> D = D(c,p,r) = {p in S^2 | dist(p, c) < r} >> >> where 0 < r <= pi/2 and dist is distance measured along >> the sphere. Every such D is contained in some hemisphere. >> >> Puzzle: >> ------- >> Suppose we have a collection of disjoint open disks on >> the unit sphere. Is is possible that for every point p >> of S^2, either p or its antipodal point -p lies inside >> some disk of the collection? >> >> —Dan >>
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