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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