2007-03-10, 16:47 | #1 |
"Richard B. Woods"
Aug 2002
Wisconsin USA
2^{2}×3×599 Posts |
Tetrahedral network problem
Inspired by Mally's triangular "Network problem" (http://mersenneforum.org/showthread.php?t=7315), here's a 3-D extension:
Suppose we dip an idealized wire tetrahedron into an idealized soap solution, bring it back out, and pop the bubbles until the idealized surface tension forms only 6 (one for each edge of the tetrahedron) flat films intersecting at a single interior point. If in the course of bubble-popping we reach a situation with fewer than 6 flat films (e.g., one on each of the 4 sides of the tetrahedron), re-dip the wire, or wave it around so the air creates more bubbles, and try again. Is that intersection the solution to the 3-D minimum-sum-of-distances-to-vertices problem? Or does it instead minimize the sum of areas of the flat films? Or does the same point minimize both? |
2007-03-10, 17:29 | #2 |
"Lucan"
Dec 2006
England
6,451 Posts |
Soap films minimize surface area (energy) and I would guess this
does not correspond to minimizing the sum of the distances in general. Is it obvious that the resulting 6 surfaces are flat? Last fiddled with by davieddy on 2007-03-10 at 17:31 |
2007-03-10, 19:06 | #3 |
"Lucan"
Dec 2006
England
6,451 Posts |
On second thoughts if the four arms to the vertices subtend equal
angles, this seems to be a recipe for stability of the 6 plane soap films AND minimize the sum of distances. |
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