This looks like fun; wish I'd had the time to play with it. Starting with a 3-4-5 triangle, for example, and clearing denominators at the end, Bill's geometric construction yeilds the hexad 108, 144, 180, 5*sqrt(373), 13*sqrt(229), sqrt(44749) which I'm sure Fred's program will agree has three volume-zero tetrahedra and no degeneracy. I wonder whether this might be a universal construction for the case where there are three planar arrangements. That is, given (almost) any triangle ABC in the plane, Bill's method allows you to identify three points Z1, Z2, Z3 such that the three Z's distances to A,B,C are permutations of one another. (Bill says that sometimes he can produce a Z4 also, but that's beside the point here.) Is there only one such set of Z-distances for each ABC, or might there be more? How about for generic ABC? Anyway, now we're back to the problems Ed sent out two days ago:
Q1. Are there 6 integer road lengths that lead to distinct town configurations?
Q2. Are there 10 road lengths that lead to distinct [five-]town configurations?
--Michael Kleber On 11/27/06, Bill Thurston <wpthurston@mac.com> wrote:
On Nov 27, 2006, at 10:29 PM, Fred lunnon wrote:
I typed Bill's hexad below into my program, which sure enough reported 3 small tetrahedra (out of the 30 possible) with absolute volumes
15.32733125 , 35.63804829 , 46.20211861
However, among the remainder were another 5 with volumes around 100, followed by a steady ramp up to the largest around 1000.
Run-of-the mill integer hexads I've generated frequently show distinct volumes much closer together than these figures. So taken in the round, I have to say they're not terribly convincing.
What are you unconvinced of? These aren't purported to give integral solutions. The GSP demo is completely convincing, you can see there exists an exact solution because the two relevant intersection points of circles with edges independently move back and forth past the midpoints of the edges. I'm sure one could easily get numerically accurate solutions from these using Newton's method, but I don't see the point to doing so. If we could attach pictures, this would be much easier to talk about.
But, here's a geometric construction that gives 3 configurations with the same edge lengths.
Start with (almost) any triangle. Make a row of 3 congruent copies, by rotating the triangle around two of its edges. This gives a quadrilateral Q subdivided into 3 triangles, since the 3 angles meeting at a point add to pi.
Let z be the intersection of the perpendicular bisectors of the two diagonals of Q.
The cone of z to any of the three congruent triangles makes a 4-tuple having an identical set of edge lengths. As long as Q is not inscribed in a center (whose center would be z) these arrangements are not congruent.
The condition for Q to be inscribed in a circle is that diagonally opposite angles sum to pi. If the angles of the original triangle are p, q, r, the angles of Q are p, pi-p, pi-r, r, so if the original triangle is not isosceles, Q is not so inscribed.
Using sketchpad you can make this construction, which still has 2 degrees of freedom up to scaling. It appears you can get at least one more distinct configuration --- you can choose another flip of a vertex across an edge, with one distance-match to be made. You can draw a circle to see whether the two relevant distances are equal, and it's easy to arrange, in a way that the new 4-tuple (by visual inspection) is not equivalent to any of the others.
I don't know if you can get 5---some attempts to impose distance equalities produce symmetry.
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