On 11/29/06, Michael Kleber <michael.kleber@gmail.com> wrote:
Okay, I had another drip of time...
Oof, of course; very nice. What's more, starting with a 3-4-5 and rotating about the midpoints of the length 3 and 5 sides gets you one extra rational (integer, even!) value; you end up with {3, 4, 5, 13, 5*sqrt(145), sqrt(7081)}.
Sorry, MK --- this one is a dud, with no planar perms at all! If you're only going for double distinct planar charts (flat tetrahedra) with unequal lengths, you might just as well choose arbitrary integers for the first four --- for example [3, 4, 5, 6, 7.885487244, 3.133542872] in the same order as usual; also with the two quartic irrationals reversed. Inter alia, we can generate triple charts along the lines of the Thurston-Kleber example given earlier, from three arbitrary integer lengths u,v,w and three rational square roots x,y,z, as follows: Let a^2 = 9*(u+v+w)*(u+v-w)*(v+w-u)*(w+u-v) = (12*ABC)^2; then [u,v,w,x,y,z], [u,v,w,z,y,x], [u,v,w,y,x,z] are all planar, where (a*x)^2 = v^2*u^4+9*w^6-u^6-u^2*v^2*w^2-v^6+5*u^4*w^2 -9*w^4*u^2+5*w^2*v^4+v^4*u^2-9*w^4*v^2, (a*y)^2 = (2*v^4+2*u^4-4*u^2*v^2-6*v^2*w^2-3*u^2*w^2) *(v^2-2*w^2-2*u^2), (a*z)^2 = (2*v^4+2*u^4-4*u^2*v^2-6*u^2*w^2-3*v^2*w^2) *(u^2-2*w^2-2*v^2). Besides the triple planar chart, this hexad also yields a triple proper tetrahedron and a two double tetrahedra; though not all these are necessarily real. Of course, if we could find u,v,w such that all four polynomials were simultaneously rational squares, we'd have a triply-planar integer chart. Over to the number-theorists!
Jim Buddenhagen likewise mentioned four integer sides. Jim, is yours any smaller than this one?
Going back further, I just reread Fred Lunnon's mail from two days ago:
I'm not sure why you decided it was somehow inferior, Fred! Yes, the first hexad has three cities collinear, but when handed a problem about distances along roads, nothing seems more likely than that some city is along the route between two others.
I suppose it's a matter of aesthetic opinion --- like Ed, I'd prefer to see a solution with unequal lengths and proper triangles.
Did anyone verify this, per Fred's request? It certainly seems like the prettiest configuration seen so far.
The program had been (badly) designed to omit such cases: I could have generated many similar but smaller examples. An efficient search needs to be refined to exclude boring cases --- at this stage I just haven't got around to trying. ------------------------------------------------------------------------------------------ I have now conducted a search for integer planar charts with unequal sides and proper triangles, and while there are some, they're pretty sparse --- for edge sum up to 215, there are just 15 (single) examples --- I'll post a list if requested. Since at a rough (and probably optimistic) guess based on program statistics, I expect on average one doubly planar example among every 360 singly planar, it's clear we'll have to wait a good deal longer than the 17 hours this run cost, for a half-decent chance of striking lucky. The two smallest such (singly planar) hexads found were [15, 19, 20, 18, 9, 13]; [15, 19, 20, 13, 26, 18]; Notice the curious quasi-symmetry relating them, which also appears amongst other examples: ABC remains the same, BD swaps with CD, only AD is new. Fred Lunnon