quickfur wrote:Your lace city with x2C looks interesting. It's basically the bipyramid of a (not necessarily regular) pentagonal prism. Here there are two constraints that we must satisfy: (1) the edges of this non-uniform prism lie on the exterior of the polytope, and so are required to be unit length. However, (2) the vertices of this prism must also be unit distance from the o2o's, otherwise it becomes non-CRF. So the vertices can only lie on a circle, since otherwise the prism is non-corealmar and you wouldn't be able to close it up the way you described it. But these two requirements: vertices on a circle and unit length apart, can only have one solution: a regular pentagon. So C must be equal to f, and we get the (regular) pentagonal prism bipyramid.

Perhaps the middle lace city with x2A might yield something interesting?

I'm sorry, I'm not that good with the 4D visualizing, with the x2C, I was trying to take a side-view orientation of the x2A-polytope

I don't know if side-viewing is possible on 4D with lace cities as it is in 3D, because you already squash two dimensions in a point.

I came around to the math today, and wolframalpha gave A = (something really difficult)

x2o-x2A lacing distance = 1/2*sqrt(4-A^2)

o2o-x2A lacing distance = 1/2*sqrt(2-A^2)

x2o-x2o lacing distance = 1

x2o-o2o lacing distance = 1/2*sqrt(3)

so, we get this construction:

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` d----c a = x2o b = x2A c = x2o d = o2o`

/\ / | a/d = d-c = sqrt(3)/2 a-b = b/c = 1/2*sqrt(4-A^2) d\b = 1/2*sqrt(2-A^2)

a---b | a = (-1/2*sqrt(4-A^2), 0) b = (0, 0) c = (sqrt((1/2*sqrt(4-A^2)^2-1/4, 1/2) = (1/2*sqrt(3-A^2), 1/2)

now, we can make circles with the lacing length around the coordinates, to find A (and obtain the coordinates for d for free

)

a: (x+ sqrt( 4 -A^2)/ 2)^2+ y^2 = 3/4

b: x^2+ y^2 = 1/2-1/4*A^2

c: (x- sqrt(3- A^2)/2)^2+ (y- 1/2)^2 = 3/4

putting this in wolframalpha gives 2 real and positive, really ugly exact solutions, but hey, they are solutions

exact form:

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`A = sqrt(17/11-((1-i sqrt(3)) (-2169+11 i sqrt(8391))^(1/3))/(44 3^(2/3))-(31 (1+i sqrt(3)))/(11 (3 (-2169+11 i sqrt(8391)))^(1/3)))`

x = -((-1)^(1/4) sqrt(-1344 i+(46128 i 3^(1/3))/(-2169+11 i sqrt(8391))^(2/3)+(46128 3^(5/6))/(-2169+11 i sqrt(8391))^(2/3)-(4464 3^(1/6))/(-2169+11 i sqrt(8391))^(1/3)+(1488 i 3^(2/3))/(-2169+11 i sqrt(8391))^(1/3)+12 3^(5/6) (-2169+11 i sqrt(8391))^(1/3)-3 3^(1/6) (-2169+11 i sqrt(8391))^(2/3)+12 i (3 (-2169+11 i sqrt(8391)))^(1/3)+i (3 (-2169+11 i sqrt(8391)))^(2/3)))/(60 sqrt(22))

y = (264-(46128 3^(1/3))/(-2169+11 i sqrt(8391))^(2/3)+(46128 i 3^(5/6))/(-2169+11 i sqrt(8391))^(2/3)+(1116 i 3^(1/6))/(-2169+11 i sqrt(8391))^(1/3)+(372 3^(2/3))/(-2169+11 i sqrt(8391))^(1/3)-3 i 3^(5/6) (-2169+11 i sqrt(8391))^(1/3)-3 i 3^(1/6) (-2169+11 i sqrt(8391))^(2/3)+3 (3 (-2169+11 i sqrt(8391)))^(1/3)-(3 (-2169+11 i sqrt(8391)))^(2/3))/3960

approximate form:

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`A≈0.9834232572780998913369702481822330581806+0.×10^-40 i the 0.*10^-40 i gets smaller as I ask for more digits, so I guess this becomes 0?`

x≈-0.1529927470946949461587582865691637298282+0.×10^-40 i

y≈0.4845749617938918523245446506490979027648+0.×10^-40 i

solution 2, exact:

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`A = sqrt(17/11-((1+i sqrt(3)) (-2169+11 i sqrt(8391))^(1/3))/(44 3^(2/3))-(31 (1-i sqrt(3)))/(11 (3 (-2169+11 i sqrt(8391)))^(1/3)))`

x = -1/60 (-1)^(3/4) sqrt(1/22 (1344 i-(46128 i 3^(1/3))/(-2169+11 i sqrt(8391))^(2/3)+(46128 3^(5/6))/(-2169+11 i sqrt(8391))^(2/3)-(4464 3^(1/6))/(-2169+11 i sqrt(8391))^(1/3)-(1488 i 3^(2/3))/(-2169+11 i sqrt(8391))^(1/3)+12 3^(5/6) (-2169+11 i sqrt(8391))^(1/3)-3 3^(1/6) (-2169+11 i sqrt(8391))^(2/3)-12 i (3 (-2169+11 i sqrt(8391)))^(1/3)-i (3 (-2169+11 i sqrt(8391)))^(2/3)))

y = (264-(46128 3^(1/3))/(-2169+11 i sqrt(8391))^(2/3)-(46128 i 3^(5/6))/(-2169+11 i sqrt(8391))^(2/3)-(1116 i 3^(1/6))/(-2169+11 i sqrt(8391))^(1/3)+(372 3^(2/3))/(-2169+11 i sqrt(8391))^(1/3)+3 i 3^(5/6) (-2169+11 i sqrt(8391))^(1/3)+3 i 3^(1/6) (-2169+11 i sqrt(8391))^(2/3)+3 (3 (-2169+11 i sqrt(8391)))^(1/3)-(3 (-2169+11 i sqrt(8391)))^(2/3))/3960

approximate:

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`A≈1.3272160632106629924127375166040478892325+0.×10^-40 i`

x≈0.08739116013663362659489301265096308769828+0.×10^-41 i

y≈-0.2280069418216704466332648296378194445083+0.×10^-40 i

now is this CRF?

(this has got to be a CRF with one of the ugliest coordinates around if it is

)

Edit: on second thought, the second solution can't be correct, since it gives a negative pyramid height

Edit: I just realized, that wolframalpha did give right solutions, it was just that it didn't at first, because I messed up my calculations a bit, the notion that A = -1 is therefore incorrect.