[math-fun] Tetrahedon test-case
A useful minimal "Heronian" tetrahedron with integer edge-lengths, face-areas, volume etc (due to Rathbun?) is given at http://mathworld.wolfram.com/HeronianTetrahedron.html Warning: the sides have been shuffled on that page: for consistency with other examples given there, they should instead be re-ordered as 51 52 53 80 117 84 . Furthermore the areas and volume given there are bizarrely garbled [although correct for other examples, with the ordering repaired] --- they should actually read 1170, 1890, 2016, 1800 (in some order); 18144 [is Eric Weisstein online?]. To utilise this as a test-case requires coordinates: one pose has vertices [w,x,y,z] = [1,0,0,0], [1,51,0,0], [17,416,780,0], [1105,-43680,63756,-51408], where a point has Cartesian coordinate [x/w, y/w, z/w] ; and faces [d,a,b,c] = [0,0,0,1], [0,0,204,-253], [0,45,24,-68], [835380,-16380,9471,20128], where a plane has equation a x + b y + c z + d w = 0 . Almost certainly, a cunningly chosen isometry would reduce the size of these components substantially; but I don't have any ideas about a good algorithm for finding the optimal pose. Any ideas, anybody? Fred Lunnon
Minor error in coordinate lists: the faces below are listed in reverse order --- the last face corresponds to the first vertex, etc. Caused much headscratching until noticed! WFL On 11/2/11, Fred lunnon <fred.lunnon@gmail.com> wrote:
A useful minimal "Heronian" tetrahedron with integer edge-lengths, face-areas, volume etc (due to Rathbun?) is given at http://mathworld.wolfram.com/HeronianTetrahedron.html
Warning: the sides have been shuffled on that page: for consistency with other examples given there, they should instead be re-ordered as 51 52 53 80 117 84 . Furthermore the areas and volume given there are bizarrely garbled [although correct for other examples, with the ordering repaired] --- they should actually read 1170, 1890, 2016, 1800 (in some order); 18144 [is Eric Weisstein online?].
To utilise this as a test-case requires coordinates: one pose has vertices [w,x,y,z] = [1,0,0,0], [1,51,0,0], [17,416,780,0], [1105,-43680,63756,-51408], where a point has Cartesian coordinate [x/w, y/w, z/w] ; and faces [d,a,b,c] = [0,0,0,1], [0,0,204,-253], [0,45,24,-68], [835380,-16380,9471,20128], where a plane has equation a x + b y + c z + d w = 0 .
Almost certainly, a cunningly chosen isometry would reduce the size of these components substantially; but I don't have any ideas about a good algorithm for finding the optimal pose. Any ideas, anybody?
Fred Lunnon
Same minimal Heronian tetrahedron as before, now re-posed with vertices on Cartesian lattice --- Side lengths QP,RP,RQ,SP,SQ,SR: [51,52,53,84,117,80]; Vertices P,Q,R,S: [w,x,y,z] = [1,84,0,0], [1,108,27,36], [1,64,0,48], [1,0,0,0], where a point has Cartesian coordinate [x/w, y/w, z/w]; Faces SRQ,PRS,SQP,PQR: [d,a,b,c] = [0,9,-20,-12]/25, [0,0,1,0], [0,0,-4,3]/5, [3024,-36,52,-15]/65, oriented inwards and normalised, where a plane has equation a x + b y + c z + d w = 0 . Centroid, circumcentre, Monge, incentre points: [w,x,y,z] = [4,256,27,84], [18,756,1283,192], [36,720,4889,12], [191,13608,1512,4536]; Quadric P' A P = 0 containing altitudes & orthocentric perpendiculars: A = [[ 0, 11088, -1296, -31104], [ 11088, -264, -21, 332], [ -1296, -21, 0, 99], [-31104, 332, 99, 264]]; Dropping the first row and column, the eigenvalues are approx. 438.915, -433.179, -5.736 with sum zero, showing that A represents an "equilateral" one-sheeted hyperboloid: albeit very eccentric in this case, and nearly a rectangular hyperbolic cylinder. More gruesome detail available on request. Fred Lunnon On 11/3/11, Fred lunnon <fred.lunnon@gmail.com> wrote:
Minor error in coordinate lists: the faces below are listed in reverse order --- the last face corresponds to the first vertex, etc.
Caused much headscratching until noticed! WFL
On 11/2/11, Fred lunnon <fred.lunnon@gmail.com> wrote:
A useful minimal "Heronian" tetrahedron with integer edge-lengths, face-areas, volume etc (due to Rathbun?) is given at http://mathworld.wolfram.com/HeronianTetrahedron.html
Warning: the sides have been shuffled on that page: for consistency with other examples given there, they should instead be re-ordered as 51 52 53 80 117 84 . Furthermore the areas and volume given there are bizarrely garbled [although correct for other examples, with the ordering repaired] --- they should actually read 1170, 1890, 2016, 1800 (in some order); 18144 [is Eric Weisstein online?].
To utilise this as a test-case requires coordinates: one pose has vertices [w,x,y,z] = [1,0,0,0], [1,51,0,0], [17,416,780,0], [1105,-43680,63756,-51408], where a point has Cartesian coordinate [x/w, y/w, z/w] ; and faces [d,a,b,c] = [0,0,0,1], [0,0,204,-253], [0,45,24,-68], [835380,-16380,9471,20128], where a plane has equation a x + b y + c z + d w = 0 .
Almost certainly, a cunningly chosen isometry would reduce the size of these components substantially; but I don't have any ideas about a good algorithm for finding the optimal pose. Any ideas, anybody?
Fred Lunnon
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Fred lunnon