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For a tetrahedron, the circumsphere is calculated, then it's recentered at the origin and the tangent planes at the vertices are found. Plane intersections are found, then the recentering is reversed.

initial = {{0, 0, 0}, {1, 2, 0}, {3, 0, 0}, {1, 1, 2}};
cs = Circumsphere[initial];
eq = Append[(# - cs[[1]]), -cs[[2]]] . {x, y, z, cs[[2]]} & /@ initial;
tangential = # + cs[[1]] & /@ Flatten[{x, y, z} /. Solve[#[[1]] == #[[2]] == #[[3]] == 0] & /@ Subsets[eq, {3}], 1];
tan = {{3/2, 1/2, -5}, {-(17/18), 31/18, 10/9}, {3/2, -(21/2), 6}, {31/6, 25/6, 7/3}};

enter image description here

That finds the tangential tetrahedron, which is tangent to the circumsphere of the given tetrahedron and which has an insphere the same as the circumsphere.

This is the right solution, but my method feels like a B- level hack. Is there a more elegant method, with pure linear algebra or trilinear coordinate methods?

Ed Pegg
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    This isn't a slicker derivation, only a rendering of the final symbolic result. Let tetrahedron $PABC$ with $a:=|PA|$, $b:=|PB|$, $c:=|PC|$, $d:=|BC|$, $e:=|CA|$, $f:=|AB|$ have tangential tetrahedron (in the sense described) $P'A'B'C'$, with vertex $P'$ opposite the face tangent at $P$, etc. The barycentric coordinates of $P'$ are $$-2d^2e^2f^2 : d^2(-a^2d^2+b^2e^2+c^2f^2) : e^2(a^2d^2-b^2e^2+c^2f^2) : f^2(a^2d^2+b^2e^2-c^2f^2)$$ (Note that $ad$, $be$, $cf$ are pairs of opposite edges in $PABC$.) Perhaps seeing this form will inspire a slicker derivation. – Blue Oct 13 '23 at 18:40
  • Would you happen to have barycentric coordinates of the Extangents tetrahedron? – Ed Pegg Oct 13 '23 at 18:54
  • What's the definition of the extangents tetrahedron? – Blue Oct 13 '23 at 18:58
  • The outer tetrahedron tangent to the exspheres – Ed Pegg Oct 13 '23 at 19:00
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    FYI: I've updated my answer to an old question of yours about tetrahedron centers. It now includes my result above about the tangential tetrahedron, as well as the extangents tetrahedron. – Blue Oct 17 '23 at 15:48

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