When does a set of edge lengths make a triangle? Assume that the lengths are non-negative. Then we need the triangle inequality, A+B <= C. We can either check all three permutations, or require the lengths are sorted, with C maximal. When the inequality is an equality, then A+B-C is zero, and Hero(n?)'s Formula for the area, S = 1/4 * sqrt((A+B+C)(A+B-C)(A-B+C)(-A+B+C)) is 0. Moreover, if A,B,C >=0, at most one of the factors in the sqrt can be negative, so looking at the sign of the radicand signals the reality of the triangle. Puzzle: Same problem, in 3D. When do six edges make a real tetrahedron? Assume the base triangle edges are A,B,C, with opposite edges D,E,F. The four faces must be real triangles; moreover the area of the largest face must be no greater than the sum of the areas of the other three faces. This constraint is obviously true for a real tetrahedron. Necessity follows from the example A=B=C=7, D=E=F=4, where the area of the base ABC exceeds the sum of areas of the three skinny 744 triangles. (In a 777 triangle, the distance from the center to a vertex is 7/sqrt3 = 4.04...) Another possible geometry failure: Build the tetrahedron as a planar quadrilateral, with the diagonals TBD. If we choose one diagonal, its length is restricted (both min & max) by the triangle inequality. Now imagine the quadrilateral as two triangles with the (first) diagonal as a hinge: The second diagonal must be long enough to connect the two remaining vertices, but not too long. I can make a too-long example with the quadrilateral (a rhombus) having sides 2,2,2,2. If arranged as a square, the diagonals would be 2sqrt2 = 2.828...; requiring each diagonal to be 3 gives an impossible tetrahedron which satisfies the various triangle inequalities, and the sum-of- areas constraints. One puzzle is to find something analogous to the sum-of-areas constraint that captures this second-diagonal-not-too-long restriction. I think that the second-diagonal-too-short case is covered by the sum-of-areas constraint: If SOA is true, then 2nd-diag-too-short can't happen. Another puzzle is to find something analogous to Hero(n?)'s formula, for the volume. One might hope that the volume could be expressed in terms of the areas of the four faces, but I think this is false. It is true in the case of a "right-tetrahedron", where one vertex is the corner of three right triangles: The volume is the (sqrt2)/3 of the square root of the product of the areas of the three right triangles. The area of the "hypotenuse" face is fixed in this case, by a 3D version of Pythagoras: square-of-hyp-area = sum of squares of other three face areas. There's a volume formula in terms of the six edges, as the square-root of a sixth-degree polynomial. I don't know of any formula which uses the area- excesses as factors, analogous to the appearance of the edge-excesses (i.e. A+B-C &c) in Hero(n?)'s area formula. Rich