Still trying to stick my two-penn'orth in here, despite proven risk to my dignity ... anyway, you have had the opportunity to admire Henry's rapiers; now you may compare Fred's (Gröbner-basis) blunderbuss. Define a polynomial ring over rationals in 10 variables, in order of elimination z3, z2, z1, y3, y2, y1, x3, x2, x1, R2 ; a set of generators for an ideal { x1^2 + y1^2 + z1^2 - R2, #circumradius squared# x1 + x2 + x3, y1 + y2 + y3, z1 + z2 + z3, #centroid at O# ( (x1 - x2)^2 + (y1 - y2)^2 + (z1 - z2)^2 ) - ( (x1 - x3)^2 + (y1 - y3)^2 + (z1 - z3)^2 ), #equilateral# ( (x1 - x2)^2 + (y1 - y2)^2 + (z1 - z2)^2 ) - ( (x2 - x3)^2 + (y2 - y3)^2 + (z2 - z3)^2 ) } ; and compute its Gröbner-basis (I used Magma, any old CAS should serve). Near the end of the basis should appear a polynomial involving R2 , from which all z's have been eliminated: 3*R2^2 - 4*(x1^2 + x1*x2 + x2^2 + y1^2 + y1*y2 + y2^2)*R2 + 4*(x1*y2 - x2*y1)^2; Solving the quadratic for R2 = R^2 yields circumradius in terms of x-, y-components alone --- sorted! Well, not quite. For one thing, the discriminant must be positive for the roots to be real: this provides a (quartic) inequality constraint on x-, y-components. For another, there are then actually two putative values of R^2 : the physics of the problem suggests that one must always be negative, but this is not obvious from the equation. Similarly, omitting the first ideal generator yields z-components, as roots of the quadratic in z1^2 3*z1^4 - 2*(2*x2^2 + 2*y2^2 + 2*x1*x2 + 2*y1*y2 - x1^2 - y1^2)*z1^2 - (x1^4 + y1^4 + 4*x1^2*x2^2 + 4*y1^2*y2^2 + 4*y1^3*y2 + 4*x1^3*x2 + 4*x1^2*y1*y2 + 4*y1^2*x1*x2 + 8*x1*x2*y1*y2 + 2*y1^2*x1^2); etc. Discriminant and attendant ambiguity are as before, mildly compounded by requiring compatible choices of sign for the z's . Fred Lunnon On 5/8/14, Henry Baker <hbaker1@pipeline.com> wrote:
We have an _equilateral_ triangle in arbitrary position in standard 3D space with coordinate axes x,y,z.
Translate this triangle (without rotations) so that its center becomes the origin.
The 3 vertices are P1=(x1,y1,z1), P2=(x2,y2,z2), P3=(x3,y3,z3).
We are given x1,y1,x2,y2 (but _not_ the z coordinates).
Obviously, x3=-x1-x2, y3=-y1-y2, due to the center being the origin.
Find a "simple" expression for the _radius_ of the circumcircle.
This problem is trivial in the case that x1^2+y1^2 = x2^2+y2^2 = x2^2+y3^2 = R^2, in which case z1=z2=z3=0, so we are interested in the more difficult case.
(Hint: you don't actually need the z coordinates.)
(Yes, I do have an answer, which is simpler than I had expected.)
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