This ingeniously camouflaged problem had me bamboozled for a day or two, off and on --- more red herrings than a Soviet trawler! And that's before I started on the suggested solutions: dunno how high-school kids would cope, 'cos I couldn't make head nor tail of either of 'em ... So for the fainter-hearted amongst us, I propose the following rather more pedestrian approach. Given some triangle P_1, P_2, P_3 in 3-space, with centroid at the origin, and vertex x, y-components; calculate circumradius R . x_i + y_i + z_i = 0 yields z-components; (a_i)^2 = || P_(i+1) - P_(i+2) || yields square sides (a_i)^2 ; Heron formula yields squared area A^2 ; (a_1)^2 (a_2)^2 (a_3)^2 = 4 A^2 R^2 yields square radius R^2 ; and a single square root yields R . While if also given that a_1 = a_2 = a_3 , then simply z_2 = -(x_2 + y_2) , z_3 = -(x_3 + y_3) , (a_1)^2 = (x_2 - x_3)^2 + (y_2 - y_3)^2 + (z_2 - z_3)^2 , R = sqrt( (3/4)(a_1)^2 ) by Pythagoras. Mind you, as Knuth apparently remarked, I have only proved it --- I haven't tested it. Fred Lunnon Henry Baker (various dates) wrote << [High school kids could understand this, but probably not do it as an exercise.] [Hints: least squares, Siebeck-Marden Theorem, max/abs hack.] [Solution at bottom.] 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 = x3^2+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.) -------------------------------------------------- The hardest part of this problem is convincing yourself that it is even soluble. The 3 vertices are equally spaced around a ring which is rotated so that its projection onto the XY plane is an ellipse whose semimajor axis is the radius we seek. With only 3 points on the ring, it isn't a priori obvious that we have enough data to compute the parameters of the ellipse. Nevertheless, the radius can be calculated with an elegant formula, thanks to a detour through the complex numbers. Compute real sumx2 = x1^2+x2^2+x3^2 real sumy2 = y1^2+y2^2+y3^2 complex sumz2 = (x1+iy1)^2+(x2+iy2)^2+(x3+iy3)^2 real 3*R^2 = sumx2 + sumy2 + |sumz2| That's it! [snip] >>