User:ChuckHG/sandbox

While it has been mentioned that the Borromean rings can be realized as a link using three ellipses with arbitrarily small eccentricity, it is actually the case Borromean rings can be realized with two circles and one ellipse of arbitrarily small eccentricity. Here is a rather canonical example. Consider the circles (with 0 < b < a):

A: y2/a2 + z2/a2 = 1;  x = 0 in the yz-plane

B: x2/b2 + z2/b2 = 1;  y = 0 in the xz-plane

and the ellipse

C: x2/a2 + y2/b2 = 1;  z = 0 in the xy-plane

Obviously, A & B do not intersect, for they meet the z-axis at points a units and b units from the origin respectively. A & C meet the y-axis at points a units and b units from the origin, respectively. And, finally, B & C meet the x-axis at points b units and a units from the origin, respectively. Since 0 < b < a, these points are all distinct, and since a & b can be chosen as close to each other as desired, the eccentricity (1 - b2/a2)1/2 of ellipse C can be as small as desired.