Olympic Rings

The five Olympic Rings define a reducible curve of degree 10 in real 3-space.

  1. Find an ideal in $\mathbb{R}[x,y,z]$ whose variety looks like this curve.
  2. Describe the convex hull of the olympic rings. What are the faces?
  3. Determine the algebraic boundary of the convex hull of your curve.

One Solution:

We first need to find the equations of five circles in 3-space that will form our olympic rings:

(1)
\begin{align} C1 : y^2+z^2-1=0, x=0 \\ C2 : x^2+(y-3/2)^2-1, z=0\\ C3 : x^2+(y+3/2)^2-1, z=0\\ C4 : (y-5/2)^2+z^2-1, x=0\\ C5 : (y+5/2)^2+z^2-1, x=0\\ \end{align}

The circles C1, C4 and C5 lie in the $yz$-plane and form circles with radius 1 and centers $(0,0,0),(0,5/2,0),(0,-5/2,0)$ and the circles C2 and C3 lie in the $zy$-plane and form circles with radius 1 and centers $(0,3/2,0),(0,-3/2,0)$.
To find the ideal of the curve, we intersect the ideals $I1,\ldots,I5$ generated by the equations for the circles given above (we should do this in Sage, if we're young, or Macaulay2, if we're old).

The convex hull will have 2-dimensional faces, edges and extreme points, which will lie on some of the circles. The 2-dimensional faces come from trisecant planes to the curve and edges come from stationary bisecant lines. We will compute the equations for these components of the algebraic boundary of the convex hull and draw a picture:

So, let's think about the 2-dimensional faces first. We expect an edge between the three points $(0,-5/2,1),(0,0,1),(0,5/2,1)$ and between the two points $(1,-3/2,0),(1,3/2,0)$. These two edges span a 2-dimensional face. Of course, the same is true, if we replace the 1's by -1's in these points. So we get four 2-dimensional faces and these form the variety of trisecant planes. These trisecant planes are defined by the equations $z+x-1,z-x-1,-z+x-1,-z-x-1$.

To compute the edge surface, we compute the locus of stationary bisecant lines: A stationary bisecant line is a line between two points on the curve such that the tangent lines to the curve at these two points lie in a plane.
We will put this into Macaulay2 by looking at the two circles to the far right and far left. The extreme points of all edges will lie on these circles because the circle in the middle, i.e. $C1$, lies in the interior of the convex hull of the rings.
So we introduce new variables $a,b,c,d$. We take $(a,b)$ on $C4$, $(c,d)$ on $C2$ and compute the tangent line to $C4$ at $(a,b)$ and the tangent line to $C2$ at $(c,d)$. Then these points lie on a stationary bisecant line, if the vectors $(1,0,a,b),(1,c,d,0),(0,0,-2b,2a-5),(0,2c-3,-2d,0)$ are linearly dependent, i.e. if the matrix $M$ with these vectors as rows does not have full rank (note that the last two vectors are orthogonal to the gradient of the equations for the circles at the corresponding point and can be thought of as the direction vectors of the tangent lines). Therefore, the stationary bisecant line between these two points can be parametrized by $((1-t)d,(1-t)c+ta,tb)$.
In order to return to our original variables, we set up the equations $x=(1-t)d$, $y=(1-t)c+ta$ and $z=tb$ in addition to the circle equation and the determinant of the above matrix and then eliminate the variables $a,b,c,d,t$:

R = QQ[a,b,c,d,t,x,y,z]
c1=(a - 5/2)^2 + b^2 - 1; c2=(c - 3/2)^2 + d^2 - 1;

M=matrix{
{1,0,a,b},
{1,d,c,0},
{0,0,-2*b, 2*a-5},
{0,2*c-3,-2*d,0}};
K = ideal(det(M),c1,c2,(1-t)*d-x,(1-t)*c+t*a-y,t*b-z);
eliminate(K,{a,b,c,d,t})

This will return a nice polynomial of degree 8. We do the same thing for the other two circles at the end, namely $C3$ and $C5$.
So the algebraic boundary of the convex hull of the olympic rings has six components and has degree $1+1+1+1+8+8=20$.
Here's a picture:
olympicRings.png

The blue and green surfaces are the two irreducible components of the edge surface and the red planes are the trisecant planes.
Here are the polynomials defining the blue and green components (as the symmetry of the arrangement of the circles suggests, these polynomials are very similar):

768*x^8+2048*x^6*y^2+1536*x^4*y^4-256*y^8-1536*x^6*z^2-3072*x^4*y^2*z^2-1536*x^2*y^4*z^2+2304*x^4*z^4-3072*x^2*y^2*z^4+1536*y^4*z^4-1536*x^2*z^6+2048*y^2*z^6+768*z^8-5120*x^6*y-6144*x^4*y^3+3072*x^2*y^5+4096*y^7+15360*x^4*y*z^2+12288*x^2*y^3*z^2-3072*y^5*z^2+9216*x^2*y*z^4-18432*y^3*z^4-11264*y*z^6+512*x^6+1280*x^4*y^2-26624*x^2*y^4-27392*y^6-16128*x^4*z^2-22784*x^2*y^2*z^2+34816*y^4*z^2-3840*x^2*z^4+75008*y^2*z^4+12800*z^6+14848*x^4*y+86528*x^2*y^3+99328*y^5-7168*x^2*y*z^2-152064*y^3*z^2-123392*y*z^4-9376*x^4-129024*x^2*y^2-211808*y^4+30880*x^2*z^2+319488*y^2*z^2+70496*z^4+85696*x^2*y+269056*y^3-323264*y*z^2-19840*x^2-196144*y^2+126336*z^2+73920*y-11025

And
768*x^8+2048*x^6*y^2+1536*x^4*y^4-256*y^8-1536*x^6*z^2-3072*x^4*y^2*z^2-1536*x^2*y^4*z^2+2304*x^4*z^4-3072*x^2*y^2*z^4+1536*y^4*z^4-1536*x^2*z^6+2048*y^2*z^6+768*z^8+5120*x^6*y+6144*x^4*y^3-3072*x^2*y^5-4096*y^7-15360*x^4*y*z^2-12288*x^2*y^3*z^2+3072*y^5*z^2-9216*x^2*y*z^4+18432*y^3*z^4+11264*y*z^6+512*x^6+1280*x^4*y^2-26624*x^2*y^4-27392*y^6-16128*x^4*z^2-22784*x^2*y^2*z^2+34816*y^4*z^2-3840*x^2*z^4+75008*y^2*z^4+12800*z^6-14848*x^4*y-86528*x^2*y^3-99328*y^5+7168*x^2*y*z^2+152064*y^3*z^2+123392*y*z^4-9376*x^4-129024*x^2*y^2-211808*y^4+30880*x^2*z^2+319488*y^2*z^2+70496*z^4-85696*x^2*y-269056*y^3+323264*y*z^2-19840*x^2-196144*y^2+126336*z^2-73920*y-11025
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