Oh, what affine day to solve a cubic. An equilateral triangle centered at the origin in the complex plane, but at an arbitrary orientation angle theta, has 3 vertices: r*(cos(theta)+i*sin(theta)), r*(cos(theta+2pi/3)+i*sin(theta+2pi/3)), r*(cos(theta-2pi/3)+i*sin(theta-2pi/3)). This equilateral triangle comes with its circumcircle |z|=r, and its incircle |z|=r/2. We can draw these parametrically: r*(cos(t)+i*sin(t)), r/2*(cos(t)+i*sin(t)). If we now "squish" the complex plane such that the x axis is _expanded_ by a real factor a>=1, and the y axis is _contracted_ by a real factor b, our triangle becomes distorted: r*(a*cos(theta)+i*b*sin(theta)), r*(a*cos(theta+2pi/3)+i*b*sin(theta+2pi/3)), r*(a*cos(theta-2pi/3)+i*b*sin(theta-2pi/3)). This distortion squishes the circumcircle & incircle into ellipses, the second of which is the Steiner inellipse: the subject of the Siebeck/Marden theorem. r*(a*cos(t)+i*b*sin(t)) r/2*(a*cos(t)+i*b*sin(t)). If we form the complex polynomial with the roots of our distorted triangle, we get the equation: (z-r*(a*cos(theta)+i*b*sin(theta)))* (z-r*(a*cos(theta+2pi/3)+i*b*sin(theta+2pi/3)))* (z-r*(a*cos(theta-2pi/3)+i*b*sin(theta-2pi/3))) = z^3 + (3/4)r^2(b^2-a^2)z - (r^3/4)*(cos(3*theta)*a*(a^2+3b^2)+i*sin(3*theta)*b*(b^2+3a^2)) = 0. Although we could independently scale a,b,r separately, this parameterization of the space is redundant; for convenience, we choose a=s*a', b=s*b', such that a'^2-b'^2=1, and r^2*(a^2-b^2) = r^2*((s*a')^2-(s*b')^2) = r^2*s^2*(a'^2-b'^2) = (r*s)^2*(a'^2-b'^2) = (r*s)^2. Thus, without loss of generality, we could have chosen a,b,r such that a^2-b^2=1 to begin with. This choice makes the coefficient of the linear term z to be -(3/4)r^2, which is a _negative real number_. As an example, choose a=1.18034, b=0.5, so our cubic is z^3 - (3/4)r^2 - r^3(cos(3*theta)*0.559 + i*sin(3*theta)*0.5) We would now like to invert this process, and go from a monic cubic equation to a squished triangle of roots. We ask the question: given a monic cubic equation with complex coefficients, how do we determine the parameters theta, r, a, b, which describe the roots of this equation? We are given the equation P(z)=z^3+p*z+q=0. Because the coefficient of the z^2 term is zero, the centroid of the roots is at the origin. We need to _rotate_ this equation so that the principle axis is located on the real axis. Thanks to Siebeck/Marden, this principle axis goes through the roots of P'(z)/3=z^2+p/3=0. Without loss of generality, we can _rotate_ and _scale_ our given equation P(z)=0, such that P(z)=z^3-(3/4)z+q'=0, which conveniently places the roots of P'(z)/3=z^2-1/4 at +-(1/2). This rotation and scaling transforms the arbitrary complex number q into another complex number q'. We know that q' = - (1/4)*(cos(3*theta)*a*(a^2+3b^2)+i*sin(3*theta)*b*(b^2+3a^2)) but how can we ever disentangle theta, a, b from this mess? The constraint a^2-b^2=1 provides a broad hint: choose parameter phi such that a=cosh(phi), b=sinh(phi), which guarantees that a^2-b^2 = cosh(phi)^2 - sinh(phi)^2 = 1. With this substitution, our rotated & scaled P(z) now looks like: P(z) = z^3 -(3/4)z - (1/4)(cos(3*theta)*cosh(3*phi)+i*sin(3*theta)*sinh(3*phi)) = z^3 -(3/4)z - (1/4)cos(3*theta - 3*i*phi) = z^3 -(3/4)z - (1/4)cos(3*(theta-i*phi)) = z^3 -(3/4)z + q' so, q' = -(1/4) cos(3*(theta-i*phi)) and theta-i*phi = acos(-4q')/3, where acos is the _complex_ arccosine function. Thus, a single call to a complex arccosine function produces a real angle theta, and a real parameter phi, and a,b are recovered from a=cosh(phi), b=sinh(phi). Using this parameterization, we can easily recover the _eccentricity_ of our ellipses (= circles distorted by a,b). The foci f for the circumellipse are +-1 (the Steiner inellipse is 1/2 the size), so f=a*e, or e=f/a=1/a.
From our numerical example above with a=1.18034, b=0.5, we now see that phi=asinh(0.5), a=cosh(asinh(0.5))=1.18034, b=sinh(asinh(0.5))=0.5, and
e = f/a = 1/cosh(phi) = 0.8944. The _string length_ for our circumellipse is 2*(a+f)=2a(1+e)=2*a+2=4.36068. (Our string is looped around the foci at +-1.) We see that our ellipse flattens to a pancake (e=1) when phi=0, in which case we get the degenerate form of the cubic _with 3 real roots_. Our ellipse converges (very quickly) to a circle (e=0) when phi->oo; however, this is a very large circle because the string length is >> distance between the foci.