Fancy Quadrilateral

I didn't know about Fuss's Theorem, so I used a brute-force approach. I started with a circle of radius 12. Placed four point on it

with angular displacements (from the positive x-axis) of $\theta_1, \theta_2,\theta_3,\theta_4$ .

I took $\theta_1 = 0$ (fixed), and assumed that $\theta_1 \lt \theta_2 \lt \theta_3 \lt \theta_4$ .

The four vertices of the quadraliteral connecting these points on the circumference of the 12-radius

circle are $P_1 , P_2 , P_3$ and $P_4$ , where $P_i = 12 ( \cos \theta_i, \sin \theta_i )$ .

Next I set up a Newton-Raphson multivariate iteration to find a solution for the required quadrilateral.

To have an incircle, it is required (by Pitot's theorem) that,

$P_1 P_2 + P_3 P_4 = P_2 P_3 + P_4 P_1 = s$ , where $s$ is the semi-perimeter, $s = \frac{1}{2} ( P_1 P_2 + P_2 P_3 + P_3 P_4 + P_4 P_1)$

Next, for the current-iteration values of $x_1(k) = \theta_2(k), x_2(k) = \theta_3 (k), x_3(k) \theta_4(k)$ , we can compute the

area of the quadrilateral, and thus compute a nominal (or average of some sort) inradius $r$ , given by

$r = \dfrac{\text{Area} }{s}$

So that the last equation to be satisifed is $r = 7$ .

Hence for the vector $\mathbf{x}(k) = [ x_1(k), x_2(k), x_3(k) ]^t$ , we define the problem as

follows:

$\mathbf{F}(\mathbf{x}(k) ) = \mathbf{0}$

with

$F_1(\mathbf{x}(k) ) = P_1 P_2 + P_3 P_4 - s$

$F_2(\mathbf{x}(k) ) = P_2 P_3 + P_4 P_1 - s$

$F_3(\mathbf{x}(k) ) = r - 7$

The Newton-Raphson iteration generates the next guess of $\mathbf{x}(k)$ as follows:

$\mathbf{x}(k+1) = \mathbf{x}(k) - \mathbf{J}^{-1}(\mathbf{x}(k) ) \mathbf{F}(\mathbf{x}(k) )$

Where $\mathbf{J}(\mathbf{x}(k) )$ is the Jacobian matrix for the vector function $\mathbf{F}(\mathbf{x}(k) )$ ,

and is defined by $\mathbf{J}(\mathbf{x}(k) ) = [ J_{ij} ] = [ \dfrac{\partial {F_i}}{\partial x_j } ]$

I found an approximation for the Jacobian matrix using numerical differentiation, then incorporated it in the iteration.

The Newton-Raphson method converged in 10 iterations. Having found the quadrilateral, what was left is simple

geometry to find the coordinates of the incenter, and this is easy to find by taking the intersection point of the

angle bisectors of two adjacent vertices. Finally, the square of the distance between the two centers is the sum of the

squares of the coordinates of the incenter.

This is Fuss' Theorem for a bicentric quadrilateral.

I also used Fuss' Theorem for solution. It is as under.
Here R=circumradius=12, and r=inradius=7, x=distance between centers.
$\huge x^2=R^2+r^2-r\sqrt{4R^2-r^2}=12^2+7^2-7\sqrt{4*12^2+7^2}=Large~\color{#D61F06}{18}$

l use Formula 39