Inscribed Ellipse in a Parallelogram

We'll look at the given figure as a transformation of the unit square and the inscribed circle within it.

So we start with a square with vertices (0,0), (1,0), (0, 1), (1, 1).

And we find the transformation matrix $P$ , such that a point $v$ is mapped into

$Pv$ .

$P = \begin{bmatrix} a && b \\ c && d \end{bmatrix}$

Note that with this, $(0,0)$ is automatically mapped into $(0,0)$

The point $(1, 0)$ is mapped into $(20, 0)$ , therefore,

$a = 20, c = 0$

The point $(0, 1)$ is mapped into $(10 / tan(\pi /3) , 10 )$ , therefore,

$b = 10 / tan(\pi / 3), d = 10$

Hence,

$P = \begin{bmatrix} 20 && \dfrac{10}{tan(\pi / 3) } \\ 0 && 10 \end{bmatrix}$

Now the equation of the original circle (before transformation) is

$( r - r_c )^T (r - r_c) = (1/2)^2$

where $r_c = ( 0.5, 0.5)$

Let $r' = P r$ , then

$( P^{-1} r' - r_c)^T (P^{-1} r' - r_c ) = \dfrac{1}{4}$

Define $r_c' = P r_c$ , then

$(r' - r_c')^T (4 P^{-T} P^{-1} ) ( r' - r_c') = 1$

Let the matrix $G = (4 P^{-T} P^{-1} )$ then $G$ is symmetric, and we can diagnonalize, by finding an orthogonal matrix (of eigenvectors) $R$ and a diagonal matrix $D$ such that,

$R^T G R = D$

Note that $R$ can be written as

$R = \begin{bmatrix} \cos \theta && -\sin \theta \\ \sin \theta && \cos \theta \end{bmatrix}$

where $\theta$ will be specified below.

Based on this, we can introduce a new vector $w$ and define it by

$r' - r_c' = R w$

This leades to

$w^T D w = 1$

Which is the equation of an ellispe centered at the origin (of the $w-$ coordinate frame) In terms of $r'$ , this is an ellipse centered at $r_c'$ , with tilted axes which make an angle of $\theta$ with the X and Y axes. The lengths of the semi-major and semi-minor axes of the ellipse are equal to the reciprocal of the square root of the diagonal entries of the diagonal matrix $D$ . So the whole problem reduces to finding the eigenstructure of matrix G. The formula for calculating R and D is a follows:

$\theta = \dfrac{1}{2} \tan^{-1} \dfrac{2 g_{12} }{g_{11}-g_{22} }$

Thus $R$ is now fully determined. The diagonal elements of $D$ are as follows

$d_{11} = g_{11} \cos^2 \theta + g_{22} \sin^2 \theta + 2 g_{12} \sin 2 \theta$

$d_{22} = g_{22} \cos^2 \theta + g_{11} \sin^2 \theta - 2 g_{12} \sin 2 \theta$

This completely specifies the diagonalization of matrix $G$ .

Performing the calculations gives

$\theta = 9.5533^{\circ}$

$a=10.5244$

$b=4.7509$

The angle that $AC$ makes with the horizontal is $21.2060^{\circ}$

This makes the answer

$2(10.5244) + 2(4.7509) + (21.2060 - 9.5533) = \boxed{42.20}$ .