Circle Shadow

Points on the circle can be represented as

$r_C = r_0 + V u$

where $r_0$ is the center of the circle and $V = [ v_1, v_2 ]$ , where $v_1$ , $v_2$ are two orthogonal vectors of length 10 that are orthogonal to the normal to the plane of the circle, and $u = [\cos t, \sin t]^T$ .

The equation of the projection plane is

$n_p^T ( r - r_p ) = 0$

Now the projected point is generated from the original point on the circle as follows

$r' = r_C + t d$

Hence,

$n_p^T ( r_c + t d - r_p) = 0$

Therefore,

$t (n_p^T d) + n_p^T(r_C - r_p) = 0$

which implies,

$t = -\dfrac{n_p^T (r_C - r_p)}{ n_p^T d}$

Therefore, the projected point onto the plane is given by,

$r' = r_C - \dfrac{n_p^T(r_C - r_p)}{ n_p^T d} d$

$r'= r_C - \dfrac{d n_p^T (r_C - r_p) }{ n_p^T d}$

$r'= r_p + (I - \dfrac{d n_p^T }{ n_p^T d} ) (r_C - r_p )$

which can written in compact form as:

$r' = r_p + P (r_C - r_p)$

with $P = I - \dfrac{d n_p^T}{ n_p^T d}$ .

Next, we note that points on the projection plane can be represented by

$r = r_p + R_p w$

where $R_p$ is a $3 \times 3$ matrix comprising a reference frame attached to the projection plane and has its origin at $r_p$ . In particular,

$R_p = [u_1, u_2, n_p]$

where $n_p$ is the unit vector normal to the plane, and $u_1$ , $u_2$ are two orthogonal unit vectors that are also orthogonal to $n_p$ .

$w$ is the coordinate of a given point with respect to this frame. Therefore,

$r' = r_p + P (r_C - r_p) = r_p + R_p w$

Simplifying and substituting $r_C$ , we have

$P ( r_0 + V u - r_p) = R_p w$

Solving for $w$ , and noting that ${R_p}^{-1} = R_p^T$ , we get,

$w = {R_p}^T P (r_0 - r_p) + {R_p}^T P V u$

Define

$h = {R_p}^T P (r_0 - r_p)$ and $G = {R_p}^T P V$ , then

$w = h + G u$

One can verify that the third row of ${R_p}^T P$ is a row of zeros, making the third row of $h$ and $G$ a row of zeros as well. This is consistent with the fact that the third element of $w$ should be zero, because the projected point lies in the projection plane. We will therefore consider only the first two rows of the above equation, thus obtaining,

$w' = h' + G' u$

where the primed variables are obtained from the original variables by taking only the first two rows.

Now, using the fact that $u^T u = 1$ , we get

$(w' - h')^T G'^{-T} G'^{-1} (w' - h') = 1$

At this point we can find a rotation matrix R , and a diagonal matrix D such that

$R^T G'^{-T} G'^{-1} R = D$

Therefore, by a change of variable, $w' = R p + h'$ , we obtain

$p^T D p = 1$

This is an equation of an ellipse, with semi-major and semi-minor axes given by the reciprocal of the square roots of the diagonal elements of D.

Performing the necessary calculations, we find that the semi-major axis M, and the semi-minor axis m, are given by

$M = 15.860296$

$m = 8.4067365$

Therefore, the answer is $M + m = \boxed{24.267}$