Shadow of a Circular Ring

The point $(0,5(1+\sin t),5(1 + \cos t))$ on $C$ projects to the point $(x,y,0)$ on the $xy$ -plane, where $x \; = \; 5 + \frac{3}{2 - \sin t}(-5) \; = \; -\frac{5}{2-\sin t}(1+\sin t) \hspace{1cm} y \; = \; -5 + \frac{3}{2 - \sin t}5(2+\cos t) \; = \; \frac{5}{2-\sin t}(4 + \sin t + 3\cos t)$ and so $x + 5 \; = \; \frac{5}{2 - \sin t}(1 - 2\sin t) \hspace{2cm} y - 15 \; = \; \frac{5}{2-\sin t}(-2 + 4\sin t + 3\cos t)$ and hence $7(x+5)^2 + 4(x+5)(y-15) + (y-15)^2 \; = \; 75$ (The values of $5$ and $-15$ are found by averaging the minimum and maximum values of $x$ and $y$ respectively as $t$ varies over $[0,2\pi]$ , and the coefficients $7,4,1$ are chosen to remove terms in $\cos t$ and $\sin t \cos t$ in the numerator of the expansion $7(x+5)^2 + 4(x+5)(y-15) + (y-15)^2$ ).

The eigenvalues of the matrix $\left(\begin{array}{cc} 7 & 2 \\ 2 & 1 \end{array}\right)$ are $4 \pm \sqrt{13}$ , and so we can find an $XY$ -axis system with origin at the point $(-5,15)$ such that the equation of the shadow of $C$ is $(4 + \sqrt{13})X^2 + (4 - \sqrt{13})Y^2 \; = \; 75$ which is the equation of an ellipse with semimajor and semiminor axes $a,b$ , where $a^2 \; = \; \frac{75}{4 - \sqrt{13}} \hspace{2cm} b^2 \; = \; \frac{75}{4 + \sqrt{13}}$ and hence $ab = 25\sqrt{3}$ . The area of the ellipse is therefore $\pi ab = \boxed{25\pi\sqrt{3}}$ .

The light rays between the point light source and points on the circle define an oblique cone, with its apex at the light source (point $S$ ), and its base being the given circle $C$ . The shadow is formed by the intersection of the $xy$ -plane with this oblique cone of light rays.

This problem was addressed here .

The solution to that problem (with the necessary modifications) is repeated below for convenience.

Vectors connecting the apex $S$ to points on the base are given by

$\mathbf{v}(t) = (0 , 5 \cos t + 5, 5 \sin t + 5) - (5, -5, 15) = ( -5, 10 + 5 \cos t , -10 + 5 \sin t)$

All the other points on the surface of the cone are scalar multiples of these vectors plus an offset of the apex coordinates.

We can think of this cone as a linear transformation of another arbitrary standard right circular cone.

Note first that $\mathbf{v}(t)$ can be written as

$\mathbf{v}(t) = \cos t (0 , 5, 0) + \sin t (0, 0, 5 ) + (-5, 10, -10) = \mathbf{v_1} \cos t + \mathbf{v_2} \sin t + \mathbf{v_0}$

An arbitrary standard right circular cone with apex at the origin, and a circular base centered at $(0, 0, -h)$ , and radius of base $r$ has the following form of its lateral vectors

$\mathbf{u}(t) = \cos t (r, 0, 0) + \sin t (0, r, 0) + (0, 0, -h) = \mathbf{u_1} \cos t + \mathbf{u_2} \sin t + \mathbf{u_0}$

So we can write the vectors $(0, 5, 0), (0, 0,5 ), (-5, 10, -10)$ (which are $\mathbf{v_1}, \mathbf{v_2}, \mathbf{v_0}$ ) as a linear transformation (matrix multiplication) of the vectors $(r, 0, 0), (0, r, 0),$ and $(0, 0, -h)$ , ( which are $\mathbf{u_1}, \mathbf{u_2}, \mathbf{u_0}$ ). In matrix form, this translates to,

$V = A U$

where $V = \begin{bmatrix} \mathbf{v_1} && \mathbf{v_2} && \mathbf{v_0} \end{bmatrix}$ and $U =\begin{bmatrix} \mathbf{u_1} && \mathbf{u_2} && \mathbf{u_0} \end{bmatrix}$

The solution for the $3 \times 3$ matrix $A$ is simply $A = V U^{-1}$

For computation, we can take $h = 1, r = 1$ . It follows that points on our oblique cone are given by $r = r_0 + A r'$ , where $r_0$ is the apex of the oblique cone, which is the point (5, -5, 15), and $r'$ is the position vector of a point on the right circular cone.

Now, we know that the algebraic equation of points on the right circular cone are given by $r'^T Q' r' = 0$ , where for the simple choice that we made of $h=1$ , and $r = 1$ , we have,

$Q' = \begin{bmatrix} 1 && 0 && 0 \\ 0 && 1 && 0 \\ 0 && 0 && -1 \end{bmatrix}$

Substituting for $r'$ , we obtain,

${( A^{-1} (r - r_0) )}^T Q' (A^{-1} (r - r_0) ) = 0$

which simplifies to,

$(r - r_0)^T Q (r - r_0) = 0$ where $Q = A^{-T} Q' A^{-1}$

Now, we are ready to intersect this oblique cone with the cutting plane, which is the \( x y )- plane. The vector equation of any plane is

\( r = r_1 + W u \)

where $r_1$ is any arbitrary point on the plane, and $W = [ \mathbf{w_1} , \mathbf{w_2} ]$ is a $3 \times 2$ matrix whose columns are arbitrary unit vectors that together with the normal to the plane form an orthonormal basis for $\mathbb{R}^3$ .

In the case of the $xy$ -plane, it is natural to choose $r_1 = (0, 0, 0)$ , $\mathbf{w_1} = (1, 0, 0)$ and $\mathbf{w_2} = (0, 1, 0)$ .

Substituting this vector expression into the equation of the oblique cone, one obtains,

$( r_1 - r_0 + W u )^T Q (r_1 - r_0 + W u) = 0$

Expanding this quadratic form gives,

$u^T (W^T Q W) u + 2 u^T W^T Q (r_1 - r_0) + (r_1 - r_0)^T Q ( r_1 - r_0) = 0$

Take $u_0 = - (W^T Q W)^{-1} W^T Q (r_1 - r_0 )$ , then the above equation becomes,

$(u - u_0)^T (W^T Q W) (u - u_0) = C$

where the constant $C$ is given by $C = - (r_1 - r_0)^T Q (r_1 - r_0) + {u_0}^T (W^T Q W) u_0$

Dividing by $C$ , we obtain,

$(u - u_0)^T Q_1 (u - u_0) = 1$

where $Q_1 = \dfrac{1}{C} (W^T Q W)$ .

The final step is to diagonalize $Q_1$ into $Q_1 = R D R^T$ . The diagonal elements of the diagonal matrix $D$ are the eigenvalues of $Q_1$ which are the reciprocal of the square of the semi-major and semi-minorr axes lengths $a$ and $b$ of the resulting ellipse of intersection.

The area is given by $\pi a b$ .

Let's now perform the above computations. We have

$V = \begin{bmatrix} 0 && 0 && -5 \\ 5 && 0 && 10 \\ 0 && 5 && -10 \end{bmatrix}$

$U = \begin{bmatrix} 1 && 0 && 0 \\ 0 && 1 && 0 \\ 0 && 0 && -1 \end{bmatrix}$

so that $U^{-1} = U = \begin{bmatrix} 1 && 0 && 0 \\ 0 && 1 && 0 \\ 0 && 0 && -1 \end{bmatrix}$

Therefore,

$A = V U^{-1} = \begin{bmatrix} 0 && 0 && -5 \\ 5 && 0 && 10 \\ 0 && 5 && -10 \end{bmatrix} \begin{bmatrix} 1 && 0 && 0 \\ 0 && 1 && 0 \\ 0 && 0 && -1 \end{bmatrix} = \begin{bmatrix} 0 && 0 && 5 \\ 5 && 0 && -10 \\ 0 && 5 && 10 \end{bmatrix}$

Its inverse is,

$A^{-1} = \dfrac{1}{5} \begin{bmatrix} 2 && 1 && 0 \\ -2 && 0 && 1 \\ 1 && 0 && 0 \end{bmatrix}$

Now, let's compute matrix $Q = A^{-T} Q' A^{-1}$ .

$Q = \dfrac{1}{25} \begin{bmatrix} 2 && -2 && 1 \\ 1 && 0 && 0 \\ 0 && 1 && 0 \end{bmatrix} \begin{bmatrix} 1 && 0 && 0 \\ 0 && 1 && 0 \\ 0 && 0 && -1 \end{bmatrix} \begin{bmatrix} 2 && 1 && 0 \\ -2 && 0 && 1 \\ 1 && 0 && 0 \end{bmatrix}$

Multiplying the last two matrices first, we obtain,

$Q = \dfrac{1}{25} \begin{bmatrix} 2 && -2 && 1 \\ 1 && 0 && 0 \\ 0 && 1 && 0 \end{bmatrix} \begin{bmatrix} 2 && 1 && 0 \\ -2 && 0 && 1 \\ -1 && 0 && 0 \end{bmatrix}$

so that,

$Q = \dfrac{1}{25} \begin{bmatrix} 7 && 2 && -2 \\ 2 && 1 && 0 \\ -2 && 0 && 1 \end{bmatrix}$

Next, we compute the matrix $W^T Q W$ , this is given by:

$W^T Q W = \dfrac{1}{25} \begin{bmatrix} 1 && 0 && 0 \\ 0 && 1 && 0 \end{bmatrix} \begin{bmatrix} 7 && 2 && -2 \\ 2 && 1 && 0 \\ -2 && 0 && 1 \end{bmatrix} \begin{bmatrix} 1 && 0 \\ 0 && 1 \\ 0 && 0 \end{bmatrix}$

Which is a straight forward computation. This comes to,

$W^T Q W = \dfrac{1}{25} \begin{bmatrix} 7 && 2 \\ 2 && 1 \end{bmatrix}$

Next, we compute the vector $u_0 = - (W^T Q W)^{-1} W^T Q (r_1 - r_0 )$ . Note that $r_1 = (0, 0, 0)$ and $r_0 = S = (5, -5, 15)$ . Therefore,

$u_0 = \dfrac{1}{3} \begin{bmatrix} 1 && -2 \\ -2 && 7 \end{bmatrix} \begin{bmatrix} 7 && 2 && -2 \\ 2 && 1 && 0 \end{bmatrix} \begin{bmatrix} 5 \\ -5 \\ 15 \end{bmatrix}$

This comes to,

$u_0 = \begin{bmatrix} -5 \\ 15 \end{bmatrix}$

Now, we're ready to compute $C = - (r_1 - r_0)^T Q (r_1 - r_0) + {u_0}^T (W^T Q W) u_0$

This comes to,

$C = -\dfrac{1}{25} (7(25)+ 1(25)+ 1(225)+ 4(-25)-4(75) ) + \dfrac{1}{25} (7(25)+1(225) + 4 (-75) )$

Simplifying, we get,

$C = - (7 + 1 + 9 - 4 - 12) + (7 + 9 - 12 ) = 3$

Therefore, the matrix $Q_1 = \dfrac{1}{C} (W^T Q W)$ is given by,

$Q_1 = \dfrac{1}{75} \begin{bmatrix} 7 && 2 \\ 2 && 1 \end{bmatrix}$

The product of the eigenvalues of this matrix is equal to its determinant,

$| Q_1 | = \dfrac{1}{a^2 b^2} = \dfrac{3}{75^2}$

Hence $a b = \dfrac{75}{\sqrt{3}} = 25 \sqrt{3}$ .

This means that the area is equal to $\pi a b = 25 \sqrt{3} \pi = 136.035$ when rounded to three decimal places.