Find the area of the shadow

The shadow cast is an ellipse, whose area $A=\pi ab$ , where $a$ and $b$ are the lengths of its semi-axes. Hence we need only to find $a$ and $b$ . Let us do that using 2-dimensional diagrams.

Looking normal to the $yz$ -plane, the major axis of the ellipse is $BC=2a$ . Let $\angle BAH = \angle CAH = \alpha$ and $\angle OAH = \beta$ . We note that $\sin \alpha = \dfrac 5{\sqrt{10^2+(16-5)^2}} = \dfrac 5{\sqrt{221}} \implies \tan \alpha = \dfrac 5{14}$ and $\tan \beta = \dfrac {10}{16-5} = \dfrac {10}{11}$ . Then $OC = 16 \tan (\beta + \alpha) = 16 \cdot \dfrac {\frac {10}{11}+\frac 5{14}}{1-\frac {10}{11}\cdot \frac 5{14}} = 30$ and $OB = 16 \tan (\beta - \alpha) = 16 \cdot \dfrac {\frac {10}{11}-\frac 5{14}}{1+\frac {10}{11}\cdot \frac 5{14}} = \dfrac {20}3$ . Therefore, $a = \dfrac {30-\frac {20}3}2 = \dfrac {35}3$ .

To find $b$ , we look at setup normal to the $xy$ -plane. The value of $b$ is such that the ellipse is tangent to the ray of light from $A$ which is also tangent to the sphere. Swap the $x$ - and $y$ -axes for convenience. Then the equation of the ellipse is

$\frac {\left(x-\frac {20}3-a\right)^2}{a^2} + \frac {y^2}{b^2} = 1 \implies \frac {(3x-55)^2}{1225} + \frac {y^2}{b^2} = 1$

And the equation of the upper ray that is tangent to the sphere is $y = \dfrac x{\sqrt 3}$ . Let point on the ellipse which is tangent to this ray be $P(x,y)$ . Then $P(x,y)$ satisfies both equations and we have:

$\frac {(3x-55)^2}{1225} + \blue{\frac {x^2}{3b^2}} = 1 \quad ...(1)$

The gradient at point $P$ is also equal to the gradient of the ray which is $\dfrac 1{\sqrt 3}$ , and we have:

$\begin{aligned} \frac {2\cdot 3(3x-55)}{1225} + \frac {2y}{b^2} \cdot \frac {dy}{dx} & = 0 & \small \blue{\text{Putting }y = \frac x{\sqrt 3} \text{ and }\frac {dy}{dx} = \frac 1{\sqrt 3}} \\ \frac {3(3x-55)}{1225} + \frac x{3b^2} & = 0 & \small \blue{\text{Multiply both sides by }x} \\ \frac {3x(3x-55)}{1225} + \frac {x^2}{3b^2} & = 0 & \small \blue{\text{Subtract from }(1)} \\ \frac {(3x-55)(3x-55-3x)}{1225} & = 1 \\ 11(55-3x) & = 245 \\ \implies x & = \frac {120}{11} & \small \blue{\text{Substitute it in }(1)} \\ \implies b & = 10\sqrt{\frac 23} \end{aligned}$

Therefore the area of the shadow $A = \pi ab = \pi \cdot \dfrac {35}3 \cdot 10\sqrt{\dfrac 23} \approx \boxed{299}$ .

By the distance equation, $AH = \sqrt{(10 - 0)^2 + (0 - 0)^2 + (5 - 16)^2} = \sqrt{221}$ .

Let a tangent point on the sphere be $T(p, q, r)$ . Then $\angle ATH$ is a right triangle and by the Pythagorean Theorem, $AT = \sqrt{AH^2 - HT^2} = \sqrt{\sqrt{221}^2 - 5^2} = 14$ .

By the distance equation on $AH$ , $p^2 + q^2 + (r - 16)^2 = 196$ , and by the distance equation on $HT$ , $(p - 10)^2 + q^2 + (r - 5)^2 = 25$ . These two equations solve to $p = \frac{11}{10}r + 2$ and $q = \sqrt{-\frac{221}{100}r^2 + \frac{138}{5}r - 64}$ .

The vector $AT$ has an equation of $(x, y, z) = (p, q, r - 16)t + (0, 0, 16)$ . Setting $z = 0$ gives $t = \cfrac{16}{16 - r}$ , so that $x = \cfrac{16p}{16 - r}$ and $y = \cfrac{16q}{16 - r}$ .

Substituting $p = \frac{11}{10}r + 2$ and $q = \sqrt{-\frac{221}{100}r^2 + \frac{138}{5}r - 64}$ into $x = \cfrac{16p}{16 - r}$ and $y = \cfrac{16q}{16 - r}$ and eliminating $r$ gives $24x^2 - 880x + 49y^2 + 4800 = 0$ , which rearranges to $\cfrac{(x - \frac{55}{3})^2}{(\frac{35}{3})^2} + \cfrac{y^2}{(\frac{10\sqrt{6}}{3})^2} = 1$ , an ellipse on the $xy$ -plane where $a = \cfrac{35}{3}$ and $b = \cfrac{10\sqrt{6}}{3}$ .

The area of the ellipse is then $A = \pi ab = \pi \cdot \cfrac{35}{3} \cdot \cfrac{10\sqrt{6}}{3} = \cfrac{350 \pi \sqrt{6}}{9} \approx \boxed{299.26}$ .

Let $\vec{P} = (x,y,(z=0))$ be a point in the $xy$ plane. Let $\vec{u} = (u_x, u_y, u_z)$ be a unit vector going from $\vec{P}$ to $\vec{A}$ . From any point within the shadow, traveling a distance $\alpha$ from $\vec{P}$ in the direction of $\vec{u}$ results in an intersection with the sphere. This can be expressed mathematically as:

$(x + \alpha u_x - H_x)^2 + (y + \alpha u_y - H_y)^2 + (z + \alpha u_z - H_z)^2 = R^2$

Scan through the $xy$ plane with reasonably high spatial resolution and solve for $\alpha$ for each $(x,y)$ . If there are real-valued solutions for $\alpha$ (meaning that there is an intersection with the sphere), increment the area sum by $dA = dx \, dy$ . Some care must be taken to ensure that the full shadow has been accounted for.

The boundary of the shadow is the intersection of the cone whose apex is point $A$ (the light source), and which is tangential to the sphere centered at $H$ . So first, we'll find the semi-vertical angle of this cone.

To that end, we'll find the distance $\overline{AH}$ :

$\overline{AH} = \sqrt{ 10^2 + 11^2 } = \sqrt{ 221 }$

Therefore, the semi-vertical angle is given by:

$\theta = \sin^{-1} ( \dfrac{5} { \sqrt{221} })$

from which, $\cos \theta = \sqrt{ 1 - \dfrac{25}{221} } = \dfrac{\sqrt{ 196 } }{ \sqrt{221}}$

Next, we want to find the equation of the cone. Let $\mathbf{r} =(x, y, z)$ be a point on the surface of the cone, then the vector extending from point $A$ to this point makes an angle $\theta$ with the vector $\vec{AH}$ . This means that,

$( \vec{r} - \vec{A} ) \cdot \dfrac{ \vec{AH} } { | \vec{A H} |} = \cos \theta | \vec{r} - \vec{A} |$

Now, the vector $\vec{AH} = (10, 0,-11) = [10 , 0, -11 ]^T$ , so in matrix notation, the above dot product is written as,

$( \mathbf{r} - \mathbf{A} )^T \begin{bmatrix} 10 \\ 0 \\ -11 \end{bmatrix} = \sqrt{196} | \mathbf{r} - \mathbf{A} |$

Squaring,

$(\mathbf{r} - \mathbf{A})^T \begin{bmatrix} 100 && 0 && -110 \\ 0 && 0 && 0 \\ -110 && 0 && 121 \end{bmatrix} (\mathbf{r} - \mathbf{A} ) = 196 (\mathbf{r} - \mathbf{A} )^T(\mathbf{r} - \mathbf{A} )$

subtracting the right hand side from the left hand side, we obtain,

$(\mathbf{r} - \mathbf{A})^T \begin{bmatrix} -96 && 0 && -110 \\ 0 && -196 && 0 \\ -110 && 0 && -75 \end{bmatrix} (\mathbf{r} - \mathbf{A} ) = 0$

this is the equation of the cone.

Plugging in $z = 0$ yields the equation of the shadow in the $xy$ plane:

$[x, y, -16]^T \begin{bmatrix} -96 && 0 && -110 \\ 0 && -196 && 0 \\ -110 && 0 && -75 \end{bmatrix} \begin{bmatrix} x \\ y \\ -16 \end{bmatrix} = 0$

Expanding this quadratic form, we obtain,

$-96 x^2 - 220 (-16)(x) - 196 y^2 - 75 (-16)^2 = 0$

which simplifies to,

$24 x^2 - 880 x + 49 y^2 + 4800 = 0$

Completing the square in $x$ ,

$24 (x - \dfrac{880}{48} )^2 - 24 \dfrac{(880)^2}{(48)^2} + 49 y^2 + 4800 = 0$

So that,

$24 ( x - \dfrac{55}{3} )^2 + 49 y^2 = \dfrac{9800}{3}$

This is the equation of the shadow ellipse in standard form; its semi-major and semi-minor axes lengths are

$a = \sqrt{\dfrac{9800}{3(24)} } = \dfrac{35}{3}$ and $b = \sqrt{ \dfrac{ 9800}{3(49)} } = 10 \sqrt{\dfrac{2}{3}}$

Therefore, the area of the shadow is $\pi a b = \pi \dfrac{350}{3} \sqrt{\dfrac{2}{3}} \approx \boxed{299.26}$