Dark shadow

Let $O$ be the apex of the cone whose curved sides are tangent to the spheres $S_1$ and $S_2$ . Then the umbra will be the intersection of this cone and plane $P$ which by definition of a conic section will be an ellipse. Let $F$ be the intersection of the cone's axis and plane $P$ , and let plane $Q$ be the plane perpendicular to the cone's axis through point $F$ . Then an umbra on plane $Q$ , also by definition of a conic section, would be a circle.

Since the cone and spheres have rotational symmetry around the cone's axis, we can rotate the whole system so that the line of intersection between planes $P$ and $Q$ is orthogonal to the picture below (so it appears as a point at $F$ ).

Then in this rotational cross-section, let $B$ be the center of sphere $S_1$ , $D$ be the center of sphere $S_2$ , $A$ be the point of tangency between $O$ and sphere $S_1$ , $B$ be the point of tangency between $O$ and sphere $S_2$ , $E$ be the intersection of $OA$ and plane $Q$ , $G$ be the intersection of the other tangent line and plane $Q$ , $H$ be the intersection of $OA$ and plane $P$ , and $I$ be the intersection of the other tangent line and plane $P$ .

$\triangle OAB$ and $\triangle OCD$ are similar by angle-angle similarity, so $\frac{AB}{OB} = \frac{CD}{OD}$ . $AB$ is the radius of sphere $S_1$ so $AB = 1$ , and $CD$ is the radius of sphere $S_2$ so $CD = 3$ . Since $B$ is $(0, 0, 0)$ and $D$ is $(10, 2, 1)$ , $BD = \sqrt{105}$ by the distance formula. Finally, since $OD = OB + BD$ , $OD = OB + \sqrt{105}$ . Therefore, the ratio becomes $\frac{1}{OB} = \frac{3}{OB + \sqrt{105}}$ , and solving this gives $OB = \frac{\sqrt{105}}{2}$ .

Since $OB$ is half of $BD$ , and $B$ is at the origin, the coordinates of $O$ are the negative half of the coordinates of $D$ , so $O$ is $-\frac{1}{2}(10, 2, 1) = (-5, -1, -\frac{1}{2})$ .

$F$ is on the vector $BD$ , which can be defined by the vector equation $(x, y, z) = (10t, 2t, t)$ . $F$ is also on plane $P$ , which has an equation of $9x - 10z = 260$ . Therefore, at $F$ , $9(10t) - 10(t) = 260$ , so $t = \frac{13}{4}$ , and its coordinates are $(10(\frac{13}{4}), 2(\frac{13}{4}), \frac{13}{4}) = (\frac{65}{2}, \frac{13}{2}, \frac{13}{4})$ .

Since $O$ is $(-5, -1, -\frac{1}{2})$ and $F$ is $(\frac{65}{2}, \frac{13}{2}, \frac{13}{4})$ , $OF = \frac{15\sqrt{105}}{4}$ by the distance formula.

$\triangle OAB$ and $\triangle OFE$ are similar by angle-angle similarity, so $\frac{OA}{AB} = \frac{OF}{EF}$ . Since $AB = 1$ and $OB = \frac{\sqrt{105}}{2}$ (from above), then by Pythagorean's Theorem on $\triangle OAB$ , $OA = \frac{\sqrt{101}}{2}$ , and since $OF = \frac{15\sqrt{105}}{4}$ , the ratio becomes $\frac{\frac{\sqrt{101}}{2}}{1} = \frac{\frac{15\sqrt{105}}{4}}{EF}$ , and solving this gives $EF = \frac{15\sqrt{10605}}{202}$ . Therefore, a circle projected from the cone onto plane Q would have a radius of $\frac{15\sqrt{10605}}{202}$ .

$\angle EFH$ is the angle between plane $P$ and plane $Q$ , which is the same measurement of the angle between a vector perpendicular to plane $P$ and a vector perpendicular to plane $Q$ . A vector perpendicular to plane $Q$ is vector $BD = (10, 2, 1)$ , and since the equation of plane $P$ is $9x - 10z = 260$ , a vector perpendicular to it is $(9, 0, -10)$ . The angle between these two vectors can be found by solving $\cos \theta = \frac{(9, 0, -10) \bullet (10, 2, 1)}{|(9, 0, -10)||(10, 2, 1|}$ , and so $\angle EFH = \theta = \cos^{-1} (\frac{16 \sqrt{19005}}{3801})$ .

Looking at $\triangle OFH$ , we have $\angle FOH = \angle OAB = \sin^{-1}(\frac{2\sqrt{105}}{105})$ from $\triangle OAB$ , $OF = \frac{15\sqrt{105}}{4}$ , and $\angle OFH$ $=$ $\angle OFE + \angle EFH$ $=$ $\frac{\pi}{2} + \cos^{-1} (\frac{16 \sqrt{19005}}{3801})$ . Then using the law of sines we can determine that $FH = \frac{200\sqrt{18281} + 5\sqrt{2281505}}{1892}$ .

Similarly, looking at $\triangle OFI$ , we have $\angle FOI = \angle OAB = \sin^{-1}(\frac{2\sqrt{105}}{105})$ from $\triangle OAB$ , $OF = \frac{15\sqrt{105}}{4}$ , and $\angle OFI$ $=$ $\angle OFG - \angle GFI$ $=$ $\frac{\pi}{2} - \cos^{-1} (\frac{16 \sqrt{19005}}{3801})$ . Then using the law of sines we can determine that $FI = \frac{200\sqrt{18281} - 5\sqrt{2281505}}{1892}$ .

Now here is a picture of this ellipse projected onto plane $P$ , where the points $F$ , $H$ , and $I$ are now represented as vertical lines, and the ellipse's center is redefined at the origin.

The horizontal length of the ellipse is $IF + FH$ , so its major axis $a$ is $a = \frac{1}{2}(IF + FH) = \frac{1}{2}(\frac{200\sqrt{18281} - 5\sqrt{2281505}}{1892} + \frac{200\sqrt{18281} + 5\sqrt{2281505}}{1892}) = \frac{50\sqrt{18281}}{473}$ .

Since we know that the projection of the cone on plane $Q$ is a circle with a radius of $r = \frac{15\sqrt{10605}}{202}$ , and $F$ is on plane $P$ and plane $Q$ , the ellipse's height at $x = F$ is also $r$ . The $x$ -value of $F$ is $x = \frac{1}{2}(IF + FH) - IF$ $=$ $\frac{1}{2}(\frac{200\sqrt{18281} - 5\sqrt{2281505}}{1892} + \frac{200\sqrt{18281} + 5\sqrt{2281505}}{1892}) - \frac{200\sqrt{18281} - 5\sqrt{2281505}}{1892}$ $=$ $\frac{5\sqrt{2281505}}{1892}$ and so the ellipse passes through the coordinate $(\frac{5\sqrt{2281505}}{1892}, \frac{15\sqrt{10605}}{202})$ . Using the standard horizontal ellipse formula $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ , the above $a$ value, and the above coordinate as $x$ and $y$ values, we can find that the minor axis $b$ is $b = \frac{100\sqrt{1419}}{473}$ .

Finally, the area of an ellipse is $A = \pi a b$ $=$ $\pi \cdot \frac{50\sqrt{18281}}{473} \cdot \frac{100\sqrt{1419}}{473}$ $=$ $\frac{5000\pi\sqrt{25940739}}{223729} \approx \boxed{357.59280144}$ .

The shadow is the elliptical intersection of the cone defined by the two spheres and the given plane.

This problem has been addressed here . In the solution of that problem, it is shown that the major and minor semi-axes of the ellipse of intersection are given by:

$a = \frac { z_0 \tan \theta \cos \phi}{1 - \sin^2 \phi \sec^2 \theta}$

$b = \frac { z_0 \tan \theta \cos \phi}{\sqrt{1 - \sin^2 \phi \sec^2 \theta} }$

To apply these formulas, we need to calculate $z_0$ , $\theta$ and $\phi$ , where $z_ 0$ is the distance along the axis of the cone between the apex and the

intersection of the cone axis with the cutting plane, and $\theta$ is the semi-vertical angle of the cone, and $\phi$ is the angle between

the axis of the cone and the normal to the cutting plane.

First, we notice that the semi-vertical angle is given by $\theta = \sin^{-1} \dfrac{3 - 1}{\sqrt{10^2 + 2^2 + 1^2}} = \sin^{-1} \dfrac{2}{\sqrt{105}}$

Next, the distance between the apex and the center of $S_1$ is simply $d = \dfrac{1}{\sin \theta } = \dfrac{1}{2} \sqrt{105}$

The unit vector along the axis of the cone is $\mathbf{u} = \dfrac{1}{\sqrt{105}}(10, 2, 1)$

Therefore the apex is located at

$(0, 0, 0) - d \mathbf{u} = (0, 0, 0) - \dfrac{1}{2} (10, 2, 1) = (-5, -1, -\dfrac{1}{2} )$

Now we find the intersection of the cone axis with the cutting plane. The parametric equation of the axis is thus

$\mathbf{p} = (-5, -1, -\dfrac{1}{2} ) + t \mathbf{u}$

Since $\mathbf{u}$ is a unit vector then t represents the distance from the apex $(-5, -1, -\dfrac{1}{2} )$ .

Substituting the point $\mathbf{p}$ into the equation of the plane, results in,

$9 ( -5 + \dfrac{10}{\sqrt{105}} t ) - 10 ( -\dfrac{1}{2} + \dfrac{1}{\sqrt{105}} t ) = 260$

From which $z_0 = t = \dfrac{30 \sqrt{105}}{8}$

Finally, we need to find $\phi$ the angle between the axis of the cone and the normal to the cutting plane.

$\phi = \cos^{-1} \dfrac{ (9, 0, -10) \cdot (10, 2, 1) }{\sqrt{9^2 + 10^2} \sqrt{105} } = \cos^{-1} \dfrac{80}{\sqrt{181}\sqrt{105} }$

We're done. We now just substitute the values of $z_0$ , $\theta$ and $\phi$ into the formulas to get the lengths of the semi-major and semi-minor axes.

This gives us,

$a = 14.29252096$ and $b = 7.963978098$ , resulting in an ellipse area of

$\text{Area} = \pi a b = 357.593$

If we consider a cone of semi-vertical angle $\alpha$ intersected by a plane $P$ whose normal makes an angle $\beta$ with the axis of the cone, then $U \; \left(\frac{OX\,\cos\alpha}{\cos(\alpha-\beta)},\frac{OX\,\sin\alpha}{\cos(\alpha-\beta)}\right) \hspace{2cm} V \; \left(\frac{OX\,\cos\alpha}{\cos(\alpha+\beta)},-\frac{OX\,\sin\alpha}{\cos(\alpha+\beta)}\right)$ (using a coordinate system with $O$ as origin, with the positive $x$ -axis along the axis of the cone, and with the normal to the plane $P$ being perpendicular to $\mathbf{k}$ ) and hence $OU \; = \; \frac{OX}{\cos(\alpha-\beta)} \hspace{1cm} OV \; = \; \frac{OX}{\cos(\alpha+\beta)} \hspace{1cm} UV \; = \; \frac{OX\,\sin2\alpha}{\cos(\alpha-\beta)\cos(\alpha+\beta)}$

The cone intersects with the plane in an ellipse provided that $\alpha + \beta < \tfrac12\pi$ . Then $UV$ will be the major axis of the ellipse, and the point of intersection $F$ of the sphere that is inscribed inside the cone and is tangential to the plane $P$ (show in the diagram as the incircle to the triangle $OUV$ ) will be one of the foci of the ellipse. If the ellipse has semi-major axis $A$ and eccentricity $E$ , this means that its area is $\pi A^2 \sqrt{1-E^2}$ and that $\begin{aligned} 2A & = \; UV \; = \; \frac{OX\,\sin2\alpha}{\cos(\alpha-\beta)\cos(\alpha+\beta)} \\ A(1-E) & = \; UF \; = \; s - OV \; = \; \frac{OX\sin\alpha(\cos\alpha - \sin\beta)}{\cos(\alpha-\beta)\cos(\alpha+\beta)} \end{aligned}$ where $s$ is the semiperimeter of the triangle $OUV$ , and hence $E \; = \; \frac{\sin\beta}{\cos\alpha}$ In this case, we are interested in the cone that is tangent to both spheres, which has its vertex $O$ at (in the original coordinate system) $(-5,-1,-\tfrac12)$ . The vector $\mathbf{u} \; = \; \left(\begin{array}{c} 10 \\ 2 \\ 1 \end{array} \right)$ points along the cone's axis. Since the radius of the two spheres differ by $2$ , while their centres are a distance $\sqrt{105}$ apart, we see that the semi-vertical angle $\alpha$ of the cone is $\alpha \; = \; \sin^{-1}\tfrac{2}{\sqrt{105}}$ The point $X$ has coordinates $(10\lambda,2\lambda,\lambda)$ where $90\lambda - 10\lambda = 260$ , so that $\lambda = \tfrac{13}{4}$ . Thus $OX \; =\; (\lambda + \tfrac12)\sqrt{105} \; = \; \tfrac{15}{4}\sqrt{105}$ Finally $\mathbf{n} \; =\; \left(\begin{array}{c} 9 \\ 0 \\ -10 \end{array} \right)$ is normal to the plane $P$ . Considering the scalar product $\mathbf{u} \cdot \mathbf{n} = 80$ , we deduce that $\beta \; = \; \cos^{-1}\tfrac{80}{\sqrt{181 \times 105}}$ Thus we calculate the desired area to be $\pi A^2\sqrt{1-E} \; = \; \tfrac{5000}{473} \sqrt{\tfrac{54843}{473}} \pi \; = \; 357.593$