Elliptical disc in the corner

The elliptical disc is tangent to the three sides of the triangle formed by the extension of the plane that is on the $xy$ , $xz$ , and $yz$ planes:

Since the unit normal vector to the plane of the disc is $(\sin \frac{\pi}{3} \cos \frac{\pi}{6}, \sin \frac{\pi}{3} \sin \frac{\pi}{6}, \cos \frac{\pi}{3}) = (\frac{3}{4}, \frac{\sqrt{3}}{4}, \frac{1}{2})$ , the equation of the plane is $\frac{3}{4}x + \frac{\sqrt{3}}{4}y + \frac{1}{2}z = d$ , and its $x$ -, $y$ -, and $z$ -intercepts ( $A$ , $B$ , and $C$ ) solve to $(\frac{4}{3}d, 0, 0)$ , $(0, \frac{4}{\sqrt{3}}d, 0)$ , and $(0, 0, 2d)$ , which by Pythagorean's Theorem makes the sides of the triangle $\frac{8}{3}d$ , $\frac{2\sqrt{21}}{3}d$ , and $\frac{2\sqrt{13}}{3}d$ .

Now move $\triangle ABC$ to a Cartesian coordinate system with $A$ at $(0, 0)$ and $B$ at $(\frac{8}{3}d, 0)$ :

$C$ is $\frac{2\sqrt{13}}{3}d$ away from $A$ so by the distance formula $\sqrt{C_x^2 + C_y^2} = \frac{2\sqrt{13}}{3}d$ , and $C$ is $\frac{2\sqrt{21}}{3}d$ away from $B$ so $\sqrt{(C_x - \frac{8}{3}d)^2 + C_y^2} = \frac{2\sqrt{21}}{3}d$ , and this solves to $(C_x, C_y) = (\frac{2}{3}d, \frac{4}{\sqrt{3}}d)$ . This means the equation of the line $AC$ is $y = 2\sqrt{3}x$ and the equation of the line $BC$ is $y = -\frac{2}{\sqrt{3}}(x - \frac{8}{3}d)$ .

The steepest descent of the plane is the altitude of $\triangle ABC$ , so the ellipse is rotated $\frac{\pi}{6}$ counterclockwise from the vertical. Since the semi-major axis is $5$ and the semi-minor axis is $2$ , the equation of the ellipse before the rotation is $\frac{(x - h)^2}{4} + \frac{(y - k)^2}{25} = 1$ for center $(h, k)$ , and is $\frac{79(x - h)^2}{400} + \frac{21\sqrt{3}(x - h)(y - k)}{200} + \frac{37(y - k)^2}{400} = 1$ after the rotation. Using implicit differentiation, the slope at any point on the ellipse solves to $m = \frac{-79(x - h) - 21\sqrt{3}(y - k)}{21\sqrt{3}(x - h) + 37(y - k)}$ .

The ellipse is tangent to $y = 0$ which has a slope of $m = 0$ at $y = 0$ . Then $\frac{79(x - h)^2}{400} + \frac{21\sqrt{3}(x - h)(y - k)}{200} + \frac{37(y - k)^2}{400} = 1$ and $m = \frac{-79(x - h) - 21\sqrt{3}(y - k)}{21\sqrt{3}(x - h) + 37(y - k)}$ with $m = 0$ and $y = 0$ solves to $k = \frac{\sqrt{79}}{2}$ .

The ellipse is also tangent to $y = 2\sqrt{3}x$ which has a slope of $m = 2\sqrt{3}$ . Then $\frac{79(x - h)^2}{400} + \frac{21\sqrt{3}(x - h)(y - k)}{200} + \frac{37(y - k)^2}{400} = 1$ and $m = \frac{-79(x - h) - 21\sqrt{3}(y - k)}{21\sqrt{3}(x - h) + 37(y - k)}$ with $k = \frac{\sqrt{79}}{2}$ , $m = 2\sqrt{3}$ , and $y = 2\sqrt{3}x$ solves to $h = \frac{1}{12}(5\sqrt{93} + \sqrt{237})$ .

The ellipse is also tangent to $y = -\frac{2}{\sqrt{3}}(x - \frac{8}{3}d)$ which has a slope of $m = -\frac{2}{\sqrt{3}}$ at $T$ . Then $\frac{79(x - h)^2}{400} + \frac{21\sqrt{3}(x - h)(y - k)}{200} + \frac{37(y - k)^2}{400} = 1$ and $m = \frac{-79(x - h) - 21\sqrt{3}(y - k)}{21\sqrt{3}(x - h) + 37(y - k)}$ with $k = \frac{\sqrt{79}}{2}$ , $h = \frac{1}{12}(5\sqrt{93} + \sqrt{237})$ , $m = -\frac{2}{\sqrt{3}}$ and $y = -\frac{2}{\sqrt{3}}(x - \frac{8}{3}d)$ solves to $x = T_x = \frac{11}{2\sqrt{3}} + \frac{5\sqrt{93}}{12} + \frac{\sqrt{237}}{12}$ , $y = T_y = \frac{37\sqrt{3}}{2\sqrt{133}} + \frac{\sqrt{79}}{2}$ , and $d = \frac{1}{32}(5\sqrt{93} + 3\sqrt{133} + 4\sqrt{237})$ . We now know the coordinates of $B$ are $(\frac{1}{12}(5\sqrt{93} + 3\sqrt{133} + 4\sqrt{237}), 0)$ , and by the distance formula, $BT^2 = \frac{1}{152}(7307 + 37\sqrt{31521})$ .

Moving back to the three-dimensional picture, the equation of the line $BC$ is $z = \frac{\sqrt{3}}{2}y + 2d$ . Since $T$ is on $BC$ , $T_z = \frac{\sqrt{3}}{2}T_y + 2d$ and by the distance formula $(T_y - \frac{4}{\sqrt{3}}d)^2 + T_z^2 = BT^2$ , and since $d = \frac{1}{32}(5\sqrt{93} + 3\sqrt{133} + 4\sqrt{237})$ and $BT^2 = \frac{1}{152}(7307 + 37\sqrt{31521})$ , this solves numerically to $(T_y, T_z) \approx (3.198, 6.255)$ .

Therefore, $a = T_y \approx 3.198$ and $b = T_z \approx 6.255$ , and $a + b \approx 3.193 + 6.255 \approx \boxed{9.453}$ .

The unit normal specified by the problem is

$\hat{n} = ( \sin \theta \cos \phi, \sin \theta \sin \phi , \cos \theta)$

where $\theta = \frac{\pi}{3}$ and $\phi = \frac{\pi}{6}$ . The unit vector in the direction of the steepest descent is given by,

$\hat{u}_1 = ( \cos \theta \cos \phi, \cos \theta \sin \phi, -\sin \theta)$

In addition, let

$\hat{u}_2 = ( -\sin \phi, \cos \phi, 0 )$

Then the three unit vectors $\hat{u}_1 , \hat{u}_2$ and $\hat{n}$ make a right-handed reference frame.

The major axis of the ellipse makes an angle of $\psi = \dfrac{\pi}{6}$ with $\hat{u}_1$ . Thus, it is along the unit vector $\hat{w}_1$ , and the minor axis is along $\hat{w}_2$ , where

$[ \hat{w}_1 , \hat{w}_2 ] = [ \hat{u}_1 , \hat{u}_2 ] R(\frac{\pi}{6})$

and where,

$R(\frac{\pi}{6}) = \begin{bmatrix} \cos \frac{\pi}{6} && -\sin \frac{\pi}{6} \\ \sin \frac{\pi}{6} && \cos \frac{\pi}{6} \end{bmatrix}$

Now the points on the perimeter of the ellipse can be parameterized as:

$\vec{r}(t) = \vec{r_C} + a \cos t \hspace{4pt} \hat{w}_1 + b \sin t \hspace{4pt} \hat{w}_2$

where $a = 5$ is the semi-major axis length and $b = 2$ is the semi-minor axis length, and where $\vec{r_C} = (x_C, y_C, z_C)$ is the position vector of the center of the ellipse. Tangency to the three coordinate planes implies that, for the $xy$ plane, we're looking for $t_1 \in [0, 2\pi )$ , such that,

$0 = z_C + a \cos t_1 \hspace{4pt} {w_{1z}} + b \sin t_1 \hspace{4pt} {w_{2z}}$

We also require that $\dfrac{d\vec{r}}{dt}$ has a zero $z$ -component at $t=t_1$ , that is,

$0 = -a \sin t_1 \hspace{4pt} {w_{1z}} + b \cos t_1 \hspace{4pt} {w_{2z}}$

The last equation can be solved quite easily for two possible values of $t_1$ that are separated by $\pi$ .

To choose the correct one, we use the fact that $z_C$ is positive, and after some trigonometry, we deduce that,

$z_C = \sqrt{ a^2 {w_{1z}}^2 + b^2 {w_{2z}}^2 }$

Similar formulas apply to $x_C$ and $y_C$ . Finally, to find the tangency point with the $yz$ plane, we simply plug in

the value of $t_3$ into the parametric equation of the ellipse.

${\vec{r}}^* = \vec{r_C} + a \cos t_3 \hspace{4pt} \hat{w}_1 + b \sin t_3 \hspace{4pt} \hat{w}_2$

where, $-a \sin t_3 \hspace{4pt} {w_{1x}} + b \cos t_3 \hspace{4pt} {w_{2x}} = 0$

And this gives the required result, ${\vec{r}}^* = (0,3.19825, 6.25493)$ making the answer $(3.19825+6.25493) = \boxed{9.45318}$