Disc in the corner

The circular disc is the incircle of the triangle formed by the extension of the plane that is on and the $xy$ , $xz$ , and $yz$ planes.

Since the unit normal vector to the plane of the disc is $(\sin 60° \cos 30°, \sin 60° \sin 30°, \cos 60°) = (\frac{3}{4}, \frac{\sqrt{3}}{4}, \frac{1}{2})$ , 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 $(C_x, C_y) = (\frac{2}{3}d, \frac{4}{\sqrt{3}}d)$ . Using the incenter equation $(\frac{ax_1 + bx_2 + cx_3}{a + b + c}, \frac{ay_1 + by_2 + cy_3}{a + b + c})$ , the incenter of $\triangle ABC$ solves to $(\frac{8\sqrt{13} + 8}{3(\sqrt{21} + \sqrt{13} + 4)}d, \frac{16}{\sqrt{3}(\sqrt{21} + \sqrt{13} + 4)}d)$ . Since the radius of the circle is $5$ , $\frac{16}{\sqrt{3}(\sqrt{21} + \sqrt{13} + 4)}d = 5$ , so $d = \frac{5\sqrt{3}(\sqrt{21} + \sqrt{13} + 4)}{16}$ , which makes $AT$ and the $x$ -coordinate of the incenter $\frac{5\sqrt{3}}{6}(\sqrt{13} + 1)$ .

Moving back to the three-dimensional picture with $d = \frac{5\sqrt{3}(\sqrt{21} + \sqrt{13} + 4)}{16}$ , we can calculate that $A$ is $(\frac{5\sqrt{3}}{12}(\sqrt{21} + \sqrt{13} + 4), 0, 0)$ , $B$ is $(0, \frac{5}{4}(\sqrt{21} + \sqrt{13} + 4), 0)$ , and the equation of the line $AB$ is $y = -\sqrt{3}x + \frac{5}{4}(\sqrt{21} + \sqrt{13} + 4)$ . Since $T$ is on $AB$ , $T_y = -\sqrt{3}T_x + \frac{5}{4}(\sqrt{21} + \sqrt{13} + 4)$ and since $AT$ is $\frac{5\sqrt{3}}{6}(\sqrt{13} + 1)$ , $\sqrt{(T_x - \frac{5\sqrt{3}}{12}(\sqrt{21} + \sqrt{13} + 4))^2 + T_y^2} = \frac{5\sqrt{3}}{6}(\sqrt{13} + 1)$ , and this solves to $(T_x, T_y) = (\frac{5}{4}(\sqrt{3} + \sqrt{7}), \frac{5}{4}(1 + \sqrt{13}))$ .

Therefore, $a = T_x = \frac{5}{4}(\sqrt{3} + \sqrt{7})$ and $b = T_y = \frac{5}{4}(1 + \sqrt{13})$ , and $a + b = \frac{5}{4}(1 + \sqrt{3} + \sqrt{7} + \sqrt{13}) \approx \boxed{11.229}$ .

The unit normal specified by the problem is

$\hat{n} = ( \sin \frac{\pi}{3} \cos \frac{\pi}{6}, \sin \frac{\pi}{3} \sin \frac{\pi}{6} , \cos \frac{\pi}{3})$

Let $\hat{u}_1$ and $\hat{u}_2$ be two mutually orthogonal unit vectors that are orthogonal to $\hat{n}$ .

Then points on the perimeter of the disc can be parameterized as:

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

where $\vec{r_C} = (x_C, y_C, z_C)$ is the position vector of the center of the circle. Tangency to the three coordinate planes implies that, for the $xy$ plane, we're looking for $t_1 \in [0, 2\pi )$ , such that,

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

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

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

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,

$z_C = \sqrt{ {u_{1z}}^2 + {u_{2z}}^2 }$

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

the value of $t_1$ into the parametric equation of the circle.

${\vec{r}}^* = \vec{r_C} + \cos t_1 \hspace{4pt} \hat{u}_1 + \sin t_1 \hspace{4pt} \hat{u}_2$

And this gives the required result, ${\vec{r}}^* = (5.4722526, 5.756939, 0)$ making the answer $(5.4722526+5.756939) = \boxed{11.2291916}$