Reflecting off the walls of a tetrahedron

Start with the tetrahedron $ABCD$ where $A\;(4\sqrt{3},0,0)$ , $B\;(-2\sqrt{3},6,0)$ , $C\; (-2\sqrt{3},-6,0)$ and $D\;(0,0,4\sqrt{6})$ , and the points $P\;(1,1,2)$ and $Q\;(0,0,5)$ .

Reflect $D$ and $Q$ in $ABC$ to obtain $D_1$ and $Q_1$ , and the tetrahedron $ABCD_1$ , where $D_1 \; (0,0,-4\sqrt{6}) \hspace{1cm} Q_1\;(0,0.-5)$ Reflect $C$ and $Q_1$ in $ABD_1$ to obtain $C_1$ and $Q_2$ , and the tetrahedron $ABC_1D_1$ , where $C_1 \; \big(\tfrac{10}{\sqrt{3}}, 10, -\tfrac{8\sqrt{2}}{\sqrt{3}}\big) \hspace{1cm} Q_2 \; \big(\tfrac{2}{27}(5\sqrt{2}-8\sqrt{3}), \tfrac29(5\sqrt{6} - 24), -\tfrac19(35+8\sqrt{6})\big)$ Then reflect $A$ and $Q_2$ in $BC_1D_1$ to obtain $A_1$ and $Q_3$ , and the tetrahedron $A_1BC_1D_1$ , where $A_1 \; \big(-\tfrac{28}{3\sqrt{3}}, \tfrac{32}{3}, -\tfrac{40\sqrt{2}}{3\sqrt{3}}\big) \hspace{1cm} Q_3 \; \big(-\tfrac{2}{27}(8\sqrt{3}-5\sqrt{2}), \tfrac{10}{27}(24-5\sqrt{6}), -\tfrac{1}{27}(55 + 64\sqrt{6})\big)$ Finally reflect $B$ and $Q_3$ in $A_1C_1D_1$ to obtain $B_1$ and $Q_4$ , and the tetrahedron $A_1B_1C_1D_1$ , where $B_1 \; \big(\tfrac{58}{9\sqrt{3}}, \tfrac{70}{9}, -\tfrac{200\sqrt{2}}{9\sqrt{3}}\big) \hspace{1cm} Q_4 \; \big(\tfrac{8}{81}(8\sqrt{3}-5\sqrt{2}), \tfrac{32}{81}(24 - 5\sqrt{6}), -\tfrac{1}{81}(392\sqrt{6}-85)\big)$ The length of the laser beam is therefore

$PQ_4 \;=\; \tfrac19\sqrt{19451 + 80\sqrt{2} - 128\sqrt{3} - 2032\sqrt{6}} \; \approx \; \boxed{13.3172}$

The vertices of the tetrahedron as specified in the problem are $A( 4 \sqrt{3}, 0, 0) , B ( -2 \sqrt{3}, 6, 0 ) , C( -2 \sqrt{3} , -6, 0 ), D ( 0, 0, 4 \sqrt{6} )$ .

The light starting at $P(1, 1, 2)$ travels in sequence to points $P_1$ on $ABC$ , $P_2$ on $ABD$ , $P_3$ on $BCD$ , $P_4$ on $CAD$ , and finally reflects from $P_4$ to point $Q(0, 0, 5)$ . We want to determine the length

$d = \overline{PP_1} + \overline{P_1 P_2} + \overline{P_2 P_3} + \overline{P_3 P_4 } + \overline{P_4 Q }$

Since we don't know $P_1, P_2, P_3, P_4$ ahead of time, what we can do is reflect $Q$ about plane $CAD$ to the point $Q_1$ , then the vector $P_3 P_4$ will be along the vector $P_4 Q_1$ , and therefore,

$\overline{P_3 P_4 } + \overline{P_4 Q } = \overline{P_3 P_4 } + \overline{P_4 Q_1 } = \overline{P_3 Q_1 }$

applying the same reasoning, by reflecting $Q_1$ into plane $BCD$ , to obtain $Q_2$ , it follows that

$\overline{P_2 P_3 } + \overline{P_3 Q_1 } = \overline{P_2 P_3 } + \overline{P_3 Q_2 } = \overline {P_2 Q_2 }$

and similarly, reflecting $Q_2$ into plane $ABD$ to $Q_3$ , leads to

$\overline{P_1 P_2} + \overline{P_2 Q_2} = \overline{P_1 Q_3}$

and one final time, reflecting $Q_3$ into plane $ABC$ to $Q_4$ , leads to

$\overline{P P_1} + \overline{P_1 Q_3} = \overline{P Q_4}$

The right hand side is the desired distance travelled by the laser beam.

Now, how do you reflect a point about a plane ? If the plane passes through a point $r_p$ and has a normal vector $n_p$ then the reflection of a point $p_0$ is the point $p_1$ given by

$p_1 = r_p + M (p_0 - r_p)$

where $M$ is the $3 \times 3$ matrix defined by $\displaystyle M = I_3 -2 \dfrac{n_p n_p^T }{ n_p^T n_p}$ , where $I_3$ is the $3 \times 3$ identity matrix.

What remains is to find the normals of planes $ABC, ABD, BCD, CAD$ . For plane $ABC$ , the normal is $n_1 = (0, 0, 1)$

and for planes $ABD, BCD, CAD$ the normal vector is $n_i = ( \sin \theta \cos \phi_i, \sin \theta \sin \phi_i, \cos \theta ), \hspace{6pt} i = 2, 3, 4$ .

where $\theta = \cos^{-1}(\frac{1}{3} )$ and $\phi_i = \dfrac{\pi}{3} + (i- 2)\dfrac{2 \pi}{3}$ , and $i = 2$ corresponds to plane $ABD$ , $i=3$ corresponds to plane $BCD$ and $i = 4$ corresponds to plane $CAD$ . Points on these planes can be taken as:

$r_1 = A , r_2 = r_3 = r_4 = D$

To recap, we'll reflect point $Q (0, 0, 5)$ in about $CAD, BCD, ABD, ABC$ in this order, to obtain $Q_1, Q_2, Q_3, Q_4$ respectively,

If $Q_0 = Q$ then

$Q_{k+1} = r_{4 - k} + M_{4-k} ( Q_k - r_{4 - k} ), \hspace{12pt} k = 0, 1, 2, 3$

and our desired answer will be $\overline{P Q_4 }$ .