Rectangular room $A B C D$ has mirrors on walls $A B$ and $D C$. A laser is placed at $B$. It is aimed at $E$ and the beam reflects off of the mirrors at $E, F$ and $G$, arriving at $H$. The laws of physics tell us that $\angle B E C=\angle F E G$ and $\angle B F E=\angle A F G$ and $\angle F G E=\angle H G D$. If $A B=18 \mathrm{~m}, B C=10 \mathrm{~m}$ and $H D=6 \mathrm{~m}$, what is the total length of the path $B E F G H$ travelled by the laser beam?

![](https://cdn.mathpix.com/cropped/2024_04_17_a167ac2f1051780ff733g-2.jpg?height=387&width=699&top_left_y=476&top_left_x=1293)
Let points $P$ and $Q$ be on $A B$ so that $G P$ and $E Q$ are perpendicular to $A B$.

Let point $R$ be on $D C$ so that $F R$ is perpendicular to $D C$.

![](https://cdn.mathpix.com/cropped/2024_04_30_be08db732f430ab57efag-03.jpg?height=377&width=699&top_left_y=928&top_left_x=775)

Note that each of $\triangle B C E, \triangle E Q B, \triangle E Q F, \triangle F R E, \triangle F R G$, and $\triangle F P G$ is right-angled and has height equal to $B C$, which has length $10 \mathrm{~m}$.

Since $\angle B E C=\angle F E G$, then the angles of $\triangle B C E$ and $\triangle F R E$ are equal. Since their heights are also equal, these triangles are congruent.

Since $F R E Q$ and $B C E Q$ are rectangles (each has four right angles) and each is split by its diagonal into two congruent triangles, then $\triangle E Q F$ and $\triangle E Q B$ are also congruent to $\triangle B C E$. Similarly, $\triangle F P G$ and $\triangle F R G$ are congruent to these triangles as well.

Let $C E=x \mathrm{~m}$. Then $E R=R G=E C=x \mathrm{~m}$.

Since $\triangle H D G$ is right-angled at $D$ and $\angle H G D=\angle F G E$, then $\triangle H G D$ is similar to $\triangle F R G$.

Since $H D: F R=6: 10$, then $D G: R G=6: 10$ and so $D G=\frac{6}{10} R G=\frac{3}{5} x \mathrm{~m}$.

Since $A B=18 \mathrm{~m}$ and $A B C D$ is a rectangle, then $D C=18 \mathrm{~m}$.

But $D C=D G+R G+E R+E C=\frac{3}{5} x+3 x=\frac{18}{5} x \mathrm{~m}$.

Thus, $\frac{18}{5} x=18$ and so $x=5$.

Since $x=5$, then by the Pythagorean Theorem in $\triangle B C E$,

$$
B E=\sqrt{B C^{2}+E C^{2}}=\sqrt{(10 \mathrm{~m})^{2}+(5 \mathrm{~m})^{2}}=\sqrt{125 \mathrm{~m}^{2}}=5 \sqrt{5} \mathrm{~m}
$$

Now $F G=E F=B E=5 \sqrt{5} \mathrm{~m}$ and $G H=\frac{6}{10} F G=\frac{3}{5}(5 \sqrt{5} \mathrm{~m})=3 \sqrt{5} \mathrm{~m}$.

Therefore, the length of the path $B E F G H$ is $3(5 \sqrt{5} \mathrm{~m})+(3 \sqrt{5} \mathrm{~m})$ or $18 \sqrt{5} \mathrm{~m}$.

ANSWER: $18 \sqrt{5} \mathrm{~m}$