Problem 10. In a triangular pyramid $ABCD$ with base $ABC$, the lateral edges are pairwise perpendicular, $DA=DB=2, DC=5$. A light ray is emitted from a point on the base. After reflecting exactly once from each lateral face (the ray does not reflect from the edges), the ray hits a point on the base of the pyramid. What is the shortest distance the ray could have traveled?
Answer: $\frac{10 \sqrt{6}}{9}$.

Solution. Let $N$ be the point from which the ray is emitted, and $K$ be the point on the base of the pyramid where the ray will finally hit.

We will sequentially "straighten" the path of the ray as follows: if at some moment the ray should reflect off a certain plane, we will assume that it continues its path, but in a "mirrored" copy of the pyramid. Note also that due to the perpendicularity of the edges $D A, D B$, and $D C$, the planes $(A B D), (B C D)$, and $(C A D)$ remain in place when reflected relative to each other. Therefore, the result of our "straightening" will be the segment $N K_{1}$, where $K_{1}$ is the point obtained from $K$ after sequential reflections relative to the planes $(A B D), (B C D)$, and $(C A D)$ in some order. The length of the segment $N K_{1}$ is exactly equal to the total distance traveled by the ray.

We introduce coordinates: $D$ is the origin, and $D A, D B$, and $D C$ are the directions of the axes. Then the symmetry relative to each of the specified planes changes the sign of exactly one of the coordinates of the point. Consequently, the point $K_{1}$ is obtained from the point $K$ by changing the signs of all coordinates, i.e., by symmetry relative to the point $D$. Therefore, the point $K_{1}$ lies on the image of the plane $A B C$ under symmetry relative to the point $D$.

![](https://cdn.mathpix.com/cropped/2024_05_06_a567280d18f9fc82386fg-09.jpg?height=711&width=691&top_left_y=1118&top_left_x=702)

The length of any such segment does not exceed the distance between these planes. It is easy to see that this distance is equal to twice the distance from the point $D$ to the plane $A B C$. We can reason in different ways, but we will present only one of them.

The distance from the point $D$ to the plane $A B C$ is equal to the height $h$ of the pyramid $A B C D$ dropped from the point $D$. The volume of the pyramid $A B C D$ is then, on the one hand, $V_{A B C D} = \frac{1}{3} S_{A B C} h$, and on the other hand, $V_{A B C D} = \frac{1}{6} \cdot D A \cdot D B \cdot D C = \frac{10}{3}$. It remains to find $S_{A B C}$. This can also be done in several ways.

For example: by the Pythagorean theorem, $C A = C B = \sqrt{2^2 + 5^2} = \sqrt{29}$, $A B = \sqrt{2^2 + 2^2} = 2 \sqrt{2}$. In the isosceles triangle $A C B$, it is easy to compute $\cos \angle C A B = \frac{A B / 2}{A C} = \sqrt{\frac{2}{29}}$, from which $\sin \angle C A B = \sqrt{\frac{27}{29}} = \frac{3 \sqrt{3}}{\sqrt{29}}$, and

$$
S_{A B C} = \frac{1}{2} \cdot A C \cdot A B \cdot \sin \angle C A B = 3 \sqrt{6}
$$

Therefore, $h = \frac{3 V_{A B C D}}{S_{A B C}} = \frac{10}{3 \sqrt{6}} = \frac{5 \sqrt{6}}{9}$, from which it follows that the length of the path of any ray is not less than $\frac{10 \sqrt{6}}{9}$.

It remains to provide an example where the length we found is achieved. For this, it is sufficient to present a point on the plane $A B C$ whose perpendicular from $D$ falls inside the triangle symmetric to $A B C$ relative to the point $D$.

Indeed, then the coordinates of the point $K_{1}$ will be negative, so the perpendicular will intersect all coordinate planes. As the point $N$, almost any point in the neighborhood of the base of the perpendicular from the point $D$ to the plane $(A B C)$ will do. Almost any, since we need the perpendicular $N K_{1}$ not to pass through the coordinate axes.

Comment. The curious reader may notice that the situation described in the problem is taken from real life - it refers to a corner reflector, also known as a cataphote.

## Variant III