A regular tetrahedron has four vertices labeled $1, 2, 3$, and $4$. An arbitrary point $P$ is reflected over vertex $1$, resulting in the image $P_{1}$. This image is then reflected over vertex $2$, resulting in $P_{12}$. Next, $P_{12}$ is reflected over vertex $3$, resulting in $P_{123}$. Finally, $P_{123}$ is reflected over vertex $4$, resulting in $P_{1234}$. How can the result of the four reflections be obtained more simply? How does the result change with the 24 permutations of the reflections?
I. solution. a) According to a well-known theorem from the I. textbook, $P_{12}$ can be obtained from $P$ more simply, in a single step, by translating it by twice the vector $\overrightarrow{12}$ (read: one, two vector) (because in general the segment 12 is the midline of the triangle $P \quad P_{1} P_{12}$, but this is also valid if $P$ is chosen on the line 12).

Similarly, $P_{1234}$ can be obtained from $P_{12}$ by translating it by twice the vector $\overrightarrow{34}$, so $P_{1234}$ can be obtained from $P$ in a single step by translating it by twice the vector sum $\overrightarrow{12}+\overrightarrow{34}$.

b) Let the permutation of the 4 reflection centers be $A, B, C, D$, and let the vector $t_{A B C D}=2(\overrightarrow{A B}+\overrightarrow{C D})$ (Fig. 1) characterize the transformation that takes $P$ to $P_{A B C D}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_b53fc13f8d3f78afe8bdg-1.jpg?height=1098&width=979&top_left_y=497&top_left_x=562)

The lengths of the two components are always equal because the edges of the regular tetrahedron are equal, and they are always perpendicular to each other because $C A=C B$ and $D A=D B$ imply that $C$ and $D$ lie in the perpendicular bisector plane of the edge $A B$. Thus, the vector parallelogram specializes to a square, and the magnitude of $t$ is $2 \sqrt{2} a$, where $a$ is the edge length of the tetrahedron. - Since addition is commutative, it is immediately clear that $t_{C D A B}=t_{A B C D}$.

Turning to the direction of the vector, it is also easy to see that the vectors for the center sequences $D, C, B, A$ and $B, A, D, C$ are obviously the negative of the previous one. Let $F_{X Y}$ denote the midpoint of the edge $X Y$, and let $X$ and $Y$ run through the letters $A, B, C, D$. Thus, $2 \cdot \overrightarrow{A B}=4 \cdot \overrightarrow{F_{A C} F_{B C}}, 2 \cdot \overrightarrow{C D}=4 \cdot \overrightarrow{F_{A C} F_{A D}}$, and $t_{A B C D}=4 \cdot \overrightarrow{F_{A C} F_{B D}}=8 \cdot \overrightarrow{F_{A C} O}=8 \cdot \overrightarrow{0 F_{B D}}$, where $O$ is the center of the tetrahedron, which is also its centroid, since these 4 midpoints are the 4 vertices of a parallelogram (specifically a square).

From this, we see that the resulting translation vector can take 6 different values: $8 \cdot \overrightarrow{O F_{X} Y}$, and by swapping $A, C$ or $B, D$ in the sequence, or both, the result is the same. In the original notation, for example, $P_{1234}, P_{1432}, P_{3412}$, and $P_{3214}$ are identical.

II. solution. Reflect our tetrahedron $T$ onto its center $O$ (which, as shown in the first solution, bisects the segments connecting the midpoints of opposite edges). Let the vertices of the image $T^{\prime}$ be 5, 6, 7, 8 (Fig. 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_b53fc13f8d3f78afe8bdg-2.jpg?height=797&width=749&top_left_y=119&top_left_x=685)

Fig. 2

These 8 vertices together form the vertices of a cube. Break down the translation vectors $2 \cdot \overrightarrow{12}$ and $2 \cdot \overrightarrow{34}$ into components in the directions of the cube edges, so

$$
t_{1234}=2(\overrightarrow{12}+\overrightarrow{34})=2(\overrightarrow{18}+\overrightarrow{82}+\overrightarrow{36}+\overrightarrow{64})=4 \cdot \overrightarrow{82}
$$

since $\overrightarrow{36}=-\overrightarrow{18}$ and $\overrightarrow{64}=\overrightarrow{82}$. Finally, "symmetrize" the result so that on the one hand, the auxiliary vertex 8 does not appear in it, and on the other hand, the center 2 does not appear as an endpoint: take the vector equal to $\overrightarrow{82}$ but passing through $O$:

$$
t_{1234}=4 \cdot \overrightarrow{F_{13} F_{24}}=8 \cdot \overrightarrow{F_{13} O}=8 \cdot \overrightarrow{O F_{24}}
$$

the translation is four times the directed edge of the cube, or four times the directed edge of the tetrahedron, or four times the directed vertex axis of the regular octahedron, which is the common part of $T$ and $T^{\prime}$.

From this, we see that the independent or simultaneous swapping of the center pairs 1 and 3, and 2 and 4 in the reflection sequence does not change the result, and that the 24 permutations of the centers give 6 different results for the reflection sequence.

Remark. It is tempting to use this theorem: reflecting a figure successively over the vertices of the parallelogram $A B C D$, it returns to its original position (since $\overrightarrow{A B}+\overrightarrow{C D}=\overrightarrow{0}$). According to this, $P_{123}$ is identical to the reflection $P_{E}$ of $P$ over $E$, where $123 E$ is a parallelogram, so $\overrightarrow{P P_{E 4}}=2 \cdot \overrightarrow{E 4}$. We leave the symmetrization of the result, for example, in the perpendicular bisector plane of the edge 13, to the reader.