Library / Brazilian Mathematical Olympiad /3 of 120

Brazilian Mathematical Olympiad

Geometry Difficulty 4.5 AIME Prove it Brazil

The triangle with vertices $(0, 0)$ , $(0, 1)$ , $(2, 0)$ is repeatedly reflected in the three lines $AB$ , $BC$ , $CA$ where $A$ is $(0, 0)$ , $B$ is $(3, 0)$ , $C$ is $(0, 3)$ . Show that one of the images has vertices $(24, 36)$ , $(24, 37)$ and $(26, 36)$ .

Solution

Let $R$ be reflection in $BC$ : $(x, y) \rightarrow (3 - y, 3 - x)$ . Let $S$ be reflection in $AC$ : $(x, y) \rightarrow (-x, y)$ . Let $T$ be reflection in $AB$ : $(x, y) \rightarrow (x, -y)$ . Let $RS$ denote the reflection $R$ followed by the reflection $S$ . Then we have $Q = STR$ is $(x, y) \rightarrow (y + 3, x + 3)$ . So $Q^2$ is $(x, y) \rightarrow (x + 6, y + 6)$ , and $Q^{10}$ is $(x, y) \rightarrow (x + 30, y + 30)$ .
$P = TRS$ is $(x, y) \rightarrow (x, -y) \rightarrow (y+3, -x+3) \rightarrow (-y-3, -x+3)$ . So $P^2$ is $(x, y) \rightarrow (x-6, y+6)$ . Hence $Q^{10}P^2$ is $(x, y) \rightarrow (x+24, y+36)$ , which is the required translation.

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.