Brazilian Mathematical Olympiad

Observe first that the images of equilateral triangles of side $1$ are equilateral triangles of side $1$. Consider two equilateral triangles $ABC$ and $A'B'C$ with side $1$ and a common side. Notice that $AA' = \sqrt{3}$ and $f(A)f(A') = 0$ or $\sqrt{3}$. If $f(A) = f(A')$, then consider $B$ such that $AB = 1$ and $A'B = \sqrt{3}$, so $f(A)f(B) = 1 \Leftrightarrow f(A')f(B) = 1$, contradiction. So $f(A)f(A') = \sqrt{3}$ and all points $\sqrt{3}$ apart are taken to two points $\sqrt{3}$ apart.

So any triangular lattice is taken to a triangular lattice. In particular, any triangle with sides $1$, $\sqrt{n^2-n+1}$ and $\sqrt{n^2+n+1}$ are preserved by $f$.

Consider two different triangles $ABC$ and $AB'C$ such that $B'C = BC = 1$, $AB = AB' = \sqrt{n^2-n+1}$ and $AC = \sqrt{n^2+n+1}$. Notice that $BB' = \sqrt{\frac{3}{n^2+n+1}}$. Let $\epsilon_n = \sqrt{\frac{3}{n^2+n+1}}$. Again, $f(B)f(B') = 0$ or $\epsilon_n$. If $B = A_0$ and $B' = A_1$ and $f(B) = f(B')$, let $k$ be an integer such that $k\epsilon_n < 1 \le (k+1)\epsilon_n$ and $A_2, A_3, \dots, A_{k+1}$ points such that $A_iA_{i+1} = \epsilon_n$, $i = 0, 1, 2, \dots, k$, and $A_0A_{k+1} = 1$. We have $f(A_0)f(A_{k+1}) = 1$, so by the triangle inequality,

$$1 = f(A_0A_{k+1}) \le f(A_0)f(A_1) + f(A_1)f(A_2) + \dots + f(A_k)f(A_{k+1}) \\ \le f(A_0)f(A_1) + k\epsilon_n < 1,$$

which is a contradiction. So $f(A_0)f(A_1) = \epsilon_n$ for all points $A_0, A_1$, $\epsilon_n$ apart. Moreover, if $AB = m\epsilon_n$, $m$ positive integer, $f(A)f(B) = m\epsilon_n$.

Now let $X$ and $Y$ be two arbitrary points and suppose that $f(X)f(Y) \ne XY$. Choose $n$ such that $|XY - f(X)f(Y)| > 4\epsilon_n$ and $P$ such that $PX = m\epsilon_n$, $m$ integer, and $PY = 2\epsilon_n$. Then $f(X)f(P) = XP = m\epsilon_n$, $f(Y)f(P) = 2\epsilon_n$ and $|f(X)f(Y) - XY| \le |f(X)f(Y) - f(P)f(X)| + |f(P)f(X) - XY| =$

$|f(X)f(Y) - f(P)f(X)| + |PX - XY| \le f(Y)f(P) + PY = 4\epsilon_n$, contradiction.

So $f(X)f(Y) = XY$ for all points $X$, $Y$.