XIX OBM

Observe first that the images of equilateral triangles of side $1$ are equilateral triangles of side $1$. Consider two equilateral triangles $ABC$ and $A^{\prime }{B}^{\prime }C$ with side $1$ and a common side. Notice that $A{A}^{\prime }=\sqrt{3}$ and $f(A)f(A^{\prime })=0$ or $\sqrt{3}$. If $f(A)=f(A^{\prime })$, then consider $B$ such that $AB=1$ and $A^{\prime }B=\sqrt{3}$, so $f(A)f(B)=1\iff f(A^{\prime })f(B)=1$, contradiction. So $f(A)f(A^{\prime })=\sqrt{3}$ 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 $A{B}^{\prime }C$ such that $B^{\prime }C=BC=1$, $AB=A{B}^{\prime }=\sqrt{n^2-n+1}$ and $AC=\sqrt{n^2+n+1}$. Notice that $B{B}^{\prime }=\sqrt{\frac{3}{n^2+n+1}}$. Let ${ϵ}_n=\sqrt{\frac{3}{n^2+n+1}}$. Again, $f(B)f(B^{\prime })=0$ or ${ϵ}_n$. If $B=A_0$ and $B^{\prime }=A_1$ and $f(B)=f(B^{\prime })$, let $k$ be an integer such that $k{ϵ}_n<1\le (k+1){ϵ}_n$ and $A_2,A_3,\dots ,A_{k+1}$ points such that $A_i{A}_{i+1}={ϵ}_n$, $i=0,1,2,\dots ,k$, and $A_0{A}_{k+1}=1$. We have $f(A_0)f(A_{k+1})=1$, so by the triangle inequality,

$$\begin{gathered} 1=f(A_0{A}_{k+1})\le f(A_0)f(A_1)+f(A_1)f(A_2)+\cdots +f(A_k)f(A_{k+1}) \\ \le f(A_0)f(A_1)+k{ϵ}_n<1, \end{gathered}$$

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

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

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

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