Let $f$ be a map of the plane into itself with the property that if $d(A,B)=1$, then $d(f(A),f(B))=1$, where $d(X,Y)$ denotes the distance between points $X$ and $Y$. Prove that for any positive integer $n$, $d(A,B)=n$ implies $d(f(A),f(B))=n$.
Problem
Source: Mongolia 1999 Teachers secondary level P6
Tags: fe, functional equation, functional geometry, geometry
jasperE3
20.06.2021 07:50
Bump.$~~~~~$
rama1728
09.08.2021 19:00
jasperE3 wrote: Let $f$ be a map of the plane into itself with the property that if $d(A,B)=1$, then $d(f(A),f(B))=1$, where $d(X,Y)$ denotes the distance between points $X$ and $Y$. Prove that for any positive integer $n$, $d(A,B)=n$ implies $d(f(A),f(B))=n$. I might be wrong, please correct me if I am.
Consider a line segment \(A_1A_n\) with \(d(A_1, A_n)=1\). Mark points \(A_2\), \(A_3\), \(\hdots\), \(A_{n-1}\) such that they equally divide \(A_1A_n\) into \(n\) parts. It therefore suffices to show that \(d(f(A_1), f(A_n))=1\). We show this result by strong induction, the base case being clear. So, \(d(f(A_1), f(A_{n-1})=n-1\) and also, notice that the points \(f(A_i)\) are all collinear, for \(0<i<n\). Now, consider the segment \(A_{n-2}A_n\). Note that \(d(f(A_{n-2}), f(A_n))=2\), and also \(f(A_{n-2}\), \(f(A_{n-1}\) and \(f(A_n)\) are collinear. Thus, \(d(f(A_1), f(A_n))=n\) and we are done.
jasperE3
09.08.2021 20:48
With strong induction you'll need another base case.
rama1728
09.08.2021 20:51
jasperE3 wrote: With strong induction you'll need another base case. For what?