Question

Let D be the union of $n\ge 1$ concentric circles in the plane. Suppose that the function $f:D\to D$ satisfies $d(f\left(A\right),f\left(B\right))\ge d(A,B)$ for every A, B $ϵ$ D (d(M, N) is the distance between the points M and N). Then for every A, B$ϵ$ D

• A. f is an isometry.
• B. $d(f\left(A\right),f\left(B\right))\ge d(A,B)$
• C. $d(f\left(A\right),f\left(B\right))\le d(A,B)$
• D. None of these

Answer 1

Answer: (A), (B)

The correct options are
A f is an isometry.
B $d(f\left(A\right),f\left(B\right))\ge d(A,B)$

Let $D_1,D_2,...D_n$ be the concentric circles, with radi $r_1<r_2<...<r_n$ and center $O$. We will denote $f\left(A\right)=A^{\prime }$, for an arbitrary point $AD$.We first notice that if $A,BDn$, we have $A^{\prime }{C}^{\prime 2}+B^{\prime }{C}^{\prime 2}\ge A{C}^2+B{C}^2=A{B}^2=A^{\prime }{B}^{\prime 2}.$ Because $O{C}^{\prime }$ is a median of the triangle $A^{\prime }{B}^{\prime }{C}^{\prime }$. it follows that $O{C}^{\prime 2}=\frac{1}{2}(A^{\prime }{C}^{\prime 2}+B^{\prime }{C}^{\prime 2})-\frac{1}{4}{A}^{\prime }{B}^{\prime 2}=r_n^2,$ hence $C^{\prime }Dn$ is an isometry. Now take $A,X,Y,Z$ on$Dn$ such that $AX=AY=A^{\prime }Z$. It follows that $A^{\prime }{X}^{\prime }=A^{\prime }{Y}^{\prime }=A^{\prime }Z$, hence one of the points $X^{\prime },Y^{\prime }$ coincides with $Z$. This shows that $f\left(D_n\right)=D_n$ and since f is clearly injective it results in the same way that $f\left(D_i\right)=D_i$ ,for all $i,l\le i\le n-1,$ and that all restrictions $f\mid D_i$ are isometries. Next we prove that distances between adjacent circles, say $D_1$ and $D_2$ are preserved. Take $A,B,C,D$ on $D_1$ such that $ABCD$ is a square and let $A^{\prime },B^{\prime },C^{\prime },D^{\prime }$, be the point on $D_2$ closest to $A,B,C,D$, respectively. Then $A^{\prime }{B}^{\prime }{C}^{\prime }{D}^{\prime }$ is also a square and the distance from $Ato{C}^{\prime }$ is the maximum between any point on $D_1$ and any point on $D_2$. Hence the eight points maintain their relative position under f and this shows that $f$is an isometry.