Team selection test for 50. IMO

Denote by $D(Z,r)$ and $S(Z,r)$ the open disc and the circle with center $Z$ and radius $r$, respectively. Let $A=A_X$ be the set of points that can be reached from $X$ by finite number of jumps (possibly zero). We have to show that $A={\mathbb{R}}^2$.

First, we shall prove that $D(X,2r_X)\subset A$. We may assume that $X=O$ and $r_0=1$. Set $a_n=2^{-n}$, $b_n=2-a_n$ and $A_n=\{X\in {\mathbb{R}}^2:a_n\le |OX|\le b_n\}$, $n\in {\mathbb{N}}_0$. It is enough to show that $A_n\subset A$.

We shall proceed by induction. For $n=0$, this follows from the given condition. Suppose that $A_k\subset A$. It suffices to prove that $Y\in A$ for $Y\in [a_{k+1},a_k)\cup (b_k,b_{k+1}]$ and then apply rotation. Let $f(Z)=f_Y(Z)=|YZ|-r_Z$. It follows by the condition of the problem that $1-a_{k+1}\le r_{a_k}\le 1+a_{k+1}$ and $a_{k+1}\le r_{b_k}\le b_{k+1}$. So, if $Y\in [a_{k+1},a_k)$, then $f(a_k)<a_k-1\le 0<f(-b_k)$, and if $Y\in (b_k,b_{k+1}]$, then $f(b_k)<0<1-a_{k+1}<f(-a_k)$. Since $A_k$ is a linearly connected set and $f$ is a continuous function, we get that $f(Z)=0$ for some $Z\in A_k$ and hence $Y\in A$.

Assume now that $A\ne {\mathbb{R}}^2$. Then $\sup \{r:D(X,r)\subset A\}=R<\infty$. Let $m=\inf \{r_Y:Y\in D(X,R)\}$. Since $r$ is a continuous function, then $m>0$ and, by the proved above, $D(Y,2m)\subset A$ for $Y\in D(X,R)$. Hence $D(X,R+2m)\subset A$, a contradiction.