IMO 1982, Problem 6. Let $S$ be a square with sides of length 100. Let $L$ be a path within $S$ that does not meet itself and that is composed of linear segments $A_0A_1, A_1A_2, \ldots, A_{n-1}A_n$ with $A_0\neq A_n$. Suppose that for every point $P$ of the boundary of $S$ there is a point of $L$ at a distance from $P$ not greater than $1/2$. Prove that there are two points $X$ and $Y$ in $L$ such that the distance between $X$ and $Y$ is not greater than $1$ and the length of the part of $L$ that lies between $X$ and $Y$ is not smaller than 198.
Attempt.
I am able to prove a weaker result. What I can show is that there are two points $X$ and $Y$ in $L$ such that the distance between $X$ and $Y$ is at most $2$ (the problem asks for $1$) and the length of the part of the path between $X$ and $Y$ is at least $50$ (this '50' is probably improvable).
Assume that the path has unit speed. Let $P=(0, 0)$, $Q=(100, 0)$, $R=(100, 100)$ and $S=(0, 100)$ be the four corners of the square. For simplicity, assume that $A_0=P$, that is, the path starts at $P$. Assume on the contrary that for any two points $X$ and $Y$ on the path with $d(X, Y)\leq 2$ we have the property that the length of the path between $X$ and $Y$ is less than $50$.
Assume without loss of generality that the path reaches the $1/2$-neighborhood of $(50, 0)$ before it reaches the $1/2$-neighborhood of $(0, 50)$. Let $0\leq v\leq 100$ be such that $(0, v)$ is a point closest to $(0, 50)$ with the property that the path has reached the $1/2$-neighborhood of $(0, v)$ prior to reaching the $1/2$-neighborhood of $(50, 0)$.
Arguing somewhat colloquially, let us look at the journey starting from a point $P$ near $(0, v)$ (within a distance of $1/2$). At some point of time the path must reach withing $1/2$ distance of a point (on the left vertical side of the square) that is a distance $1$ away from $v$. Thus the path reaches near $(50, 0)$ and must comes back to a point near $P$ within a distance of at most $2$ in time less than $50$ units (because of our assumption). But this is impossible by the Pythagoras theorem.
One can give a similar argument without assuming that the path starts at a corner. This argument does not use the hypothesis that the path does not self-intersect.
