Problem

Source: MOP 2005 Homework - Red Group #14

Tags: geometry, combinatorics unsolved, combinatorics



A segment of length $2$ is divided into $n$, $n\ge 2$, subintervals. A square is then constructed on each subinterval. Assume that the sum of the areas of all such squares is greater than $1$. Show that under this assumption one can always choose two subintervals with total length greater than $1$.