International Mathematics Competition
for University Students
2026

IMC2026: Day 2, Problem 7

Problem 7. For a continuous function \(\displaystyle f\colon[0,1]\to\RR\), let \(\displaystyle S(f)\) be the union of all straight segments in the plane joining points \(\displaystyle (x,0)\) and \(\displaystyle (f(x),1)\), where \(\displaystyle x\in [0,1]\). Let \(\displaystyle A(f)\) be the area of \(\displaystyle S(f)\). Find the infimum of \(\displaystyle A(f)\) over all continuous \(\displaystyle f\).

David Preiss, University of Warwick, UK

Solution. For \(\displaystyle s\in[0,1]\) define \(\displaystyle f_s(x) = x + s(f(x)-x)\). Then \(\displaystyle (f_s(x),s)\) belongs to the straight segment joining \(\displaystyle (x,0)\) and \(\displaystyle (f(x),1)\), so it is in \(\displaystyle S(f)\). Since \(\displaystyle f_s\) is continuous, by the intermediate value theorem the straight segment joining \(\displaystyle (f_s(0),s)\) and \(\displaystyle (f_s(1),s)\) (which could be just a point) lies in \(\displaystyle S(f)\). Hence the area of \(\displaystyle S(f)\) is at least \(\displaystyle \int_0^1 |f_s(1)-f_s(0)|\,ds = \int_0^1 |1- cs|\,ds\) where \(\displaystyle c=f(0)-f(1) +1\). The latter integral is \(\displaystyle 1-c/2\) if \(\displaystyle c\le 1\) and \(\displaystyle c/2+ 1/c -1\) when \(\displaystyle c\ge 1\). In the first case the minimum is \(\displaystyle 1/2\) (for \(\displaystyle c=1\)) and in the second \(\displaystyle \sqrt{2} -1<1/2\) (for \(\displaystyle c=\sqrt{2}\)). So the minimum of the areas of \(\displaystyle S(f)\) is \(\displaystyle \sqrt{2}-1\) which is attained, for example, when \(\displaystyle f(x) = (1 - \sqrt{2})x\) since in this case \(\displaystyle S(f)\) is exactly the union of the straight segments joining \(\displaystyle (f_s(0),s)\) and \(\displaystyle (f_s(1),s)\).