IMC 2026 · Day 2 · P7

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