Let $\mathcal{P}$ be a compact polygon in $\mathbb{R}^2$, and let $\mathcal{R}$ be a line.
- Show by topological arguments (i.e. using concepts and theorems from general topology) that there exists a line $\mathcal{R'}$ $\mathit{parallel}$ to $\mathcal{R}$ such that $\mathcal{R'}$ divides $\mathcal{P}$ in two parts with $\mathit{equal}$ $\mathit{area}$.
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Remarks:
- Specifying that $\mathcal{P}$ is compact may be superfluous since it is closed and bounded in $\mathbb{R}^2$
- The main difficulty I find to even try sketching a proof is that I don't know how to translate in topological language the fact that the two parts of $\mathcal{P}$ must have the same area.
(Edit) Sketched proof attempt:
Choose a cartesian system of axes with the $X$ axis superimposed on $\mathcal{R}$, and the $Y$ axis directed in such a way that the polygon lies above the $X$ axis. Now define a function $f:\mathbb{R} \to \mathbb{R}$ such that for $y \in \mathbb{R}$, $f(y)$ is the difference between the two areas in which the line parallel to the $X$ axis at height $y$ divides the polygon. Now $f$ is continuous and assumes the values $A$ and $-A$, where $A$ is the area of the polygon, before and after reaching it. $\mathbb{R}$ is connected, therefore the Intermediate Value Theorem holds and $f$ must be zero somewhere in between.