> [!NOTE] Lemma > The union of nine planar surfaces, each of area equal to $1,$ has total area equal to $5.$ Then the overlap of some two of these surfaces has an area greater than or equal to $1/9.$ Proof: Enumerate the surfaces as $S_{1},S_{2},\dots,S_{9}.$ Applying inclusion-exclusion we have $\begin{align} \text{Area}\left( \bigcup_{1\leq i\leq 9} S_{i}\right) &= \sum_{1\leq i\leq 9}\text{Area}( S_{i}) - \sum_{1\leq i<j\leq 9} \text{Area}(S_{i} \cap S_{j}) +\dots \\ &\geq \sum_{1\leq i\leq 9}\text{Area}( S_{i}) - \sum_{1\leq i<j\leq 9} \text{Area}(S_{i} \cap S_{j}) \end{align} $Thus $\sum_{1\leq i<j\leq 9} \text{Area}(S_{i} \cap S_{j}) \geq \sum_{1\leq i\leq 9}\text{Area}( S_{i}) - \text{Area}\left( \bigcup_{1\leq i\leq 9} S_{i}\right) = 9-5=4. $ BWOC, suppose the overlap of any two of the $S_{i}$ has an area strictly less that $1/9.$ Noting that there ${9 \choose 2}=36$ choices for $1\leq i<j\leq 9,$ we have $\sum_{1\leq i<j\leq 9} \text{Area}(S_{i} \cap S_{j}) < 36 \times \frac{1}{9} = 4$- a contradiction.