User:Anghel/Sandbox

Theorem
Let $P$ be a polygon that is not a triangle.

Then $P$ has at least two ears that do not overlap each other.

Proof
Let $P$ have $n$ sides, where $n \in \N_{ \ge 3 }$.

Polygon Triangulation Theorem shows that there exists a triangulation $\family { \triangle_i }_{i \mathop = 1}^{n-2}$ of $P$, and the sides of each triangle $\triangle_i$ are either sides of $P$ or chords of $P$.

Suppose $\triangle_i$ has $0$ or $1$ sides in common with $P$.

In this case, $\triangle_i$ cannot be an ear of $P$.

Suppose $\triangle_i$ has $2$ sides in common with $P$.

In this case, $\triangle_i$ is an ear of $P$.

Suppose $\triangle_i$ has $3$ sides in common with $P$.

In this case, we have $\triangle_i = P$, which would contradict our assumption that $P$ is not a triangle.

Now, suppose that at most one of the triangles $\family { \triangle_i }_{i \mathop = 1}^{n-2}$ is an ear.

The maximal number of sides that the $n-2$ triangles have in common with $P$ is:


 * $2 + 1 \paren { n-2-1} = n-1$

As the triangulation covers all $n$ sides of $P$, this is a contradiction.

It follows that at least two of the triangles $\family { \triangle_i }_{i \mathop = 1}^{n-2}$ are ears.

As the two ears are part of a triangulation, they do not overlap.