Competition problem — Combinatorics

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Combinatorics Difficulty 5.5 AIME, harder Prove it JBMO

Problem:

A polygon having $n$ sides is arbitrarily decomposed in triangles having all the vertices among the vertices of the polygon. We paint in black the triangles that have two sides that are also sides of the polygon, in red if only one side of the triangle is side of the polygon and white those triangles that have in common with the polygon only vertices.
Prove that there are 2 more black triangles than white ones.

Solution

Solution:

Denote by $b, r, w$ the number of black, red, and white triangles respectively.
It is easy to prove that the polygon is divided into $n-2$ triangles, hence
$b + r + w = n - 2$
Each side of the polygon is a side of exactly one triangle of the decomposition, and thus
$2b + r = n$
Subtracting the two relations yields $w = b - 2$ , as needed.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.