2004 JBMO Problems/Problem 4
Consider a convex polygon having vertices, . We arbitrarily decompose the polygon into triangles having all the vertices among the vertices of the polygon, such that no two of the triangles have interior points in common. 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 also a side of the polygon and in white those triangles that have no sides that are sides of the polygon.
Prove that there are two more black triangles that white ones.
First we shall state a :
: A convex polygon with sides can be arbitrarily decomposed into exactly triangles having all the vertices among the vertices of the polygon, such that no two of the triangles have interior points in common.
: Sum of internal angles of a convex polygon with sides is . Since triangles are using all of the vertices of the polygon, the aggregate sum of angles of all the triangles must equal the sum of internal angles of a convex polygon = . Thus, number of triangles = .
: Let there be a total of triangles in sided convex polygon .
Now consider the polygon formed after removing all the black triangles from polygon . It's not hard to see that has exactly sides.
Now, in polygon , we are left with exactly sides which are still part of the original Polygon .
No two adjacent sides of these can be joined to form another triangle (since we have assumed a total of triangles exist).
So that leaves us with triangles in total.
So, now all the triangles that remain must be triangles.
From lemma, has exactly triangles satisfying the conditions of the problem, so the number of triangles =
Thus total number of triangles = which is less than total number of triangles.