Use the dissection of the ngon into triangles to show that t

Use the dissection of the n-gon into triangles to show that the angle sum of a convex n-gon is (n - 2)k. Use Exercise 2.1.3 to find the angle at each vertex of a regular n-gon (an n-gon with equal sides and equal angles). Deduce from Exercise 2.1.4 that copies of a regular n-gon can tile the plane only for n = 3.4.6.

Solution

1) Every triangulation of an n-gon has (n-2)-triangles formed by (n-3) diagonals.

The proof is by induction. If n = 3, the assertion is trivially true. Assume the statement holds for all n < K. Given a K-gon, find - as in the proof of the existence of a triangulation - a diagonal that splits the polygon into smaller two, say n-gon and m-gon such that n + m = K + 2 and both are less than K. Then, by the induction hypothesis, the n-gon consists of (n-2) triangles, while the m-gon consists of (m-2) triangles. In all, there are

(n - 2) + (m - 2) = (K + 2) - 4 = K - 2

triangles, as required. The number of the diagonals is

(n - 3) + (m - 3) + 1 = K + 2 - 5 = K - 3.

Now, for the proof of the main statement. Consider a triangulation of an n-gon, with n > 3. The triangulation consists of n-2 triangles. Each of the triangles in the triangulation shares at most 2 edges with the polygon. Since the latter has n edges but there are only two triangles, by the Pigeonhole Principle, there are at least two triangles with two polygon\'s edges. These are the ears.

2)    interior angles, since they all have the same values. So for example the interior angles of a pentagon always add up to 540°, so in a regular pentagon (5 sides), each one is one fifth of that, or 108°. Or, as a formula, each interior angle of a regular polygon is given by:

180(n-2)/n

degrees

where
n  is the number of sides