Counting Triangles and Segments in a Polygon

This is an extension of the problem of diagonal count in a convex polygon. Now, we have an N-gon, i.e. a polygon with N vertices and N sides and M additional points in its interior. Connect all the given points without intersection until no segments could be added. Count the number of added segments and the triangles formed along the way. What can you say about the two numbers?

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(Points are connected by dragging the cursor from one to the other.)

The number of segments and triangles depends only on the numbers M and N.

Let, when no additional segments could be drawn, there be K triangles. Angles in K triangles sum up to 180°K. We may find the same quantity in a different way. Internal angles of the N-gon add up to 180°(N - 2). The angles around each of the internal points add up to 360°M. We thus obtain the relationship

180°K = 180°(N - 2) + 360°M, or K = N - 2 + 2M.

So that the number of triangles is always K = N - 2 + 2M, which depends only on N and M, but not on the manner or order in which the points have been connected.

K triangles have the total of 3K sides, of which N belong only to one triangle, while the rest belong to two triangles each. The latter are exactly the segments that have been added by construction and whose number we are looking for. It's therefore equals (3K - N)/2.

(3K - N)/2 = (3N - 6 + 6M - N)/2 = N - 3 + 3M.

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