Question on counting the number of triangles formed by 1999 points in a square

Question

I was reading an explanation for a solution to an Olympiad problem as follows: Let $S$ be a square with the side length 20 and let $M$ be the set of points formed with the vertices of $S$ and another 1999 points lying inside $S$. Prove that there exists a triangle with vertices in $M$ and with area at most equal with $\frac 1{10}$.

In the solution it states that if you want to cover the square S with triangles formed by the points in M and the vertices of S, you will need 4000 triangles for which this calculation is provided: 2 * (1999 + 1) = 4000. How was this figure obtained/what is the explanation behind this calculation?

Answer 1

For a much simpler verification, double count the angle sum.

Consider all of the vertex angles of the triangles. What is their sum?

• Each triangle has an angle sum of $180^\circ$.
• The 1999 points each contribute $360^\circ$ (Why?). Add in the 4 corners which have total of $360^\circ$.
• Hence, there are a total of $ \frac { (1999+1) \times 360^\circ } { 180^\circ} = 4000$ triangles.

Answer 2

The Euler's formula tells that $v-e+f=2$, where $v$ is the number of vertices, $e$ is the number of edges, $f$ is the number of faces (including the outer infinite face). We know that $$v=4+1999=2003.$$

Let us calculate $e$. Each triangle face has $3$ edges surrounding it. The outer face has $4$ edges. It would give $3(f-1)+4$ edges. But every edge is counted twice here, since every edge splits two faces. So, $$e= \frac{3(f-1)+4}2.$$

Put it into the Euler’s formula: $$2003-\frac{3(f-1)+4}2+f=2$$

$$4006-3f+3-4+2f=4$$

$$f=4001.$$

So the number of triangles is $f-1=4000.$