Solution to Exercise 3: Choosing such a triangle amounts to choosing six distinct vertices of the n-gon: indeed, if we are given six distinct vertices, we order them in a clockwise direction and connect two points with a diagonal if the arc connecting these two points contains exactly two vertices, which are opposite. This provides a triangle that is not reduced to a single point (since three diagonals never intersect at a single point), and none of its vertices are common with the n-gon.

Conversely, if we have a triangle with no vertices in common with the n-gon, considering its three sides, which are segments of diagonals of the n-gon, we indeed obtain six distinct vertices (the six intersections of these diagonals with the n-gon, which are distinct because two sides of the triangle already intersect at a vertex of the triangle). Moreover, if we cyclically order these six vertices, those that form the sides of the triangle are indeed opposite, because given a side, extending the other two sides finds two vertices on each side of it.

Since there are $\binom{n}{6}$ ways to choose six vertices of the n-gon, there are $\binom{n}{6}$ such triangles in the figure.

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FIGURE 1 - Six vertices on the circle and their associated triangle

Graders' Comments: Students generally have the right ideas for this problem. The main errors come either from a poor understanding of the statement, where students consider degenerate triangles, or from approaches that do not lead to a solution, for example by counting the diagonals involved in the triangles. Some justifications of the correspondence between "choosing six vertices" and "choosing a triangle" are very incomplete.