# Polygon written as intersection of triangles

Find $n$ such that any convex polygon with $100$ sides can be obtained as an intersection of $n$ triangles.

First, $n \geq 34$ since if the polygon is the intersection of $n$ triangles, each of the $100$ sides of the polygon must be contained in at least one of the triangles, therefore $3n \geq 100$. Now, I must prove that $n=34$, i.e. any polygon with 100 sides can be obtained intersecting $34$ triangles. These triangles should be formed by extending some sides of the polygon. The only problem would be that there is possible that three sides of the polygon form a triangle which does not contain the polygon, or do not form a triangle at all (if two sides are parallel).

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Your first problem, that each triangle contains the polygon, should follow from whatever definition of convexity you are using. For the second, show that any side is parallel to at most one other. – deinst Aug 13 '11 at 20:17

I'm pretty sure that $n \ge 50$.
If I'm not mistaken, there's no way for the sides of any triangle to match more than two of the short sides. Thus, this 100-gon cannot be obtained as the intersection of less than $\left\lceil \frac{99}{2} \right\rceil = 50$ triangles.
Edit: It's also easy to see that $n = 50$ is sufficient for any bounded convex 100-gon: Divide the edges into 50 pairs of adjacent sides, and for each pair extend the sides into rays starting from their common endpoint.