Problem (Combinatorics):
Five equilateral triangles are drawn in the plane so that no two sides of any of the triangles are parallel. Compute the maximum number of points of intersection among all five triangles.

Solution:
Any two of the triangles intersect in at most six points, because each side of one triangle can intersect the other triangle in at most two points. To count the total number of intersections among the five triangles, note that there are $\left(\begin{array}{l}5 \\ 2\end{array}\right)=10$ ways to select a pair of triangles, and each pair may result in 6 intersections. Thus $10 \times 6=60$ is an upper bound.

This can be achieved, for example, by taking six equilateral triangles of equal size, centered at a single point, and rotating them different amounts so that no three sides intersect at a single point. Thus the answer is 60.

Final answer: 60