Library / /646 of 740

, 2014

Geometry Difficulty 5.5 AIME, harder Prove it United States

Problem:

Let $\mathcal{P}_1$ , $\mathcal{P}_2$ , $\mathcal{P}_3$ be pairwise distinct parabolas in the plane. Find the maximum possible number of intersections between two or more of the $\mathcal{P}_i$ . In other words, find the maximum number of points that can lie on two or more of the parabolas $\mathcal{P}_1$ , $\mathcal{P}_2$ , $\mathcal{P}_3$ .

Solution

Solution:

Note that two distinct parabolas intersect in at most 4 points, which is not difficult to see by drawing examples. Given three parabolas, each pair intersects in at most 4 points, for at most $4 \cdot 3 = 12$ points of intersection in total. It is easy to draw an example achieving this maximum, for example, by slanting the parabolas at different angles.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.