Track / Stage 5 / 23 of 400 #623 of 1964

Problem 623

AIME late

Geometry Difficulty 5.0 Prove it Ukrainian Mathematical Olympiad · Ukraine

In the plane 5 circles are given such that no three of them have a common point. Can it happen that they have exactly:

a) 12;

b) 24 different intersection points?

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

a) Yes, it is possible. It is enough to take 4 circles, each pair of which intersects in two points, and a fifth circle such that it does not intersect the others.

b) No, it is not possible. The first circle can intersect the others in at most 8 points, the second adds at most 6 intersection points, and so on. In total, we have at most $8 + 6 + 4 + 2 = 20$ different intersection points.

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