2002 AMC 10B Problem 18

Upper bound: two distinct circles share at most $2$ points. Four circles form $\binom42 = 6$ pairs, so there are at most $6 \cdot 2 = 12$ intersection points.

Construction: start with one circle and draw three more copies of it, each shifted by a small amount in a different direction. Any two of these congruent circles overlap and cross in exactly $2$ points, and with generic shifts no three circles pass through a common point, so the $12$ crossing points are all distinct.

The bound is achieved, so the maximum is $12$.

The answer is $\boxed{\textbf{(D)}\ 12}$.

"Maximum number of intersections" problems have two halves: count the most each pair can contribute, then exhibit a configuration that reaches that total with no coincidences. The pair count $\binom{n}{2}$ times $2$ is the general bound for $n$ circles; the only subtlety is making sure no intersection point is shared by three circles, which a small generic perturbation guarantees.

Add circles one at a time: the second circle can cross the first in $2$ points, the third crosses each of the previous two for $4$ new points, and the fourth crosses three earlier circles for $6$ more. Total $2 + 4 + 6 = 12$.

• Double counting by taking 4 3 2 = 24 without dividing by 2, or assuming a new circle can cross an old one only once (bound 6).
• Stopping at the bound without checking it is attainable, or choosing a symmetric picture where several circles pass through the same point and losing intersections.