2001 AMC 10 Problem 4

This asks for the greatest possible number, so there are two halves: show that more than $6$ is impossible, then show $6$ can happen.

Part 1: never more than $6$.

1. Draw a straight line and a circle. The line can miss the circle ($0$ points), just touch it ($1$ point), or cut through it ($2$ points). It can never meet the circle $3$ times, because after going in and coming out, a straight line cannot turn back.
2. Each side of a triangle is a piece of a straight line, so each side meets the circle at most $2$ times.
3. There are $3$ sides, so the total is at most $3 \times 2 = 6$.

Part 2: $6$ is possible.

1. Draw a triangle and a circle inside it that touches all three sides. That gives $3$ points.
2. Now make the circle a little bigger, keeping the same center. Each side now cuts across the circle instead of just touching it, so each side crosses it twice, near where it used to touch.
3. That gives $2 + 2 + 2 = 6$ points.

The answer is $\boxed{\textbf{(E)}\ 6}$.

For "maximum number of crossings" questions, count piece by piece: each side can add at most $2$ points, which gives a ceiling. Then draw a picture that reaches the ceiling. You need both halves: the picture alone does not prove you cannot do better, and the ceiling alone does not prove it can be reached.

A straight line can cross a circle in at most two points, and each side of a triangle is part of a line. With three sides, the total number of intersection points cannot exceed $3 \cdot 2 = 6$.

To see that $6$ is actually possible, start with a triangle whose incircle touches all three sides, then enlarge the circle very slightly (keeping its center). Each former tangency point splits into two crossing points, one on each side of it, and all six lie on the sides of the triangle.

The answer is $\boxed{\textbf{(E)}\ 6}$.