2002 AMC 10B Problem 18

P16-20~3 minPrint

Circles Geometric Optimization Bound the quantity above/below or estimate to pin it down

Four distinct circles are drawn in a plane. What is the maximum number of points where at least two of the circles intersect?

Show solution without answering

Key insight

Two circles cross in at most 2 points. Four circles make 6 pairs, so at most 12 points, and slightly shifted copies of one circle reach 12.

Solution

This is a "greatest possible" question, so there are two halves: show $12$ cannot be beaten, then show $12$ can actually happen.

Part 1: at most 12.

1. Two different circles can cross in at most $2$ points.
2. Count the pairs of circles. Call the circles $1, 2, 3, 4$ . The pairs are
$\{1,2\},\ \{1,3\},\ \{1,4\},\ \{2,3\},\ \{2,4\},\ \{3,4\},$
which is $6$ pairs.
3. Every crossing point belongs to at least one pair, and each pair gives at most $2$ points. So there are at most $6 \times 2 = 12$ points.

Part 2: 12 is possible.

Draw one circle. Then draw three more circles of the same size, each moved a little from the first in a different direction (for example up, down-left, down-right).
Any two of these overlap heavily, so they cross in exactly $2$ points. If the shifts are chosen a bit unevenly, no point lies on three circles at once, so all $12$ crossing points are different.

At most $12$ , and $12$ is possible, so the maximum is $12$ .

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

Why this works

For "maximum number of crossings" questions, find the most each pair can give, multiply by the number of pairs, and then build a picture where no two crossings land on the same spot. The same count works for any number of circles: each new circle can cross every earlier one twice.

Alternative approach

Add the circles one at a time and count only new points. The $2$ nd circle crosses the $1$ st in $2$ points. The $3$ rd crosses the first two in $2 + 2 = 4$ new points. The $4$ th crosses the first three in $6$ new points. Total: $2 + 4 + 6 = 12$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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 $\boxed{\textbf{(D)}\ 12}$ .

Review in the AoPS books

Introduction to Geometry, Chapter 11: Circles
Introduction to Geometry, Chapter 10: Geometric Inequalities
Introduction to Geometry, Chapter 12: Circles and Angles
Introduction to Geometry, Chapter 13: Power of a Point