AMC 10 · 2002 · #14

Grade 7 arithmetic

Archetype Overlap / Complement Counting └ Sub-archetype Complement by Subtracting Forbidden Arrangements

combinations-basicpair-counting extremal-construction ↑ Prerequisites: combinations-basic

📏 Medium solution 💡 2 insights

Problem

Four different circles are drawn on a plane. Placing them as cleverly as possible, what is the largest total number of points that lie on two or more of the circles at once?

Pick an answer.

(A)

$\ 8$

(B)

$\ 9$

(C)

$\ 10$

(D)

$\ 12$

(E)

$\ 16$

How to solve

Strategy Identify Subproblems

Four circles crossing all at once looks tangled, but the crossings never happen between three circles — every crossing point belongs to exactly one pair of circles. Tool #7 (Identify Subproblems) uses that to split the whole count into a sum over pairs: the total is just (points per pair) times (number of pairs). Tool #1 (Draw a Diagram) pins down the first factor — two circles cross at most twice. Tool #2 (Make a Systematic List) pins down the second — carefully list the pairs of four circles so none is missed or double-counted. Tool #14 (Extreme Principle) closes the argument: the total is largest when every pair truly meets twice and no three circles share a point, and we check that this best case is actually drawable.

1STEP 1

Two circles cross at most twice

Two distinct circles can meet at most twice — never three times.

points from one pair ≤ 2

2STEP 2

Every crossing belongs to one pair

Each crossing lies on exactly two circles, so it is owned by one pair.

total crossings = Σ_pairs (crossings of that pair)

3STEP 3

List the pairs of circles

Listing pairs of four circles without repeats gives 6 pairs.

{1,2},{1,3},{1,4},{2,3},{2,4},{3,4} → 6 pairs

4STEP 4

Multiply, then check it is reachable

Six pairs at two points each gives 12, reachable with no triple point, choice (D).

6 × 2 = 12 → (D)

Answer

Check the pattern on smaller cases. Two circles: 1 pair, up to 2 points. Three circles: 3 pairs, up to 6 points. Four circles: 6 pairs, up to 12 points. In general n circles give 2C(n, 2)=n(n-1) points, and 4 × 3 = 12 agrees. The answer must beat 8 (choice A only counts 2 points for each single circle, forgetting that a new circle crosses every earlier one), and it cannot reach 16 (choice E) because 16 would need some pair to cross more than twice or a point shared without penalty — both impossible. So 12 is the honest maximum.

💡Key takeaway

Two circles can cross at most twice, so count how many pairs of circles there are and double it.

Two circles cross at most twice
Every crossing belongs to one pair
List the pairs of circles
Multiply, then check it is reachable

Two circles cross at most twice

Every crossing belongs to one pair

List the pairs of circles

Multiply, then check it is reachable

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