AMC 10 · 2002 · #18

Grade 7 arithmetic
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)
$\ 16$

AMC 10 2002 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

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 miss, touch once, or cross twice — never three times, so one pair gives at most 2 points.

points from one pair ≤ 2
2STEP 2

Every crossing belongs to one pair

A crossing sits on two circles, so it is owned by exactly one pair — the grand total is the pair-by-pair counts added up.

total crossings = Σ_pairs (crossings of that pair)
3STEP 3

List the pairs of circles

Label the circles 1, 2, 3, 4 and list each pair once, smaller first: {1,2},{1,3},{1,4},{2,3},{2,4},{3,4} — 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 2 points each caps the count, and four equal circles centered on a small square reach it: 6 × 2 = 12, choice (D).

6 × 2 = 12 → (D)
Answer
12
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