AMC 10 · 2002 · #18
Grade 7 arithmeticFour distinct circles are drawn in a plane. What is the maximum number of points where at least two of the circles intersect?
Pick an answer.
AMC 10 2002 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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?
Givens: There are exactly four circles; The circles are distinct (no two are the same circle); We count points where at least two circles cross; Answer choices: (A) $8$, (B) $9$, (C) $10$, (D) $12$, (E) $16$
Unknowns: The maximum possible number of crossing points, over every way of drawing the four circles
Understand
Restated: 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?
Givens: There are exactly four circles; The circles are distinct (no two are the same circle); We count points where at least two circles cross; Answer choices: (A) $8$, (B) $9$, (C) $10$, (D) $12$, (E) $16$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #1 Draw a Diagram, #2 Make a Systematic List, #14 Extreme Principle
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.
Execute — Answer: D
4.OA.A.3 Step 1 Two circles cross at most twice
- Draw two circles and slide one across the other.
- They can miss entirely (0 points), touch at a single point (1 point), or pass through each other (2 points) — but there is no way to make two distinct circles meet at 3 points.
- So each pair of circles contributes at most $2$ crossing points.
- To make the grand total as big as possible, we want every pair to hit that maximum of $2$.
💡 A circle is fixed once you know its center and radius, so two circles just can't wrap around each other enough to cross a third time.
4.OA.A.3 Step 2 Every crossing belongs to one pair
- Here is the key move: a crossing point sits on two of the circles, so it is owned by exactly one pair of circles.
- That means the total number of crossings is nothing more than the number of crossings added up pair by pair.
- Instead of untangling all four circles at once, we only need two easy numbers: how many points a single pair makes, and how many pairs there are.
💡 Splitting a messy count into one tidy count per pair turns a tangle into simple arithmetic.
7.SP.C.8 Step 3 List the pairs of circles
- Label the circles $1, 2, 3, 4$ and list every pair once, keeping the smaller label first so nothing repeats: $\{1,2\}, \{1,3\}, \{1,4\}, \{2,3\}, \{2,4\}, \{3,4\}$.
- Counting them gives $6$ pairs.
- (This is $\binom{4}{2}$: circle $1$ pairs with $3$ others, circle $2$ with $2$ new ones, circle $3$ with $1$ new one — $3+2+1=6$.)
💡 Listing pairs in order — smaller number first — guarantees you count each pair exactly once.
3.OA.A.1 Step 4 Multiply, then check it is reachable
- Each of the $6$ pairs gives at most $2$ points, so the total can be at most $6 \times 2 = 12$.
- This top value is actually reachable: draw the four circles so that every pair genuinely crosses twice and no three circles pass through the same point — for example four circles of equal size whose centers form a small square all overlap two-by-two with no shared triple point.
- Nothing is wasted, so the maximum is exactly $12$, which is choice (D).
💡 The most points appears when every pair is pushed to its limit of two and no crossing is shared away.
4.OA.A.3 Draw two circles and slide one across the other. They can miss entirely (0 point 4.OA.A.3 Here is the key move: a crossing point sits on two of the circles, so it is owne 7.SP.C.8 Label the circles $1, 2, 3, 4$ and list every pair once, keeping the smaller lab 3.OA.A.1 Each of the $6$ pairs gives at most $2$ points, so the total can be at most $6 \ Review
Reasonableness: 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 $2\binom{n}{2}=n(n-1)$ points, and $4\times 3 = 12$ 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.
Alternative: Add the circles one at a time. The first circle makes no crossing yet. The second crosses the first in at most $2$ points. The third can cross each of the two earlier circles twice, adding at most $4$. The fourth crosses each of the three earlier circles twice, adding at most $6$. Total: $2+4+6 = 12$, again choice (D). Mis-adding this running total (say $2+3+4$) is exactly what produces the near-miss decoys $9$ and $10$.
CCSS standards used (min grade 7)
7.SP.C.8Represent sample spaces and count outcomes using organized lists, tables, and tree diagrams (Listing every pair of the four circles once and counting that there are $6$ of them.)4.OA.A.3Solve multistep word problems using the four operations (Reasoning that each pair of circles meets in at most two points and that the whole count splits into a sum over pairs.)3.OA.A.1Interpret products of whole numbers as total number of objects in groups (Reading $6$ pairs of $2$ points as the product $6 \times 2 = 12$.)
⭐ Two circles can cross at most twice, so count how many pairs of circles there are and double it.
⭐ Two circles can cross at most twice, so count how many pairs of circles there are and double it.
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