AMC 8 · 2001 · #24

Grade 6 countinglogic
Archetype Overlap / Complement Counting └ Sub-archetype Two-Attribute Table: Fill or Optimize a Cell
complementary-countingset-partitionsystematic-enumerationreflection-symmetry complementary-countingcaseworkidentify-subproblems ↑ Prerequisites: set-partitionsystematic-enumeration
📏 Long solution 💡 5 insights 📊 Diagram
Problem
A symmetric figure has an upper half and a lower half. Each half contains 3 red, 5 blue, and 8 white triangles, for 16 triangles per half. When the upper half is folded onto the lower half along the centerline, each upper triangle lands on exactly one lower triangle, producing 16 pairs. You are told that 2 pairs are red-red, 3 pairs are blue-blue, and 2 pairs are red-white. How many of the 16 pairs are white-white?

Pick an answer.

(A)
4
(B)
5
(C)
6
(D)
7
(E)
9

AMC 8 2001 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Make a Systematic List

There are only six possible pair types (R-R, B-B, W-W, R-B, R-W, B-W). Tool #2 (Make a Systematic List) lets us write each color's total (6 red, 10 blue, 16 white) and assign the known pair counts. Tool #16 (Change Focus / Count the Complement) is the move that finishes it: instead of counting white-white pairs directly, we account for every red and blue triangle first, see how many whites are spoken for as partners to non-whites, and the leftover whites must pair with each other. We avoid Tool #13 (Algebra) because the bookkeeping is short enough to do in a table.

1STEP 1

Double each half: both halves together hold 6 red, 10 blue, and 16 white triangles.

Reds = 6, Blues = 10, Whites = 16
2STEP 2

The 2 red-red pairs use 4 reds and the 2 red-white pairs use 2 more — all 6 reds are gone, so there are no red-blue pairs.

2 · 2_R-R + 2 · 1_R-W = 6 = total reds
3STEP 3

The 2 red-white pairs consume 2 whites, leaving 14 whites still to place.

16 - 2 = 14 whites still to place
4STEP 4

The 3 blue-blue pairs use 6 blues; the remaining 4 blues can only pair with white, making 4 blue-white pairs.

Blues left = 10 - 6 = 4 → 4 pairs of type B-W
5STEP 5

Those 4 blue-white pairs take 4 more whites, leaving 10 whites that pair among themselves — 5 white-white pairs.

14−42\frac{14 - 4}{2} = 102\frac{10}{2} = 5 → (B)
Answer
5
Total pair check: R-R + B-B + R-W + B-W + W-W = 2 + 3 + 2 + 4 + 5 = 16. That matches the 16 pairs the fold creates. Color check: reds used = 2(2) + 2(1) = 6; blues used = 2(3) + 4(1) = 10; whites used = 2(5) + 1(2) + 1(4) = 16. All three totals match the inventory exactly, so 5 white-white pairs is consistent and answer (B) holds.
💡Key takeaway

Account for the loud colors first. All 6 reds are eaten by the 2 red-red and 2 red-white pairs. Of 10 blues, the 3 blue-blue pairs eat 6, leaving 4 blues that have to pair with whites. Whites used so far: 2 + 4 = 6, leaving 16 - 6 = 10 whites — and 10 whites make 5 white-white pairs, answer (B).

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