← Sensim Lab Sensim Math

AMC 8 · 2001 · #24

Grade 6 countinglogic
Archetype Overlap / Complement Counting
complementary-countingset-partitionsystematic-enumerationreflection-symmetry complementary-countingcaseworkidentify-subproblems ↑ Prerequisites: set-partitionsystematic-enumeration
📏 Long solution 💡 5 insights 📊 Diagram

Problem

Each half of this figure is composed of 3 red triangles, 5 blue triangles and 8 white triangles. When the upper half is folded down over the centerline, 2 pairs of red triangles coincide, as do 3 pairs of blue triangles. There are 2 red-white pairs. How many white pairs coincide?

Pick an answer.

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

Toolkit + CCSS Solution

Understand

Restated: 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?

Givens: Each half has $3$ red ($R$), $5$ blue ($B$), $8$ white ($W$) triangles; Folding produces $16$ pairs in total; $2$ red-red pairs ($R$-$R$); $3$ blue-blue pairs ($B$-$B$); $2$ red-white pairs ($R$-$W$, in either orientation); Answer choices: (A) $4$, (B) $5$, (C) $6$, (D) $7$, (E) $9$

Unknowns: The number of white-white ($W$-$W$) pairs

Understand

Restated: 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?

Givens: Each half has $3$ red ($R$), $5$ blue ($B$), $8$ white ($W$) triangles; Folding produces $16$ pairs in total; $2$ red-red pairs ($R$-$R$); $3$ blue-blue pairs ($B$-$B$); $2$ red-white pairs ($R$-$W$, in either orientation); Answer choices: (A) $4$, (B) $5$, (C) $6$, (D) $7$, (E) $9$

Plan

Primary tool: #2 Make a Systematic List

Secondary: #16 Change Focus / Count the Complement

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.

Execute — Answer: B

#2 Make a Systematic List 6.EE.A.2 Step 1
  • Make a triangle inventory.
  • The two halves together hold $2 \times 3 = 6$ red, $2 \times 5 = 10$ blue, and $2 \times 8 = 16$ white triangles, for $32$ triangles total — exactly twice the $16$ pairs.
$$\text{Reds} = 6, \quad \text{Blues} = 10, \quad \text{Whites} = 16$$

💡 Each pair eats $2$ triangles, so totals work cleanly in pair units.

#16 Change Focus / Count the Complement 6.EE.B.6 Step 2
  • Account for every red triangle.
  • The $2$ red-red pairs use $2 \times 2 = 4$ reds.
  • The $2$ red-white pairs use $2$ more reds.
  • That is $4 + 2 = 6$ reds — every red triangle is accounted for.
  • None is left over, so there are zero red-blue pairs.
$$\underbrace{2 \cdot 2}_{R\text{-}R} + \underbrace{2 \cdot 1}_{R\text{-}W} = 6 = \text{total reds}$$

💡 Sealing the red column closes off one whole color and forces the remaining colors to pair among themselves.

#2 Make a Systematic List 6.EE.A.2 Step 3
  • Account for the whites already used.
  • The $2$ red-white pairs use $2$ whites.
  • Whites remaining to be placed: $16 - 2 = 14$.
$$16 - 2 = 14 \text{ whites still to place}$$

💡 Whites used as red partners are off the table; only the rest can still form $B$-$W$ or $W$-$W$ pairs.

#16 Change Focus / Count the Complement 6.EE.B.6 Step 4
  • Account for every blue triangle.
  • The $3$ blue-blue pairs use $3 \times 2 = 6$ blues.
  • The other $10 - 6 = 4$ blues have nowhere to go but white partners (reds are sealed), giving $4$ blue-white pairs.
$$\text{Blues left} = 10 - 6 = 4 \;\Rightarrow\; 4 \text{ pairs of type } B\text{-}W$$

💡 Once reds are sealed, every remaining blue has only white to pair with.

#16 Change Focus / Count the Complement 6.NS.B.2 Step 5
  • Count the leftover whites.
  • The $4$ blue-white pairs swallow $4$ more whites, leaving $14 - 4 = 10$ whites.
  • These can only pair with each other, producing $10 \div 2 = 5$ white-white pairs.
$$\dfrac{14 - 4}{2} = \dfrac{10}{2} = 5 \;\Rightarrow\; \textbf{(B)}$$

💡 Whatever isn't paired with another color must pair within its own color.

[1] #2 6.EE.A.2 Make a triangle inventory. The two halves together hold $2 \times 3 = 6$ red, $2
[2] #16 6.EE.B.6 Account for every red triangle. The $2$ red-red pairs use $2 \times 2 = 4$ reds.
[3] #2 6.EE.A.2 Account for the whites already used. The $2$ red-white pairs use $2$ whites. Whi
[4] #16 6.EE.B.6 Account for every blue triangle. The $3$ blue-blue pairs use $3 \times 2 = 6$ bl
[5] #16 6.NS.B.2 Count the leftover whites. The $4$ blue-white pairs swallow $4$ more whites, lea

Review

Reasonableness: Total pair check: $R\text{-}R + B\text{-}B + R\text{-}W + B\text{-}W + W\text{-}W = 2 + 3 + 2 + 4 + 5 = 16$. 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.

Alternative: Tool #11 (Work Backwards) on one half: in the upper half, $2$ reds are matched red-red below and $1$ red is matched red-white below, so all $3$ upper reds are placed. By symmetry, the lone red-white pair using a lower red leaves $1$ white in the upper half paired with that lower red — so $1$ upper white is spoken for. Similarly, $3$ upper blues match blue-blue and the remaining $2$ upper blues must match white below; the $2$ lower blues in the unused blue-white slot match $2$ more upper whites. Upper whites used as non-white partners: $1 + 2 = 3$. Upper whites left for white-white pairs: $8 - 3 = 5$. Same answer (B).

CCSS standards used (min grade 6)

  • 6.EE.A.2 Write, read, and evaluate expressions in which letters stand for numbers (Naming the per-color totals ($6$ red, $10$ blue, $16$ white) so each pair type can be tracked as a contribution to those totals.)
  • 6.EE.B.6 Use variables to represent numbers and write expressions when solving a real-world or mathematical problem (Setting up the accounting equations $\text{reds used} = 6$ and $\text{blues used} = 10$, then solving for the unknown pair counts.)
  • 6.NS.B.2 Fluently divide multi-digit numbers using the standard algorithm (Dividing the leftover $10$ whites by $2$ to count the white-white pairs they form.)

⭐ 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).

⭐ 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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