AMC 8 · 2001 · #16

Grade 4 geometry-2drate-ratio

Archetype Paper Fold-and-Cut Symmetry └ Sub-archetype Fold-Cut-Unfold Shape Identification

paper-foldingspatial-visualizationperimeterratio-proportion physical-representationidentify-subproblems ↑ Prerequisites: perimeterratio-proportion

📏 Long solution 💡 4 insights 📊 Diagram

📘 View easy version →

Problem

A 4 inch by 4 inch square is folded in half along a vertical line. Both layers of the folded paper are then cut in half by a single vertical cut parallel to the fold. Unfolding produces three rectangles: one large and two small. Find the ratio of the perimeter of one small rectangle to the perimeter of the large rectangle.

Pick an answer.

(A)

$\dfrac{1}{3}$

(B)

$\dfrac{1}{2}$

(C)

$\dfrac{3}{4}$

(D)

$\dfrac{4}{5}$

(E)

$\dfrac{5}{6}$

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

How to solve

Strategy Use a Physical Model

The whole problem is a sequence of physical actions on paper — fold, cut, unfold. Tool #10 (Use a Physical Model) is the most honest way in: take any rectangular sheet, perform the two actions, and read the resulting dimensions directly. Tool #1 (Draw a Diagram) replaces the actual paper with a labelled sketch when paper is not handy. The key insight that decides everything is that the cut piece containing the fold stays connected when unfolded (it doubles in width), while the cut piece made of the two outer edges falls apart into two separate sheets. We avoid Tool #13 (Algebra) because once the dimensions are read off, the perimeter formula is a Grade 3 one-line computation.

1STEP 1

Folding the 4 × 4 square along the vertical line halves the width, leaving a two-layer rectangle 2 inches wide and 4 inches tall.

after fold: 2 in (wide) × 4 in (tall), two layers

2STEP 2

The vertical cut halves the 2-inch width through both layers, giving two 1×4 strips; one holds the fold, the other the two outer edges.

two strips, each 1 in × 4 in, two layers

3STEP 3

Unfolding the fold-strip doubles its width into the large 2 × 4 rectangle; the other strip separates into two small 1 × 4 rectangles.

large: 2 × 4 small: 1 × 4 (two of them)

4STEP 4

Apply P = 2 × (length + width): the 1 × 4 small gives perimeter 10 and the 2 × 4 large gives perimeter 12.

P_small = 2(1+4) = 10 P_large = 2(2+4) = 12

5STEP 5

Form the ratio $\frac{10}{12}$ ; dividing top and bottom by 2 gives $\frac{5}{6}$ — answer (E).

\frac{P_small}{P_large}

\frac{10}{12}

\frac{5}{6}

→ (E)

Answer

\frac{5}{6}

Check by total area. The three pieces should together account for the original 4 × 4 = 16 square inches. Large 2 × 4 = 8 plus two smalls each 1 × 4 = 4 gives 8 + 4 + 4 = 16. The pieces are accounted for. The ratio

\frac{5}{6}

also passes a size check: the small rectangle is narrower than the large one, so its perimeter should be a little smaller —

\frac{5}{6}

is less than 1 but not by much, which matches. Trap answers fit common slips: (B)

\frac{1}{2}

is the ratio of widths 1/:2, not perimeters; (A)

\frac{1}{3}

comes from comparing one small piece to all three pieces combined; (C)

\frac{3}{4}

comes from forgetting that the unfolded fold-strip doubles in width.

💡Key takeaway

Imagine doing the fold and cut yourself: the strip that holds the fold opens up to a 2 × 4 large rectangle, and the outer-edge strip falls apart into two 1 × 4 small rectangles. Perimeters 10 and 12 give the ratio $\frac{5}{6}$ , answer (E).

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