AMC 8 · 2001 · #16
Grade 4 geometry-2drate-ratioProblem
A square piece of paper, 4 inches on a side, is folded in half vertically. Both layers are then cut in half parallel to the fold. Three new rectangles are formed, a large one and two small ones. What is the ratio of the perimeter of one of the small rectangles to the perimeter of the large rectangle?
Pick an answer.
AMC 8 2001 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: 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.
Givens: The starting paper is a square of side $4$ inches; First action: fold in half vertically (so the fold runs top-to-bottom); Second action: cut both layers in half, with the cut parallel to the fold; After unfolding, the pieces are one "large" rectangle and two "small" rectangles; Answer choices: (A) $\dfrac{1}{3}$, (B) $\dfrac{1}{2}$, (C) $\dfrac{3}{4}$, (D) $\dfrac{4}{5}$, (E) $\dfrac{5}{6}$
Unknowns: $\dfrac{\text{perimeter of one small rectangle}}{\text{perimeter of the large rectangle}}$
Understand
Restated: 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.
Givens: The starting paper is a square of side $4$ inches; First action: fold in half vertically (so the fold runs top-to-bottom); Second action: cut both layers in half, with the cut parallel to the fold; After unfolding, the pieces are one "large" rectangle and two "small" rectangles; Answer choices: (A) $\dfrac{1}{3}$, (B) $\dfrac{1}{2}$, (C) $\dfrac{3}{4}$, (D) $\dfrac{4}{5}$, (E) $\dfrac{5}{6}$
Plan
Primary tool: #10 Use a Physical Model
Secondary: #1 Draw a Diagram
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.
Execute — Answer: E
4.MD.A.3 Step 1 - Track the paper after the vertical fold.
- Folding the $4 \times 4$ square in half along a vertical line halves the width but keeps the height.
- The result is a two-layer rectangle that is $2$ inches wide and $4$ inches tall, with the fold running along one of the $4$-inch sides.
💡 Folding in half along a line that runs in one direction always halves the perpendicular dimension and keeps the parallel one. Width was $4$, now it is $2$.
4.MD.A.3 Step 2 - Make the vertical cut.
- The cut is parallel to the fold, so it is also vertical, and it cuts the $2$-inch width in half.
- It slices through both layers at once, producing two two-layer strips, each $1$ inch wide and $4$ inches tall.
- One strip contains the fold (the inner edge); the other strip is made of the two outer edges that were never connected to each other.
💡 A cut halfway across the $2$-inch width gives two $1$-inch-wide strips. Which one contains the fold matters for the next step.
4.MD.A.3 Step 3 - Unfold each strip.
- The strip that contains the fold opens up to a single sheet whose width doubles from $1$ inch to $2$ inches — this is the large rectangle, $2 \times 4$.
- The other strip has no fold to hold its two layers together, so when picked up it separates into two identical small rectangles, each $1 \times 4$.
💡 Only the fold edge keeps two layers connected after unfolding. The outer-edge strip was held together by nothing, so it falls into two sheets.
3.MD.D.8 Step 4 Apply the perimeter formula to a small rectangle ($1 \times 4$) and to the large rectangle ($2 \times 4$).
💡 Perimeter $= 2 \times (\text{length} + \text{width})$ — the standard Grade 3 rectangle rule.
4.NF.A.1 Step 5 Form the requested ratio and simplify by dividing numerator and denominator by their common factor $2$.
💡 $\tfrac{10}{12}$ and $\tfrac{5}{6}$ are equivalent fractions — divide top and bottom by $2$.
4.MD.A.3 Track the paper after the vertical fold. Folding the $4 \times 4$ square in half 4.MD.A.3 Make the vertical cut. The cut is parallel to the fold, so it is also vertical, 4.MD.A.3 Unfold each strip. The strip that contains the fold opens up to a single sheet w 3.MD.D.8 Apply the perimeter formula to a small rectangle ($1 \times 4$) and to the large 4.NF.A.1 Form the requested ratio and simplify by dividing numerator and denominator by t Review
Reasonableness: Check by total area. The three pieces should together account for the original $4 \times 4 = 16$ square inches. Large $2 \times 4 = 8$ plus two smalls each $1 \times 4 = 4$ gives $8 + 4 + 4 = 16$. The pieces are accounted for. The ratio $\tfrac{5}{6}$ also passes a size check: the small rectangle is narrower than the large one, so its perimeter should be a little smaller — $\tfrac{5}{6}$ is less than $1$ but not by much, which matches. Trap answers fit common slips: (B) $\tfrac{1}{2}$ is the ratio of widths $1{:}2$, not perimeters; (A) $\tfrac{1}{3}$ comes from comparing one small piece to all three pieces combined; (C) $\tfrac{3}{4}$ comes from forgetting that the unfolded fold-strip doubles in width.
Alternative: Tool #1 alone (Draw a Diagram): on grid paper, draw the $4 \times 4$ square, then draw the vertical fold line at $x=2$ and the two cut lines at $x=1$ and $x=3$ (the cut, made on the folded paper, lands in two places when unfolded). The three pieces appear directly: one $2 \times 4$ strip in the middle ($1 \le x \le 3$) and two $1 \times 4$ strips on the sides ($0 \le x \le 1$ and $3 \le x \le 4$). Compute perimeters $10$ and $12$ and form $\tfrac{10}{12} = \tfrac{5}{6}$.
CCSS standards used (min grade 4)
3.MD.D.8Solve real world and mathematical problems involving perimeters of polygons (Applying $P = 2(\text{length} + \text{width})$ to the small rectangle ($1 \times 4 \Rightarrow 10$) and the large rectangle ($2 \times 4 \Rightarrow 12$).)4.MD.A.3Apply the area and perimeter formulas for rectangles in real world and mathematical problems (Tracking the rectangle's dimensions through the fold ($4 \times 4 \to 2 \times 4$ two-layer) and the cut ($2 \times 4 \to$ two $1 \times 4$ two-layer strips).)4.NF.A.1Explain why a fraction $a/b$ is equivalent to a fraction $(n \times a)/(n \times b)$ (Simplifying $\dfrac{10}{12}$ to $\dfrac{5}{6}$ by dividing numerator and denominator by the common factor $2$.)
⭐ Imagine doing the fold and cut yourself: the strip that holds the fold opens up to a $2 \times 4$ large rectangle, and the outer-edge strip falls apart into two $1 \times 4$ small rectangles. Perimeters $10$ and $12$ give the ratio $\tfrac{5}{6}$, answer (E).
⭐ Imagine doing the fold and cut yourself: the strip that holds the fold opens up to a $2 \times 4$ large rectangle, and the outer-edge strip falls apart into two $1 \times 4$ small rectangles. Perimeters $10$ and $12$ give the ratio $\tfrac{5}{6}$, answer (E).
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