AMC 10 · 2018 · #15

Grade 8 geometry-2d
Archetype Compound Area by Decomposition / Difference └ Sub-archetype Polygon Triangulation
pythagorean-theoremarea-rectanglespaper-foldingspatial-visualization reflection-unfoldingidentify-subproblems ↑ Prerequisites: pythagorean-theorem
📏 Medium solution 💡 3 insights 📊 Diagram
Problem
A closed box has a square base of side w and height h. It sits in the center of a square sheet of wrapping paper, turned so that the four corners of the base land on the two midlines of the sheet. The four corners of the paper are folded up over the sides and meet at point A, the center of the top of the box. Find the area of the square sheet.

Pick an answer.

(A)
$2(w+h)^2$
(B)
$frac{(w+h)^2}2$
(C)
$2w^2+4wh$
(D)
$2w^2$
(E)
$w^2h$

AMC 10 2018 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Visualize Spatial Relationships

Tool #17 (Visualize Spatial Relationships): the hard part is the fold. A flat triangular flap of paper bends up the side of the box and then lies over the top to land on A. If we mentally unfold that path, the whole journey becomes one straight perpendicular distance measured on the flat sheet, which we can compute. Tool #1 (Draw a Diagram): putting the sheet on xy-coordinates with the box base vertices on the axes turns the tilt into clean numbers. Tool #7 (Identify Subproblems): split the unknown distance into two easy pieces — the climb up the side (h) and the reach across the top from a top edge to its center (w/2) — then add them.

1STEP 1

Put the sheet on coordinates

Put the sheet on axes; the tilted base has corners (d,0),(0,d),(-d,0),(0,-d). One edge runs (d,0) to (0,d), so by Pythagoras d√(2)=w.

d√(2)=w → d=w/√(2)
2STEP 2

Unfold the fold into one distance

Unfolded flat, the corner at (S/2,S/2) lies a straight distance h+w/2 from the crease: climb h up the side plus reach w/2 across the top.

(perpendicular distance from corner to base edge) = h+w/2
3STEP 3

Measure that distance on the flat sheet

The crease lies on x+y=d, so the corner's perpendicular distance to it is (S-d)/√(2). Set that equal to h+w/2.

(S-d)/√(2) = h+w/2
4STEP 4

Solve for the side of the sheet

Clear the √(2) and substitute d=w/√(2); the two w/√(2) pieces combine to w√(2), giving S=√(2)(w+h).

S = w√(2)+√(2) h = √(2) (w+h)
5STEP 5

Square the side to get the area

The area is S²=(√(2)(w+h))²; squaring turns √(2) into 2 and keeps (w+h)², giving 2(w+h)² — choice (A).

S² = (√(2))² (w+h)² = 2(w+h)² → (A)
Answer
2(w+h)²
Test with w=2,h=1. Then S=√2(3)=3√2 and the area is 2·3²=18. Check it a different way by accounting for what the paper covers: the bottom of the box (w²=4), the four sides (4wh=8), and the top (w²=4) total 16, and the four little corner triangles that overlap going up the sides add the remaining 2, giving 18 — a match. The answer also passes unit sense: every choice except (E) is an area in square length units, and (A) grows correctly when either w or h grows, while (D) 2w² ignores the height entirely and (C) 2w²+4wh is exactly the box's surface area with the corner triangles left out, so both fall short.
💡Key takeaway

Unfold the fold: a paper corner's trip up the side and across the top is one straight distance h+w/2 on the flat sheet, which forces the sheet's side to be √2(w+h) and its area 2(w+h)².

  • Put the sheet on coordinates
  • Unfold the fold into one distance
  • Measure that distance on the flat sheet
  • Solve for the side of the sheet
  • Square the side to get the area

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