AMC 10 · 2018 · #11

Grade 8 geometry-2d
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 centre 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 the centre 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)
$w^2h$
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

The box is turned forty-five degrees, so the diagonal rules.

d√(2)=w → d=w/√(2)
2STEP 2

Unfold the fold into one distance

Unfold the fold into one distance.

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

Measure that distance on the flat sheet

Measure that distance on the flat sheet.

(S-d)/√(2) = h+w/2
4STEP 4

Solve for the side of the sheet

That fixes the sheet's side.

S = w√(2)+√(2) h = √(2) (w+h)
5STEP 5

Square the side to get the area

Squaring gives two times w plus h, squared.

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