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2018 AMC 10B Problem 15

Problem 15 of 25IntermediateGeometry

A closed box with a square base is to be wrapped with a square sheet of wrapping paper. The box is centered on the wrapping paper with the vertices of the base lying on the midlines of the square sheet of paper, as shown in the figure on the left. The four corners of the wrapping paper are to be folded up over the sides and brought together to meet at the center of the top of the box, point AA in the figure on the right. The box has base length ww and height h.h. What is the area of the sheet of wrapping paper?

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Solution

Let the sheet have side s.s. The base sits as a square of side ww turned 45∘,45^\circ, so the center is w2\tfrac{w}{2} from each base edge. A corner of the sheet lies s2\tfrac{s}{\sqrt2} from the center. Folding that corner up to the top center traces a straight line: w2\tfrac{w}{2} out to the base edge, then hh up the side, then w2\tfrac{w}{2} across the top. So s2=w2+h+w2=w+h.\tfrac{s}{\sqrt2} = \tfrac{w}{2} + h + \tfrac{w}{2} = w + h. Then s=2 (w+h),s = \sqrt2\,(w + h), and the area is s2=2(w+h)2.s^2 = 2(w + h)^2. Thus, A is the correct answer.

More practice

Concepts: paper folding · Pythagorean Theorem · square (geometry)

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.