2018 AMC 10B Problem 15

Below is the professionally curated solution for Problem 15 of the 2018 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2018 AMC 10B solutions, or check the answer key.

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Concepts:paper foldingPythagorean Theoremsquare (geometry)

Difficulty rating: 1730

15.

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. The four corners of the wrapping paper are folded up over the sides and brought together to meet at the center of the top of the box. The box has base length ww and height h.h. What is the area of the sheet of wrapping paper?

2(w+h)22(w + h)^2

(w+h)22\dfrac{(w + h)^2}{2}

2w2+4wh2w^2 + 4wh

2w22w^2

w2hw^2 h

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.

Problem 15 in Other Years