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2023 AMC 12A Problem 15

Problem 15 of 25IntermediateGeometry

Usain is walking for exercise by zigzagging across a 100100-meter by 3030-meter rectangular field, beginning at point AA and ending on the segment BC‾.\overline{BC}. He wants to increase the distance walked by zigzagging as shown in the figure below (APQRSAPQRS). What angle θ=∠PAB\theta=\angle PAB =∠QPC=\angle QPC =∠RQB=⋯=\angle RQB=\cdots will produce a length that is 120120 meters? (Do not assume the zigzag path has exactly four segments as shown; there could be more or fewer.)

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Solution

Every segment of the zigzag makes angle θ\theta with a horizontal side of the field. Therefore a segment of length ss advances scos⁡θs\cos\theta meters horizontally. This remains true for the last segment even if it ends before crossing the full width of the field. Adding the horizontal projections over the entire 120120-meter path gives 120cos⁡θ=100.120\cos\theta=100. Therefore cos⁡θ=56,\cos\theta=\dfrac56, so θ=arccos⁡56.\theta=\arccos\dfrac56. Thus, the correct answer is A.

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Concepts: trigonometry · right triangle

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