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2025 AMC 12B Problem 12

Problem 12 of 25IntermediateGeometry

The windshield wiper on the driver’s side of a large bus is depicted below. Arm AB\overline{AB} pivots back and forth around point A,A, sweeping out an arc of 60,60^\circ, symmetric about the vertical line through A.A. The wiper blade CD\overline{CD} is attached to BB at its midpoint and stays vertical as the arm moves. The arm is 33 feet long, and the wiper blade is 3.53.5 feet tall. What is the area of the windshield cleaned by the wiper, in square feet, to the nearest hundredth? (Assume that the windshield is a flat vertical surface.)

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Solution

Put AA at the origin. Then B=(3sinθ,3cosθ)B = (3\sin\theta, 3\cos\theta) for θ[30,30],\theta \in [-30^\circ, 30^\circ], so the horizontal coordinate of BB ranges over [1.5,1.5],[-1.5, 1.5], a width of 3.3. At each horizontal position exactly one vertical blade of height 3.53.5 passes through, so by Cavalieri’s principle the cleaned area is 3.5×3=10.53.5 \times 3 = 10.5 square feet. Thus, the correct answer is C.

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Concepts: area · arc

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