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2025 AMC 12B 第 12 题

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所有题目均经美国数学协会（MAA）官方合法授权使用。

12.

下图显示了一辆大型公交车驾驶员一侧的雨刷。

雨刷臂 AB‾\overline{AB} 绕点 AA 来回转动，扫过 60∘60^\circ 的弧，并且关于过 AA 的竖直线对称。雨刷片 CD‾\overline{CD} 在其中点处连到 BB，并且在雨刷臂运动时始终保持竖直。雨刷臂长 33 英尺，雨刷片高 3.53.5 英尺。雨刷清洁的挡风玻璃面积是多少平方英尺？答案四舍五入到百分位。（假设挡风玻璃是平的竖直平面。）

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.)

9.689.68

10.1410.14

10.5010.50

11.3211.32

12.0012.00

答案：C
知识点：面积弧
难度评级：1690
小提示：

雨刷臂转动时，BB 走过半径为 33 的圆弧；雨刷片始终是以 BB 为中点、高为 3.53.5 的竖直线段。

As the arm turns, BB traces an arc of radius 3;3; the blade is always a vertical segment of height 3.53.5 centered at BB

大提示：

每条竖直线与清洁区域的交集长度恒为 3.53.5，所以面积等于这个高度乘以扫过的水平宽度。

Each vertical line meets the cleaned region in a segment of constant height 3.5,3.5, so the area is that height times the horizontal width swept

解答：

把 AA 放在原点。则 B=(3sin⁡θ,3cos⁡θ)B = (3\sin\theta, 3\cos\theta)，其中 θ∈[−30∘,30∘]\theta \in [-30^\circ, 30^\circ]，所以 BB 的水平坐标范围是 [−1.5,1.5][-1.5, 1.5]，宽度为 33。每个水平位置都有一条高度为 3.53.5 的竖直雨刷片经过，所以由卡瓦列里原理，清洁面积为 3.5×3=10.53.5 \times 3 = 10.5 平方英尺。

所以正确答案是 C。

Put AA at the origin. Then B=(3sin⁡θ,3cos⁡θ)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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