AMC 8 1997 Problem 24

AMC8 1997 年第 24 题

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Diameter ACEACE is divided at CC in the ratio 2:32:3 . The two semicircles, ABCABC and CDECDE , divide the circular region into an upper (shaded) region and a lower region. The ratio of the area of the upper region to that of the lower region is
直径 ACEACE 被点 CC 分成 2:32:3 的比例。两个半圆 ABCABC 和 CDECDE 将圆形区域分成上部（着色）区域和下部区域。上部区域与下部区域的面积之比是
AMC 8 1997 Problem 24 figure
  1. A.2:3
  2. B.1:1
  3. C.3:2
  4. D.9:4
  5. E.5:2

正确答案 · Correct Answer

C

解析 · Solution

解法1

常规解法 · 三个半圆加减 取 AC=4、CE=6（比 2:32:3），大圆半径 5。上方阴影 = 大半圆（直径 AEAE）+ 下方小半圆 CDECDE − 上方小半圆 ABCABC，三半径为 5、3、2，面积 12π(52+32−22)=15π\tfrac12\pi(5^2+3^2-2^2)=15\pi。整圆 25π25\pi，下方 =25π−15π=10π=25\pi-15\pi=10\pi，比 15:10=3:2。

Solution 1

Elementary · add and subtract three semicircles Take AC=4 and CE=6 (ratio 2:32:3), so the big circle has radius 5. The upper region equals the large semicircle on diameter AEAE plus the small semicircle CDECDE minus the small semicircle ABCABC, with radii 5, 3, 2: 12π(52+32−22)=15π\tfrac12\pi(5^2+3^2-2^2)=15\pi. The whole circle is 25π25\pi, so the lower region is 10π10\pi, and the ratio is 15:10=3:2.

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