2021 AMC 12A Fall 第 11 题

先试着解答 2021 AMC 12A Fall 第 11 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2021 AMC 12A Fall 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

11.

考虑两个同心圆,半径分别为 1717 和 1919。较大的圆有一条弦,其中一半位于较小圆内。这条较大圆中的弦长是多少?
Consider two concentric circles of radius 1717 and 19.19. The larger circle has a chord, half of which lies inside the smaller circle. What is the length of the chord in the larger circle?
12212\sqrt{2}
10310\sqrt{3}
17⋅19\sqrt{17 \cdot 19}
1818
868\sqrt{6}
答案:E
知识点:弦勾股定理
小提示:
设弦到圆心的距离为 dd。它的半长是 192−d2\sqrt{19^2 - d^2},在小圆内的部分半长是 172−d2\sqrt{17^2 - d^2}。
Let the chord be at distance dd from the center. Its half-length is 192−d2,\sqrt{19^2 - d^2}, and the part inside the small circle has half-length 172−d2\sqrt{17^2 - d^2}
大提示:
“一半位于小圆内”表示内部部分等于整条弦的一半:2172−d2=192−d22\sqrt{17^2 - d^2} = \sqrt{19^2 - d^2}。
“Half lies inside” means the inside portion equals half the whole chord: 2172−d2=192−d22\sqrt{17^2 - d^2} = \sqrt{19^2 - d^2}
解答:
设这条弦到共同圆心的距离为 dd。它的总长度为 2361−d22\sqrt{361 - d^2},位于较小圆内的部分长度为 2289−d22\sqrt{289 - d^2}。
因为弦的一半位于小圆内,2289−d2=12⋅2361−d22\sqrt{289 - d^2} = \tfrac{1}{2}\cdot 2\sqrt{361 - d^2}。平方得 4(289−d2)=361−d24(289 - d^2) = 361 - d^2,所以 3d2=7953d^2 = 795,d2=265d^2 = 265。
弦长为 2361−265=296=862\sqrt{361 - 265} = 2\sqrt{96} = 8\sqrt{6}。
所以正确答案是 E。
Let the chord lie at distance dd from the common center. Its total length is 2361−d2,2\sqrt{361 - d^2}, and the portion inside the smaller circle has length 2289−d2.2\sqrt{289 - d^2}.
Since half the chord lies inside, 2289−d2=12⋅2361−d2.2\sqrt{289 - d^2} = \tfrac{1}{2}\cdot 2\sqrt{361 - d^2}. Squaring gives 4(289−d2)=361−d2,4(289 - d^2) = 361 - d^2, so 3d2=7953d^2 = 795 and d2=265.d^2 = 265.
The chord length is 2361−265=296=86.2\sqrt{361 - 265} = 2\sqrt{96} = 8\sqrt{6}.
Thus, the correct answer is E.
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