1953 AMC 12 第 14 题

先试着解答 1953 AMC 12 第 14 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 1953 AMC 12 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。

14.

两个圆中,较大圆的圆心为 PP,半径为 pp;较小圆的圆心为 QQ,半径为 qq。连接 PQ‾\overline{PQ}。下列哪个说法是错误的?
Given the larger of two circles with center PP and radius pp and the smaller with center QQ and radius q.q. Draw PQ‾.\overline{PQ}. Which of the following statements is false?
p−qp-q 可以等于 PQ‾\overline{PQ}
p−qp-q can be equal to PQ‾\overline{PQ}
p+qp+q 可以等于 PQ‾\overline{PQ}
p+qp+q can be equal to PQ‾\overline{PQ}
p+qp+q 可以小于 PQ‾\overline{PQ}
p+qp+q can be less than PQ‾\overline{PQ}
p−qp-q 可以小于 PQ‾\overline{PQ}
p−qp-q can be less than PQ‾\overline{PQ}
以上说法均非错误
none of these
答案:E
知识点:circlescenter distancetangency
小提示:
分别考虑两圆内切、外切和相离时的圆心距
Interpret the center distance for internal tangency, external tangency, and disjoint circles
大提示:
对前四个说法逐一尝试构造一种可行的两圆位置关系
For each of the first four statements, try to choose a valid relative position of the circles
解答:
两圆内切时有 PQ=p−qPQ=p-q,外切时有 PQ=p+qPQ=p+q。两圆相离时可以有 PQ>p+qPQ\gt p+q,而许多相交或相离的情形满足 PQ>p−qPQ\gt p-q。因此 A 至 D 中的每种情况都可能发生,没有一个说法是错误的。
因此,正确答案是 E。
Internal tangency realizes PQ=p−q,PQ=p-q, and external tangency realizes PQ=p+q.PQ=p+q. Disjoint circles can have PQ>p+q,PQ\gt p+q, and many intersecting or disjoint configurations have PQ>p−q.PQ\gt p-q. Thus each of A through D can occur, so none of them is false.
Therefore, the correct answer is E.
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