1959 AMC 12 第 41 题

先试着解答 1959 AMC 12 第 41 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 1959 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。

41.

在一条直线的同一侧画三个圆:一个半径为 44 英寸的圆与该直线相切,另两个圆全等,且每个圆都与该直线及另外两个圆相切。两个全等圆的半径为:
On the same side of a straight line three circles are drawn as follows: a circle with a radius of 44 inches is tangent to the line, the other two circles are equal, and each is tangent to the line and to the other two circles. The radius of the equal circles is:
2424
2020
1818
1616
1212
答案:D
知识点:相切圆距离公式
小提示:
小圆对称地位于两个全等圆之间
The small circle lies symmetrically between the two equal circles
大提示:
若全等圆半径为 RR,比较圆心距 R2+(R−4)2\sqrt{R^2+(R-4)^2} 与 R+4R+4
If an equal circle has radius R,R, compare the center distance R2+(R−4)2\sqrt{R^2+(R-4)^2} with R+4R+4
解答:
设两个全等圆的半径为 RR。它们的圆心相距 2R2R,所以半径为 44 的圆的圆心位于两者正中间。小圆圆心与任一大圆圆心的水平距离和竖直距离分别为 RR 和 R−4R-4。由相切条件得 R2+(R−4)2=(R+4)2。 R^2+(R-4)^2=(R+4)^2\text{。}化简得 R2−16R=0R^2-16R=0,由正性知 R=16R=16。
因此,正确答案是 D。
Let the equal circles have radius R.R. Their centers are 2R2R apart, so the center of the radius-44 circle lies midway between them. The horizontal and vertical separations between its center and either large center are RR and R−4.R-4. Tangency gives R2+(R−4)2=(R+4)2. R^2+(R-4)^2=(R+4)^2. Simplifying yields R2−16R=0,R^2-16R=0, and positivity gives R=16.R=16.
Thus, the correct answer is D.
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