1954 AMC 12 第 26 题

先试着解答 1954 AMC 12 第 26 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 1954 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。

26.

点 CC 将直线段 ABAB 分割,使得 AC‾=3CB‾\overline{AC}=3\overline{CB}。分别以 AC‾\overline{AC} 和 CB‾\overline{CB} 为直径作圆,两圆的一条公切线与 AB‾\overline{AB} 的延长线交于 DD。则 BD‾\overline{BD} 等于:
The straight line ABAB is divided at CC so that AC‾=3CB‾.\overline{AC}=3\overline{CB}. Circles are described on AC‾\overline{AC} and CB‾\overline{CB} as diameters and a common tangent meets AB‾\overline{AB} produced at D.D. Then BD‾\overline{BD} equals:
较小圆的直径
the diameter of the smaller circle
较小圆的半径
the radius of the smaller circle
较大圆的半径
the radius of the larger circle
CB‾3\overline{CB}\sqrt3
两圆半径之差
the difference of the two radii
答案:B
知识点:相切圆切线位似
小提示:
点 DD 是两圆的外位似中心
The point DD is the external center of similitude of the two circles
大提示:
设 CB‾=u\overline{CB}=u,并按 3:13:1 比较 DD 到两圆心的距离
Let CB‾=u\overline{CB}=u and compare the distances from DD to the two centers in the ratio 3:13:1
解答:
设 CB‾=u\overline{CB}=u,则 AC‾=3u\overline{AC}=3u,且 AB‾=4u\overline{AB}=4u。从 AA 起量,两圆心的位置分别为 3u2\frac{3u}{2} 和 7u2\frac{7u}{2},两半径之比为 3:13:1。若 AD=zAD=z,公切线使 DD 成为两圆的外位似中心,所以 z−32uz−72u=3。 \frac{z-\frac32u}{z-\frac72u}=3\text{。}因此 z=9u2z=\frac{9u}{2},且 BD=z−4u=u2, BD=z-4u=\frac u2\text{,}这正是较小圆的半径。
因此,正确答案是 B。
Let CB‾=u,\overline{CB}=u, so AC‾=3u\overline{AC}=3u and AB‾=4u.\overline{AB}=4u. Measured from A,A, the circle centers are at 3u2\frac{3u}{2} and 7u2,\frac{7u}{2}, and their radii are in the ratio 3:1.3:1. If AD=z,AD=z, the common external tangent makes DD the external center of similitude, so z−32uz−72u=3. \frac{z-\frac32u}{z-\frac72u}=3. Hence z=9u2,z=\frac{9u}{2}, and BD=z−4u=u2, BD=z-4u=\frac u2, which is the radius of the smaller circle.
Thus, the correct answer is B.
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