1995 AIME 第 4 题

先试着解答 1995 AIME 第 4 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 1995 AIME 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。

4.

半径分别为 33 和 66 的两个圆彼此外切,并且都与一个半径为 99 的圆内切。半径为 99 的圆有一条弦,它同时是另外两个圆的公外切线。求这条弦长的平方。
Circles of radius 33 and 66 are externally tangent to each other and are internally tangent to a circle of radius 9.9. The circle of radius 99 has a chord that is a common external tangent of the other two circles. Find the square of the length of this chord.
答案:224
知识点:相切圆弦坐标几何
小提示:
三个圆心共线,因为它们两两之间的距离为 66、33 和 99
The three circle centers are collinear because their pairwise distances are 6,6, 3,3, and 99
大提示:
利用两个小圆圆心到公切线的有向距离,求出大圆圆心到该切线的距离
Use signed distances from the two smaller centers to the common tangent to find its distance from the large center
解答:
将半径为 99 的圆的圆心置于原点,并将两个小圆的圆心分别置于 (−6,0)(-6,0) 和 (3,0)(3,0)。将公外切线写成 n⋅(x,y)=c\mathbf n\cdot(x,y)=c,其中 n\mathbf n 是单位法向量。两个圆心到直线的有向距离之差为 6−3=36-3=3,所以 9nx=39n_x=3,且 nx=13n_x=\frac{1}{3}。再利用距离 33(从 (−6,0)(-6,0) 量起),得到 ∣c∣=5|c|=5。因此这条弦到大圆圆心的距离为 55,弦长的平方为 4(92−52)=4⋅56=224。4(9^2-5^2)=4\cdot56=224\text{。}
Put the radius-99 circle at the origin and the smaller centers at (−6,0)(-6,0) and (3,0).(3,0). Write the common external tangent as n⋅(x,y)=c,\mathbf n\cdot(x,y)=c, where n\mathbf n is a unit normal. Its signed distances from the two centers differ by 6−3=3,6-3=3, so 9nx=39n_x=3 and nx=13.n_x=\frac{1}{3}. Using the distance 33 from (−6,0)(-6,0) gives ∣c∣=5.|c|=5. Thus the chord lies 55 units from the large center, and its squared length is 4(92−52)=4⋅56=224.4(9^2-5^2)=4\cdot56=224.
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