2000 AMC 12 第 24 题

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24.

若圆弧 ACACBCBC 的圆心分别为 BBAA,则存在一个圆同时与圆弧 ACAC、圆弧 BCBC 以及线段 AB\overline{AB} 相切。若圆弧 BCBC 的长度为 1212,则这个圆的周长是多少?

If circular arcs ACAC and BCBC have centers at BB and A,A, respectively, then there exists a circle tangent to both arc ACAC and arc BC,BC, and to AB.\overline{AB}. If the length of arc BCBC is 12,12, then what is the circumference of the circle?

2424

2525

2626

2727

2828

答案:D
知识点:圆幂相切圆
难度评级:2390
解答:

每条圆弧的半径都是 ABAB, 且 CCAABB 的距离都为 ABAB 所以 ABC\triangle ABC 是等边三角形。因此圆弧 BCBC 对应半径为 ABAB 的圆上的 6060^\circ 圆心角,该大圆的整周长为 612=726 \cdot 12 = 72

设小圆半径为 rr,并在线段 AB\overline{AB} 的中点 DD 处相切,其中 AD=12ABAD = \tfrac12 AB。由点幂定理,AD2=AB(AB2r)AD^2 = AB(AB - 2r),所以 从而 2r=34AB2r = \tfrac34 AB,因而 r=38ABr = \tfrac38 ABAB24=AB22rAB, \frac{AB^2}{4} = AB^2 - 2r\,AB,

周长之比等于半径之比,所以小圆的周长为 3872=27\tfrac38 \cdot 72 = 27

因此,正确答案是 D

Each arc has radius AB,AB, and CC is at distance ABAB from both AA and B,B, so ABC\triangle ABC is equilateral. Thus arc BCBC subtends 6060^\circ of a circle of radius AB,AB, whose full circumference is 612=72.6 \cdot 12 = 72.

Let the small circle have radius rr and touch AB\overline{AB} at its midpoint D,D, where AD=12AB.AD = \tfrac12 AB. By Power of a Point, AD2=AB(AB2r),AD^2 = AB(AB - 2r), so AB24=AB22rAB, \frac{AB^2}{4} = AB^2 - 2r\,AB, giving 2r=34AB,2r = \tfrac34 AB, hence r=38AB.r = \tfrac38 AB.

The circumferences are in the ratio of the radii, so the small circle's circumference is 3872=27.\tfrac38 \cdot 72 = 27.

Thus, the correct answer is D.

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