1952 AMC 12 第 27 题

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所有题目均经美国数学协会(MAA)官方合法授权使用。

27.

一个等边三角形的高等于某圆的半径。该三角形的周长与内接于此圆的等边三角形周长之比为:

The ratio of the perimeter of an equilateral triangle having an altitude equal to the radius of a circle, to the perimeter of an equilateral triangle inscribed in the circle is:

1:21:2

1:31:3

1:31:\sqrt3

3:2\sqrt3:2

2:32:3

答案:E
知识点:等边三角形比与比例
难度评级:1560
小提示:

用同一个圆的半径 RR 表示两个等边三角形的边长

Express each equilateral triangle’s side length in terms of the same circle radius RR

大提示:

高为 RR 的等边三角形边长为 2R3\frac{2R}{\sqrt3},而内接等边三角形的边长为 3R\sqrt3R

An equilateral triangle of altitude RR has side 2R3\frac{2R}{\sqrt3}, while an inscribed one has side 3R\sqrt3R

解答:

边长为 ss 的等边三角形,其高为 s32\frac{s\sqrt3}{2}。因此,第一个三角形的边长为 2R3\frac{2R}{\sqrt3},周长为 23R2\sqrt3R。内接于半径为 RR 的圆的等边三角形边长为 3R\sqrt3R,周长为 33R3\sqrt3R

因此,周长之比为 2:32:3,所以正确答案是 E

An equilateral triangle with side ss has altitude s32.\frac{s\sqrt3}{2}. Thus the first triangle has side 2R3\frac{2R}{\sqrt3} and perimeter 23R.2\sqrt3R. An equilateral triangle inscribed in a circle of radius RR has side 3R\sqrt3R and perimeter 33R.3\sqrt3R.

The perimeter ratio is therefore 2:3,2:3, so the correct answer is E.

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