1992 AMC 12 第 9 题

先试着解答 1992 AMC 12 第 9 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 1992 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。

9.

五个边长均为 232\sqrt3 的等边三角形排列在同一直线的一侧,且每个三角形都有一条边在该直线上。沿着这条直线,一个三角形底边的中点是下一个三角形的一个顶点。这五个三角形区域的并集所覆盖的平面区域面积为
Five equilateral triangles, each with side 23,2\sqrt3, are arranged so they are all on the same side of a line containing one side of each. Along this line, the midpoint of the base of one triangle is a vertex of the next. The area of the region of the plane that is covered by the union of the five triangular regions is
1010
1212
1515
10310\sqrt3
12312\sqrt3
答案:E
知识点:equilateral triangle area容斥原理overlapping regions
小提示:
先把五个大三角形的面积相加,再减去重叠部分
Add the areas of the five large triangles, then subtract their overlaps
大提示:
每一对相邻三角形的重叠部分都是边长为原来一半的等边三角形
Each adjacent overlap is an equilateral triangle with half the original side length
解答:
每个大三角形的面积为 34(23)2=33。 \frac{\sqrt3}{4}(2\sqrt3)^2=3\sqrt3\text{。}四个重叠部分都是边长为 3\sqrt3 的等边三角形,面积均为 334\frac{3\sqrt3}{4}。不存在三个三角形共同重叠的区域,所以并集的面积为 5(33)−4(334)=123。 5(3\sqrt3)-4\left(\frac{3\sqrt3}{4}\right)=12\sqrt3\text{。}
因此正确答案是 E。
Each large triangle has area 34(23)2=33. \frac{\sqrt3}{4}(2\sqrt3)^2=3\sqrt3. Each of the four overlaps is an equilateral triangle of side 3,\sqrt3, hence area 334.\frac{3\sqrt3}{4}. There are no triple overlaps, so the union has area 5(33)−4(334)=123. 5(3\sqrt3)-4\left(\frac{3\sqrt3}{4}\right)=12\sqrt3.
Thus the correct answer is E.
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