2009 AMC 8 第 9 题

先试着解答 2009 AMC 8 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 2009 AMC 8 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.

在一个等边三角形的一边上作一个正方形。在正方形的一条不相邻的边上作一个正五边形,如图所示。在五边形的一条不相邻的边上作一个正六边形。用同样方式继续作正多边形,直到作出正八边形。所得图形的外边界有多少条边?
Construct a square on one side of an equilateral triangle. On one non-adjacent side of the square, construct a regular pentagon, as shown. On a non-adjacent side of the pentagon, construct a regular hexagon. Continue to construct regular polygons in the same way, until you construct an octagon. How many sides does the resulting polygon have?
2121
2323
2525
2727
2929
答案:B
知识点:正多边形基本计数
小提示:
只数每个多边形留在外边界上的边
Count only the outside sides of each polygon.
大提示:
中间的多边形各共享两条边,两端的多边形各共享一条边
Middle polygons share two sides; the two end polygons share one side each.
解答:
三角形和八边形在链的两端,所以各有一条边成为共享边而不在外边界上。正方形、五边形、六边形和七边形在中间,所以各有两条边成为共享边。
所得图形有 (3−1)+(4−2)+(5−2)+(6−2)+(7−2)+(8−1)=23 \begin{aligned} &(3-1)+(4-2) \\ &\quad {}+(5-2)+(6-2) \\ &\quad {}+(7-2)+(8-1)=23 \end{aligned} 条边。
所以正确答案是 B。
The triangle and octagon are at the ends of the chain, so each loses one side to a shared edge. The square, pentagon, hexagon, and heptagon are in the middle, so each loses two sides to shared edges.
The resulting polygon has (3−1)+(4−2)+(5−2)+(6−2)+(7−2)+(8−1)=23 \begin{aligned} &(3-1)+(4-2) \\ &\quad {}+(5-2)+(6-2) \\ &\quad {}+(7-2)+(8-1)=23 \end{aligned} sides.
Thus, B is the correct answer.
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