1999 AMC 8 第 25 题

先试着解答 1999 AMC 8 第 25 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 1999 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。

25.

BBDDJJ 是直角三角形 ACGACG 各边的中点。点 KKEEII 是三角形 JDGJDG 各边的中点,依此类推。如果这种分割并涂阴影的过程进行 100100 次(图中显示了前三次),且 AC=CG=6AC = CG = 6,那么阴影三角形的总面积最接近

Points B,B, D,D, and JJ are midpoints of the sides of right triangle ACG.ACG. Points K,K, E,E, II are midpoints of the sides of triangle JDG,JDG, etc. If the dividing and shading process is done 100100 times (the first three are shown) and AC=CG=6,AC = CG = 6, then the total area of the shaded triangles is nearest

66

77

88

99

1010

答案:A
知识点:等比数列相似
难度评级:1620
解答:

第一个阴影三角形的两条直角边为 3333,所以面积为 1233=92\dfrac12\cdot3\cdot3=\dfrac92

之后每一步,相关三角形的边长变为原来的一半,因此阴影面积乘以 (12)2=14\left(\dfrac12\right)^2=\dfrac14

许多步后的阴影总面积非常接近等比和 92(1+14+116+)=92111/4=6. \begin{gathered} \frac92\left(1+\frac14+\frac1{16}+\cdots\right) \\ {}= \frac92\cdot\frac1{1-1/4} \\ {}= 6. \end{gathered}

100100 步后被省略的尾项极小,所以总面积最接近 66

所以正确答案是 A

The first shaded triangle has legs 33 and 33, so its area is 1233=92\dfrac12\cdot3\cdot3=\dfrac92.

At each later step, the relevant triangle has half the leg length, so the shaded area is multiplied by (12)2=14\left(\dfrac12\right)^2=\dfrac14.

The total shaded area after many steps is therefore very close to the geometric sum 92(1+14+116+)=92111/4=6. \begin{gathered} \frac92\left(1+\frac14+\frac1{16}+\cdots\right) \\ {}= \frac92\cdot\frac1{1-1/4} \\ {}= 6. \end{gathered}

After 100100 steps the omitted tail is tiny, so the total area is nearest 66.

Thus, A is the correct answer.

← 第 24 题#24
完整试卷

其他年份的第 25 题

1985 AMC 8 · 1986 AMC 8 · 1987 AMC 8 · 1988 AMC 8 · 1989 AMC 8 · 1990 AMC 8 · 1991 AMC 8 · 1992 AMC 8 · 1993 AMC 8 · 1994 AMC 8 · 1995 AMC 8 · 1996 AMC 8 · 1997 AMC 8 · 1998 AMC 8 · 2000 AMC 8 · 2001 AMC 8 · 2002 AMC 8 · 2003 AMC 8 · 2004 AMC 8 · 2005 AMC 8 · 2006 AMC 8 · 2007 AMC 8 · 2008 AMC 8 · 2009 AMC 8 · 2010 AMC 8 · 2011 AMC 8 · 2012 AMC 8 · 2013 AMC 8 · 2014 AMC 8 · 2015 AMC 8 · 2016 AMC 8 · 2017 AMC 8 · 2018 AMC 8 · 2019 AMC 8 · 2020 AMC 8 · 2022 AMC 8 · 2023 AMC 8 · 2024 AMC 8 · 2025 AMC 8 · 2026 AMC 8