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.

点 BB、DD 和 JJ 是直角三角形 ACGACG 各边的中点。点 KK、EE 和 II 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
小提示:
阴影三角形的面积形成一个等比数列
The shaded triangles form a geometric area pattern
大提示:
每个新的阴影三角形面积是前一个阴影面积的四分之一
Each new shaded triangle has one-fourth the previous shaded area
解答:
第一个阴影三角形的两条直角边为 33 和 33,所以面积为 12⋅3⋅3=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+⋯ )=92⋅11−14=6。 \begin{gathered} \frac92\left(1+\frac14+\frac1{16}+\cdots\right) \\ {}= \frac92\cdot\frac1{1-\frac{1}{4}} \\ {}= 6 \end{gathered}\text{。}
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 12⋅3⋅3=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+⋯ )=92⋅11−14=6. \begin{gathered} \frac92\left(1+\frac14+\frac1{16}+\cdots\right) \\ {}= \frac92\cdot\frac1{1-\frac{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.
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