AMC8 1991

AMC8 1991 · Q25

AMC8 1991 · Q25. It mainly tests Fractions, Area & perimeter.

An equilateral triangle is originally painted black. Each time the triangle is changed, the middle fourth of each black triangle turns white. After five changes, what fractional part of the original area of the black triangle remains black?

一个等边三角形最初涂成黑色。每次改变时，每个黑色三角形中间四分之一变成白色。经过五次改变后，原始黑色三角形面积的哪一部分仍然是黑色？

(A) $\frac{1}{1024}$ $\frac{1}{1024}$

(B) $\frac{15}{64}$ $\frac{15}{64}$

(D) $\frac{1}{4}$ $\frac{1}{4}$

(E) $\frac{81}{256}$ $\frac{81}{256}$

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): Since $1/4$ turns white, it follows that $3/4$ remains black after the first change. After the second change, $3/4$ of the remaining $3/4$ stays black or $\frac{3}{4}\times\frac{3}{4}$ remains black. Thus, after the fifth change, the amount remaining black is $\frac{3}{4}\times\frac{3}{4}\times\frac{3}{4}\times\frac{3}{4}\times\frac{3}{4}=\frac{243}{1024}$ of the original area.

答案（C）：由于有 $1/4$ 变成白色，因此第一次变化后有 $3/4$ 仍为黑色。第二次变化后，剩余的 $3/4$ 中又有 $3/4$ 仍为黑色，即仍为黑色的比例是 $\frac{3}{4}\times\frac{3}{4}$。因此，第五次变化后，仍为黑色的部分为 $\frac{3}{4}\times\frac{3}{4}\times\frac{3}{4}\times\frac{3}{4}\times\frac{3}{4}=\frac{243}{1024}$ 占原始面积的比例。

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