AMC8 1999

AMC8 1999 · Q25

AMC8 1999 · Q25. It mainly tests Recursion & DP style counting (basic), Triangles (properties).

Points B, D, and J are midpoints of the sides of right triangle ACG. Points K, E, I are midpoints of the sides of triangle JDG, etc. If the dividing and shading process is done 100 times (the first three are shown) and AC = CG = 6, then the total area of the shaded triangles is nearest

点 B、D 和 J 是直角三角形 ACG 边的中点。点 K、E、I 是三角形 JDG 边的中点，等等。如果划分和着色过程进行 100 次（显示前三次），且 AC = CG = 6，则着色三角形的总面积最接近

(A) 6 6

(B) 7 7

(D) 9 9

(E) 10 10

Answer

Correct choice: (A)

正确答案：（A）

Solution

Answer (A): At each stage the area of the shaded triangle is one-third of the trapezoidal region not containing the smaller triangle being divided in the next step. Thus, the total area of the shaded triangles comes closer and closer to one-third of the area of the triangular region $ACG$ and this is $\frac{1}{3}\cdot\frac{1}{2}\cdot 6\cdot 6=6$. The shaded areas for the first six stages are: 4.5, 5.625, 5.906, 5.976, 5.994, and 5.998. These are the calculations for the first three steps. $\frac{1}{2}\cdot\frac{6}{2}\cdot\frac{6}{2}=4.5$ $\frac{1}{2}\cdot\frac{6}{2}\cdot\frac{6}{2}+\frac{1}{2}\cdot\frac{6}{4}\cdot\frac{6}{4}=4.5+1.125=5.625$ $\frac{1}{2}\cdot\frac{6}{2}\cdot\frac{6}{2}+\frac{1}{2}\cdot\frac{6}{4}\cdot\frac{6}{4}+\frac{1}{2}\cdot\frac{6}{8}\cdot\frac{6}{8}=5.625+0.281=5.906$

答案（A）：在每一个阶段，阴影三角形的面积都是梯形区域（不包含下一步要继续被分割的小三角形的那部分）面积的三分之一。因此，所有阴影三角形的总面积会越来越接近三角形区域 $ACG$ 面积的三分之一，也就是 $\frac{1}{3}\cdot\frac{1}{2}\cdot 6\cdot 6=6$。前六个阶段的阴影面积分别为：4.5、5.625、5.906、5.976、5.994 和 5.998。下面是前三步的计算。 $\frac{1}{2}\cdot\frac{6}{2}\cdot\frac{6}{2}=4.5$ $\frac{1}{2}\cdot\frac{6}{2}\cdot\frac{6}{2}+\frac{1}{2}\cdot\frac{6}{4}\cdot\frac{6}{4}=4.5+1.125=5.625$ $\frac{1}{2}\cdot\frac{6}{2}\cdot\frac{6}{2}+\frac{1}{2}\cdot\frac{6}{4}\cdot\frac{6}{4}+\frac{1}{2}\cdot\frac{6}{8}\cdot\frac{6}{8}=5.625+0.281=5.906$

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