Problem:
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
Answer Choices:
A. 6
B. 7
C. 8
D. 9
E. 10
Solution:
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 31​⋅21​⋅6⋅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.
​21​⋅26​⋅26​=4.521​⋅26​⋅26​+21​⋅46​⋅46​=4.5+1.125=5.62521​⋅26​⋅26​+21​⋅46​⋅46​+21​⋅86​⋅86​=5.625+0.281=5.906​
Answer: A​.
The problems on this page are the property of the MAA's American Mathematics Competitions
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