1988 AHSME Problems/Problem 16
Problem
and are equilateral triangles with parallel sides and the same center, as in the figure. The distance between side and side is the altitude of . The ratio of the area of to the area of is
Solution
Let have side length and have side length . Thus the altitude of is . Now observe that this altitude is made up of three parts: the distance from to , plus the altitude of , plus a top part which is equal to the length of the diagonal line from the bottom-left corner of to the bottom left corner of (as an isosceles trapezium is formed with parallel sides and , and legs and ). We drop a perpendicular from to , which meets at . has angles , , and , and the vertical side is that distance from to , which is given as , so that by simple trigonometry, the length of the diagonal line is Thus using the "altitude in three parts" idea, we get Thus the sides of are half as long as , so the area ratio is , which is .
See also
1988 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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