1987 AHSME Problems/Problem 30
Problem
In the figure, has and . A line , with on and , divides into two pieces of equal area. (Note: the figure may not be accurate; perhaps is on instead of The ratio is
Solution
First we show that is on , as in the given figure, by demonstrating that if , then has more than half the area, so is too far to the right. Specifically, assume and drop an altitude from to that meets at . Without loss of generality, assume that , so that (as they have the same height, and area is times the product of base and height), which is , as required. Thus we must move to the left, scaling by a factor of such that . Thus , which is answer .
See also
1987 AHSME (Problems • Answer Key • Resources) | ||
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