1987 AIME Problems/Problem 6
Since , and the areas of the trapezoids and are the same, then the heights of the trapezoids are the same. Thus both trapezoids have area . This number is also equal to one quarter the area of the entire rectangle, which is , so we have .
In addition, we see that the perimeter of the rectangle is , so .
Solving these two equations gives .
Let , , , and . First we drop a perpendicular from to a point on so . Since and and the areas of the trapezoids and are the same, the heights of the trapezoids are both .From here, we have that . We are told that this area is equal to . Setting these equal to each other and solving gives . In the same way, we find that the perpendicular from to is . So
Since . Let . Since , then .Let be the midpoint of , and be the midpoint of . Since the area of and are the same, then their heights are the same, and so is equidistant from and . This means that is perpendicular to , and is perpendicular to . Therefore, , , , and are all trapezoids, and 2. This implies that Since , , and .
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