1988 AHSME Problems/Problem 20
Problem
In one of the adjoining figures a square of side is dissected into four pieces so that and are the midpoints of opposite sides and is perpendicular to . These four pieces can then be reassembled into a rectangle as shown in the second figure. The ratio of height to base, , in this rectangle is
Solution
Within , the parallelogram piece has vertical side , and diagonal side . Thus the triangle in the bottom-right hand corner (the one with horizontal side ) must have hypotenuse , and the only such triangle in the original figure is , so we deduce Now the rectangle must have the same area as the square, as the pieces were put together without gaps or overlaps, so its area is , and thus the vertical side is , so the required ratio is , which is .
See also
1988 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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