1971 AHSME Problems/Problem 28
Problem
Nine lines parallel to the base of a triangle divide the other sides each into equal segments and the area into distinct parts. If the area of the largest of these parts is , then the area of the original triangle is
Solution
Let the triangle be with base and longest parallel segment with on and on , as in the diagram.
By the properties of transversals, we have . Thus, by AA Similarity, we have (because they share ). From the problem, we know that , so, by similarity, , and so .
Now, let . Because , we know that . From the problem, , so . Solving for yields .
See Also
1971 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 27 |
Followed by Problem 29 | |
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