1961 AHSME Problems/Problem 27
Problem
Given two equiangular polygons and with different numbers of sides; each angle of is degrees and each angle of is degrees, where is an integer greater than . The number of possibilities for the pair is:
Solution
Each angle in each equiangular polygon is , where is the number of sides. As gets larger, each angle in the equiangular polygon approaches (but does not reach) . That means , so .
Recall that each angle in an equiangular triangle has , each angle in an equiangular quadrilateral has , and so on. If , then . That means the only option available is because an equiangular triangle is the only equiangular polygon to have acute angles. If , then , and there are no values of that satisfy. As increases, decreases even further, so we can for sure say that there is only 1 possibility for the pair , which is answer choice .
See Also
1961 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 26 |
Followed by Problem 28 | |
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