1989 AHSME Problems/Problem 17
Problem
The perimeter of an equilateral triangle exceeds the perimeter of a square by . The length of each side of the triangle exceeds the length of each side of the square by . The square has perimeter greater than 0. How many positive integers are NOT possible value for ?
Solution
If t is the side of the triangle, and s is the side of the square, then
Solving the first equation for t gives
Substituting into the second equation,
If s = 0, d = 663. But s has to be greater than 0, so the first 663 positive integers aren't possible
See also
1989 AHSME (Problems • Answer Key • Resources) | ||
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Followed by Problem 18 | |
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