Difference between revisions of "1989 AHSME Problems/Problem 17"
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== Solution == | == Solution == | ||
+ | If t is the side of the triangle, and s is the side of the square, then | ||
+ | <cmath>3t-4s=1989</cmath> | ||
+ | <cmath>t-s=d</cmath> | ||
+ | |||
+ | Solving the first equation for t gives | ||
+ | |||
+ | <cmath>t = \frac{4s+1989}{3}</cmath> | ||
+ | |||
+ | Substituting into the second equation, | ||
+ | |||
+ | <cmath> \frac{4s+1989}{3} - s = d </cmath> | ||
+ | <cmath> \frac{s+1989}{3} = d </cmath> | ||
+ | <cmath> s+1989 = 3d </cmath> | ||
+ | |||
+ | If s = 0, d = 663. But s has to be greater than 0, so the first 663 positive integers aren't possible <math>\to\boxed{\textbf{(D)}}</math> | ||
== See also == | == See also == |
Latest revision as of 12:33, 24 October 2014
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) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
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