Difference between revisions of "1989 AHSME Problems/Problem 17"
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The perimeter of an equilateral triangle exceeds the perimeter of a square by <math>1989 \ \text{cm}</math>. The length of each side of the triangle exceeds the length of each side of the square by <math>d \ \text{cm}</math>. The square has perimeter greater than 0. How many positive integers are NOT possible value for <math>d</math>? | The perimeter of an equilateral triangle exceeds the perimeter of a square by <math>1989 \ \text{cm}</math>. The length of each side of the triangle exceeds the length of each side of the square by <math>d \ \text{cm}</math>. The square has perimeter greater than 0. How many positive integers are NOT possible value for <math>d</math>? | ||
| − | <math>\text{(A)} \ 0 \qquad \text{(B)} \ 9 \qquad \text{(C)} \ 221 \qquad \text{(D)} \ 663 \qquad \text{(E)} \ \ | + | <math>\text{(A)} \ 0 \qquad \text{(B)} \ 9 \qquad \text{(C)} \ 221 \qquad \text{(D)} \ 663 \qquad \text{(E)} \ \infty </math> |
| + | |||
| + | == 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 == | ||
| + | {{AHSME box|year=1989|num-b=16|num-a=18}} | ||
| + | |||
| + | [[Category: Introductory Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
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
![\[3t-4s=1989\]](http://latex.artofproblemsolving.com/6/e/f/6ef7bc6fadfc610a51dec3970364166f136e9615.png)
![\[t-s=d\]](http://latex.artofproblemsolving.com/0/6/8/0684cc1c2a7bc32fd45165874d2cc3ea5a8acceb.png)
Solving the first equation for t gives
![\[t = \frac{4s+1989}{3}\]](http://latex.artofproblemsolving.com/1/9/3/1935e6a3352a64ac504b1c1aa6404887d2af9da5.png)
Substituting into the second equation,
![\[\frac{4s+1989}{3} - s = d\]](http://latex.artofproblemsolving.com/b/c/c/bcc2b38085a688e02a3cabb678b4dbc9c730f83c.png)
![\[\frac{s+1989}{3} = d\]](http://latex.artofproblemsolving.com/5/c/0/5c03b3cfd545bac78754ea8cbf119a527baf4518.png)
![\[s+1989 = 3d\]](http://latex.artofproblemsolving.com/9/0/4/9044f6e65b5dcf1e795b7409fda3a27b12a6b2b1.png)
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
