Difference between revisions of "1989 AHSME Problems/Problem 17"
Mrdavid445 (talk | contribs) (Created page with "==Problem== 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 leng...") |
(→Solution: added solution) |
||
(2 intermediate revisions by 2 users not shown) | |||
Line 3: | Line 3: | ||
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}} |
Latest revision as of 13:33, 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.