Difference between revisions of "1989 AHSME Problems/Problem 17"

m
(Solution: added solution)
 
Line 6: Line 6:
  
 
== 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 $1989 \ \text{cm}$. The length of each side of the triangle exceeds the length of each side of the square by $d \ \text{cm}$. The square has perimeter greater than 0. How many positive integers are NOT possible value for $d$?

$\text{(A)} \ 0 \qquad \text{(B)} \ 9 \qquad \text{(C)} \ 221 \qquad \text{(D)} \ 663 \qquad \text{(E)} \ \infty$

Solution

If t is the side of the triangle, and s is the side of the square, then

\[3t-4s=1989\] \[t-s=d\]

Solving the first equation for t gives

\[t = \frac{4s+1989}{3}\]

Substituting into the second equation,

\[\frac{4s+1989}{3} - s = d\] \[\frac{s+1989}{3} = d\] \[s+1989 = 3d\]

If s = 0, d = 663. But s has to be greater than 0, so the first 663 positive integers aren't possible $\to\boxed{\textbf{(D)}}$

See also

1989 AHSME (ProblemsAnswer KeyResources)
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. AMC logo.png