Difference between revisions of "1986 AJHSME Problems/Problem 18"
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However, the two corners where a 36-foot fence meets a 60-foot fence are counted twice, so there are actually <math>16-2=12</math> fence posts. | However, the two corners where a 36-foot fence meets a 60-foot fence are counted twice, so there are actually <math>16-2=12</math> fence posts. | ||
| − | <math>\boxed{\text{ | + | <math>\boxed{\text{B}}</math> |
==See Also== | ==See Also== | ||
Revision as of 11:22, 28 December 2021
Problem
A rectangular grazing area is to be fenced off on three sides using part of a 
meter rock wall as the fourth side. Fence posts are to be placed every 
meters along the fence including the two posts where the fence meets the rock wall. What is the fewest number of posts required to fence an area 
m by 
m?
![[asy] unitsize(12); draw((0,0)--(16,12)); draw((10.66666,8)--(6.66666,13.33333)--(1.33333,9.33333)--(5.33333,4)); label("WALL",(7,4),SE); [/asy]](http://latex.artofproblemsolving.com/0/3/c/03c15a06354898452ca990f3cafbdccb11317cbe.png)
Solution
The shortest possible rectangle that has sides 36 and 60 would be if the side opposite the wall was 60.
Each of the sides of length 36 contributes 
fence posts and the side of length 60 contributes 
fence posts, so there are 
fence posts.
However, the two corners where a 36-foot fence meets a 60-foot fence are counted twice, so there are actually 
fence posts.
See Also
| 1986 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 17 |
Followed by Problem 19 | |
| 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 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
