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Difference between revisions of "1986 AJHSME Problems/Problem 18"
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The shortest possible rectangle that has sides 36 and 60 would be if the side opposite the wall was 60.  
 
The shortest possible rectangle that has sides 36 and 60 would be if the side opposite the wall was 60.  
  
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Each of the sides of length 36 contribute <math>\frac{36}{12}+1=4</math> fence posts and the side of length 60 contributes <math>\frac{60}{12}+1=6</math> fence posts, so there are <math>4+4+6=14</math> fence posts.   
+
Each of the sides of length 36 contributes <math>\frac{36}{12}=3</math> fence posts and the side of length 60 contributes <math>\frac{60}{12}=5</math> fence posts, so there are <math>2(5+3)=16</math> fence posts.   
  
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However, the two corners where a 36 foot fence meets an 60 foot fence are counted twice, so there are actually <math>14-2=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.
  
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<math>\boxed{\text{B}}</math>
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<math>\boxed{\text{D}}</math>
  
 
==See Also==
 
==See Also==

Revision as of 21:59, 2 December 2021

Problem

A rectangular grazing area is to be fenced off on three sides using part of a $100$ meter rock wall as the fourth side. Fence posts are to be placed every $12$ 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 $36$ m by $60$ 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]

$\text{(A)}\ 11 \qquad \text{(B)}\ 12 \qquad \text{(C)}\ 13 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 16$

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 $\frac{36}{12}=3$ fence posts and the side of length 60 contributes $\frac{60}{12}=5$ fence posts, so there are $2(5+3)=16$ fence posts.

However, the two corners where a 36-foot fence meets a 60-foot fence are counted twice, so there are actually $16-2=12$ fence posts.

$\boxed{\text{D}}$

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

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