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Difference between revisions of "1986 AJHSME Problems/Problem 18"
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<asy>
 
<asy>
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unitsize(12);
 
draw((0,0)--(16,12));
 
draw((0,0)--(16,12));
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draw((5.33333,4)--(10.66666,8)--(6.66666,13.33333)--(1.33333,9.33333)--cycle);
+
draw((10.66666,8)--(6.66666,13.33333)--(1.33333,9.33333)--(5.33333,4));
 
label("WALL",(7,4),SE);
 
label("WALL",(7,4),SE);
 
</asy>
 
</asy>
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==Solution==
 
==Solution==
  
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The shortest possible rectangle that has sides 36 and 60 and area <math>36 \times 80</math> would be if the side opposite the wall was 80.  
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Since we want to minimize the amount of fence that we use, we should have the longer side of the rectangle have one side as the wall. The grazing area is a <math>36</math>m by <math>60</math>m rectangle, so the <math>60</math>m side should be parallel to the wall. That means the two fences perpendicular to the wall are <math>36</math>m. We can start by counting <math>60\div12+1</math> on the <math>60</math>m fence (since we also count the <math>0</math>m post). Next, we have the two <math>36</math>m fences. There are a total of <math>36\div12+1-1</math> fences on that side since the <math>0</math>m and <math>60</math>m fence posts are also part of the <math>36</math>m fences. So we have <math>6+2\cdot3=12</math> minimum fence posts needed to box a <math>36</math>m by <math>60</math>m grazing area, or answer <math>\boxed{\text{(B) 12}}</math>.
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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.
 
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However, the two corners where a 36 foot fence meets an 80 foot fence are counted twice, so there are actually <math>14-2=12</math> fence posts.
 
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<math>\boxed{\text{B}}</math>
 
  
 
==See Also==
 
==See Also==
  
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[[1986 AJHSME Problems]]
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{{AJHSME box|year=1986|num-b=17|num-a=19}}
 +
[[Category:Introductory Geometry Problems]]
 +
{{MAA Notice}}

Latest revision as of 17:54, 28 December 2023

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

Since we want to minimize the amount of fence that we use, we should have the longer side of the rectangle have one side as the wall. The grazing area is a $36$m by $60$m rectangle, so the $60$m side should be parallel to the wall. That means the two fences perpendicular to the wall are $36$m. We can start by counting $60\div12+1$ on the $60$m fence (since we also count the $0$m post). Next, we have the two $36$m fences. There are a total of $36\div12+1-1$ fences on that side since the $0$m and $60$m fence posts are also part of the $36$m fences. So we have $6+2\cdot3=12$ minimum fence posts needed to box a $36$m by $60$m grazing area, or answer $\boxed{\text{(B) 12}}$.

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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