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Difference between revisions of "1986 AJHSME Problems/Problem 18"
(3 intermediate revisions by one other user not shown)
Line 4: Line 4:
  
 
<asy>
 
<asy>
 +
unitsize(12);
 
draw((0,0)--(16,12));
 
draw((0,0)--(16,12));
−
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>
Line 13: Line 14:
 
==Solution==
 
==Solution==
  
−
The shortest possible rectangle that has sides 36 and 80 and area <math>36 \times 80</math> would be if the two sides adjacent to the wall were 36, and the other side was 80. Thus, the perimeter of the rectangle that is fence would be <math>2 \times 36 + 80</math>, or <math>72 + 80</math> or <math>152</math>.
+
The shortest possible rectangle that has sides 36 and 60 would be if the side opposite the wall was 60.  
  
−
To find how many fence posts we'll need, just divide 152 by 12 and add 2 if there's a remainder, add 1 if there isn't.
+
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.
  
−
Clearly 152 is not divisible by 12, since <math>152 = 144 + 8</math>, so it's just 12 + 2 = 14.
+
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.
  
−
D
+
<math>\boxed{\text{B}}</math>
  
 
==See Also==
 
==See Also==
  
−
[[1986 AJHSME Problems]]
+
{{AJHSME box|year=1986|num-b=17|num-a=19}}
 +
[[Category:Introductory Geometry Problems]]
 +
{{MAA Notice}}

Revision as of 21:27, 3 July 2013

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 contribute $\frac{36}{12}+1=4$ fence posts and the side of length 60 contributes $\frac{60}{12}+1=6$ fence posts, so there are $4+4+6=14$ fence posts.

However, the two corners where a 36 foot fence meets an 60 foot fence are counted twice, so there are actually $14-2=12$ fence posts.

$\boxed{\text{B}}$

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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