1986 AJHSME Problems/Problem 18

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.

Since we have the wall, we must cancel a side. This will be a 3 foot fenced side to limit loss. We have now a total of 9

However, the two corners where a 36-foot fence meets a 60-foot fence are not counted, so there are actually $9+2=11$ fence posts.

$\boxed{\text{A}}$

1986 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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1986 AJHSME Problems/Problem 18

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