Difference between revisions of "1986 AJHSME Problems/Problem 18"

Revision as of 18:35, 24 January 2009

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?

$[asy] draw((0,0)--(16,12)); draw((5.33333,4)--(10.66666,8)--(6.66666,13.33333)--(1.33333,9.33333)--cycle); label("WALL",(7,4),SE); [/asy]$

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

Solution

The shortest possible rectangle that has sides 36 and 80 and area $36 \times 80$ 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 $2 \times 36 + 80$ , or $72 + 80$ or $152$ .

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.

Clearly 152 is not divisible by 12, since $152 = 144 + 8$ , so it's just 12 + 2 = 14.

D

@@ Line 13: / Line 13: @@
 ==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>.
+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.
+Clearly 152 is not divisible by 12, since <math>152 = 144 + 8</math>, so it's just 12 + 2 = 14.
+D
 ==See Also==
 [[1986 AJHSME Problems]]

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Difference between revisions of "1986 AJHSME Problems/Problem 18"

Revision as of 18:35, 24 January 2009

Problem

Solution

See Also