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

Revision as of 18:45, 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 60 would be if the side opposite the wall was 80.

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 80 foot fence are counted twice, so there are actually $14-2=12$ fence posts.

$\boxed{\text{B}}$

@@ Line 13: / Line 13: @@
 ==Solution==
-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.
+The shortest possible rectangle that has sides 36 and 60 would be if the side opposite the wall was 80.
 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.

Page

Toolbox

Search

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

Revision as of 18:45, 24 January 2009

Problem

Solution

See Also