Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1986 AJHSME Problems/Problem 18"
m (→‎Solution)
 
(12 intermediate revisions by 6 users 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>.
+
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>.
−
 
 
−
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==
 
==See Also==
  
−
[[1986 AJHSME Problems]]
+
{{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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1986_AJHSME_Problems/Problem_18&oldid=208579"