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Difference between revisions of "2009 AMC 8 Problems/Problem 10"

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==Solution 2==
 
==Solution 2==
  
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The squares that don't touch the border are in a 6 by 6 square. Therefore, the number of squares that don't touch the border is <math>6*6=36</math>. There are 64 choices total. Therefore, the answer is <math>36/64</math>, which simplifies to <math>\boxed{\textbf{(D)}\ \frac{9}{16}}</math>.
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The squares that don't touch the border are in a 6 by 6 square. Therefore, the number of squares that don't touch the border is 6*6=36 There are 64 choices total. Therefore, the answer is 36/64 which simplifies to 9/16
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2009|num-b=9|num-a=11}}
 
{{AMC8 box|year=2009|num-b=9|num-a=11}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 13:54, 14 August 2021

Problem

On a checkerboard composed of 64 unit squares, what is the probability that a randomly chosen unit square does not touch the outer edge of the board?

[asy] unitsize(10); draw((0,0)--(8,0)--(8,8)--(0,8)--cycle); draw((1,8)--(1,0)); draw((7,8)--(7,0)); draw((6,8)--(6,0)); draw((5,8)--(5,0)); draw((4,8)--(4,0)); draw((3,8)--(3,0)); draw((2,8)--(2,0)); draw((0,1)--(8,1)); draw((0,2)--(8,2)); draw((0,3)--(8,3)); draw((0,4)--(8,4)); draw((0,5)--(8,5)); draw((0,6)--(8,6)); draw((0,7)--(8,7)); fill((0,0)--(1,0)--(1,1)--(0,1)--cycle,black); fill((2,0)--(3,0)--(3,1)--(2,1)--cycle,black); fill((4,0)--(5,0)--(5,1)--(4,1)--cycle,black); fill((6,0)--(7,0)--(7,1)--(6,1)--cycle,black); fill((0,2)--(1,2)--(1,3)--(0,3)--cycle,black); fill((2,2)--(3,2)--(3,3)--(2,3)--cycle,black); fill((4,2)--(5,2)--(5,3)--(4,3)--cycle,black); fill((6,2)--(7,2)--(7,3)--(6,3)--cycle,black); fill((0,4)--(1,4)--(1,5)--(0,5)--cycle,black); fill((2,4)--(3,4)--(3,5)--(2,5)--cycle,black); fill((4,4)--(5,4)--(5,5)--(4,5)--cycle,black); fill((6,4)--(7,4)--(7,5)--(6,5)--cycle,black); fill((0,6)--(1,6)--(1,7)--(0,7)--cycle,black); fill((2,6)--(3,6)--(3,7)--(2,7)--cycle,black); fill((4,6)--(5,6)--(5,7)--(4,7)--cycle,black); fill((6,6)--(7,6)--(7,7)--(6,7)--cycle,black); fill((1,1)--(2,1)--(2,2)--(1,2)--cycle,black); fill((3,1)--(4,1)--(4,2)--(3,2)--cycle,black); fill((5,1)--(6,1)--(6,2)--(5,2)--cycle,black); fill((7,1)--(8,1)--(8,2)--(7,2)--cycle,black); fill((1,3)--(2,3)--(2,4)--(1,4)--cycle,black); fill((3,3)--(4,3)--(4,4)--(3,4)--cycle,black); fill((5,3)--(6,3)--(6,4)--(5,4)--cycle,black); fill((7,3)--(8,3)--(8,4)--(7,4)--cycle,black); fill((1,5)--(2,5)--(2,6)--(1,6)--cycle,black); fill((3,5)--(4,5)--(4,6)--(3,6)--cycle,black); fill((5,5)--(6,5)--(6,6)--(5,6)--cycle,black); fill((7,5)--(8,5)--(8,6)--(7,6)--cycle,black); fill((1,7)--(2,7)--(2,8)--(1,8)--cycle,black); fill((3,7)--(4,7)--(4,8)--(3,8)--cycle,black); fill((5,7)--(6,7)--(6,8)--(5,8)--cycle,black); fill((7,7)--(8,7)--(8,8)--(7,8)--cycle,black);[/asy]

$\textbf{(A)}\ \frac{1}{16}\qquad\textbf{(B)}\ \frac{7}{16}\qquad\textbf{(C)}\ \frac{1}2\qquad\textbf{(D)}\ \frac{9}{16}\qquad\textbf{(E)}\ \frac{49}{64}$

Solution

There are $8^2=64$ total squares. There are $(8-1)(4)=28$ unit squares on the perimeter and therefore $64-28=36$ NOT on the perimeter. The probability of choosing one of those squares is $\frac{36}{64} = \boxed{\textbf{(D)}\ \frac{9}{16}}$.

Solution 2

The squares that don't touch the border are in a 6 by 6 square. Therefore, the number of squares that don't touch the border is 6*6=36 There are 64 choices total. Therefore, the answer is 36/64 which simplifies to 9/16

See Also

2009 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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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