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Difference between revisions of "2011 AMC 8 Problems/Problem 3"

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<math> \textbf{(A) }8:17 \qquad\textbf{(B) }25:49 \qquad\textbf{(C) }36:25 \qquad\textbf{(D) }32:17 \qquad\textbf{(E) }36:17</math>
 
<math> \textbf{(A) }8:17 \qquad\textbf{(B) }25:49 \qquad\textbf{(C) }36:25 \qquad\textbf{(D) }32:17 \qquad\textbf{(E) }36:17</math>
  
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==Solution==
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==Solution 1==
 
One way of approaching this is drawing the next circle of boxes around the current square.
 
One way of approaching this is drawing the next circle of boxes around the current square.
 
<asy>
 
<asy>
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</asy>
 
</asy>
 
We can now count the number of black and white tiles; 32 black tiles and 17 white tiles. This means the answer is <math>\boxed{\textbf{(D) }32:17}</math>.
 
We can now count the number of black and white tiles; 32 black tiles and 17 white tiles. This means the answer is <math>\boxed{\textbf{(D) }32:17}</math>.
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==Solution 2==
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If we did not want to draw a diagram though that is probably simpler in this case, we can imagine the last border of black tiles. We see that each side length (if the side length of a square is <math>1</math>) increases by <math>2</math> each time we go out one layer. Therefore, we know that the next border will have a side length of <math>7</math>. That border has <math>7^2-5^2=24</math> black squares in it. The other black border has <math>3^2-1^2=8</math> squares in it, so there are <math>32</math> black squares in all. The other ones must all be white, so there are <math>49-32=17</math> white squares. Thus, we get our answer of <math>\boxed{\textbf{(D) }32:17}</math>.
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==Solution 3 (somewhat fake)==
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Notice that in total, once we add the border there will be <math>7^2</math> squares total. So far as there is only one option that sums to 49 and we know that there are an integer number of boxes, it has to be <math>\boxed{\textbf{(D) }32:17}</math>. ~math_genius_11
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==Video Solution by WhyMath==
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https://youtu.be/6XNr5HWgMdI
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2011|num-b=2|num-a=4}}
 
{{AMC8 box|year=2011|num-b=2|num-a=4}}
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{{MAA Notice}}

Latest revision as of 10:02, 18 November 2024

Problem

Extend the square pattern of 8 black and 17 white square tiles by attaching a border of black tiles around the square. What is the ratio of black tiles to white tiles in the extended pattern?
[asy] filldraw((0,0)--(5,0)--(5,5)--(0,5)--cycle,white,black); filldraw((1,1)--(4,1)--(4,4)--(1,4)--cycle,mediumgray,black); filldraw((2,2)--(3,2)--(3,3)--(2,3)--cycle,white,black); draw((4,0)--(4,5)); draw((3,0)--(3,5)); draw((2,0)--(2,5)); draw((1,0)--(1,5)); draw((0,4)--(5,4)); draw((0,3)--(5,3)); draw((0,2)--(5,2)); draw((0,1)--(5,1)); [/asy]

$\textbf{(A) }8:17 \qquad\textbf{(B) }25:49 \qquad\textbf{(C) }36:25 \qquad\textbf{(D) }32:17 \qquad\textbf{(E) }36:17$

Solution 1

One way of approaching this is drawing the next circle of boxes around the current square. [asy] filldraw((-1,-1)--(6,-1)--(6,6)--(-1,6)--cycle,mediumgray,black); filldraw((0,0)--(5,0)--(5,5)--(0,5)--cycle,white,black); filldraw((1,1)--(4,1)--(4,4)--(1,4)--cycle,mediumgray,black); filldraw((2,2)--(3,2)--(3,3)--(2,3)--cycle,white,black); draw((4,-1)--(4,6)); draw((3,-1)--(3,6)); draw((2,-1)--(2,6)); draw((1,-1)--(1,6)); draw((-1,4)--(6,4)); draw((-1,3)--(6,3)); draw((-1,2)--(6,2)); draw((-1,1)--(6,1)); draw((0,-1)--(0,6)); draw((-1,5)--(6,5)); draw((-1,0)--(6,0)); draw((5,-1)--(5,6)); [/asy] We can now count the number of black and white tiles; 32 black tiles and 17 white tiles. This means the answer is $\boxed{\textbf{(D) }32:17}$.

Solution 2

If we did not want to draw a diagram though that is probably simpler in this case, we can imagine the last border of black tiles. We see that each side length (if the side length of a square is $1$) increases by $2$ each time we go out one layer. Therefore, we know that the next border will have a side length of $7$. That border has $7^2-5^2=24$ black squares in it. The other black border has $3^2-1^2=8$ squares in it, so there are $32$ black squares in all. The other ones must all be white, so there are $49-32=17$ white squares. Thus, we get our answer of $\boxed{\textbf{(D) }32:17}$.

Solution 3 (somewhat fake)

Notice that in total, once we add the border there will be $7^2$ squares total. So far as there is only one option that sums to 49 and we know that there are an integer number of boxes, it has to be $\boxed{\textbf{(D) }32:17}$. ~math_genius_11

Video Solution by WhyMath

https://youtu.be/6XNr5HWgMdI

See Also

2011 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
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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