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

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

@@ Line 42: / Line 42: @@
 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>.
+==Solution 3 (somewhat fake)==
+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
+==Video Solution by WhyMath==
+https://youtu.be/6XNr5HWgMdI
 ==See Also==
 {{AMC8 box|year=2011|num-b=2|num-a=4}}
 {{MAA Notice}}

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