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 "2011 AMC 8 Problems/Problem 3"
Line 17: Line 17:
 
<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>
  
−
==Solution==
+
==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>
Line 38: Line 38:
 
</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>.
 +
 +
==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 <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>.
 +
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2011|num-b=2|num-a=4}}
 
{{AMC8 box|year=2011|num-b=2|num-a=4}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 14:42, 18 October 2016

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 $\boxed{\textbf{(D) }32:17}$.


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

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=2011_AMC_8_Problems/Problem_3&oldid=80676"