Difference between revisions of "2004 AMC 8 Problems/Problem 15"

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Thirteen black and six white hexagonal tiles were used to create the figure below. If a new figure is created by attaching a border of white tiles with the same size and shape as the others, what will be the difference between the total number of white tiles and the total number of black tiles in the new figure?
 
Thirteen black and six white hexagonal tiles were used to create the figure below. If a new figure is created by attaching a border of white tiles with the same size and shape as the others, what will be the difference between the total number of white tiles and the total number of black tiles in the new figure?
  
<center>
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[asy]
 
[asy]
 
defaultpen(linewidth(1));
 
defaultpen(linewidth(1));
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draw(shift(2x,0)*p^^shift(-2x,0)*p^^shift(x,1.5)*p^^shift(-x,1.5)*p^^shift(x,-1.5)*p^^shift(-x,-1.5)*p);
 
draw(shift(2x,0)*p^^shift(-2x,0)*p^^shift(x,1.5)*p^^shift(-x,1.5)*p^^shift(x,-1.5)*p^^shift(-x,-1.5)*p);
 
[/asy]
 
[/asy]
</center>
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<math>\textbf{(A)}\ 5\qquad \textbf{(B)}\ 7\qquad \textbf{(C)}\ 11\qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 18</math>
 
<math>\textbf{(A)}\ 5\qquad \textbf{(B)}\ 7\qquad \textbf{(C)}\ 11\qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 18</math>

Revision as of 20:19, 12 October 2020

Problem

Thirteen black and six white hexagonal tiles were used to create the figure below. If a new figure is created by attaching a border of white tiles with the same size and shape as the others, what will be the difference between the total number of white tiles and the total number of black tiles in the new figure?


[asy] defaultpen(linewidth(1)); real x=sqrt(3)/2; path p=rotate(30)*polygon(6); filldraw(p^^shift(0,3)*p^^shift(4x,0)*p^^shift(3x,1.5)*p^^shift(2x,3)*p^^shift(-4x,0)*p^^shift(-3x,1.5)*p^^shift(-2x,3)*p^^shift(3x,-1.5)*p^^shift(-3x,-1.5)*p^^shift(2x,-3)*p^^shift(-2x,-3)*p^^shift(0,-3)*p, black, black); draw(shift(2x,0)*p^^shift(-2x,0)*p^^shift(x,1.5)*p^^shift(-x,1.5)*p^^shift(x,-1.5)*p^^shift(-x,-1.5)*p); [/asy]


$\textbf{(A)}\ 5\qquad \textbf{(B)}\ 7\qquad \textbf{(C)}\ 11\qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 18$

Solution

The first ring around the middle tile has $6$ tiles, and the second has $12$. From this pattern, the third ring has $18$ tiles. Of these, $6+18=24$ are white and $1+12=13$ are black, with a difference of $24-13 = \boxed{\textbf{(C)}\ 11}$.

See Also

2004 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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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