Difference between revisions of "2004 AMC 8 Problems/Problem 15"
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<center> | <center> | ||
− | [[ | + | [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] | ||
</center> | </center> | ||
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]
Solution
The first ring around the middle tile has tiles, and the second has . From this pattern, the third ring has tiles. Of these, are white and are black, with a difference of .
See Also
2004 AMC 8 (Problems • Answer Key • Resources) | ||
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.