Difference between revisions of "2003 AMC 8 Problems/Problem 13"
(Created page with "==solution== We can see that there are only $ \textbf{(B)6} cubes by counting.") |
Erichappy77 (talk | contribs) (→Problem) |
||
| (10 intermediate revisions by 7 users not shown) | |||
| Line 1: | Line 1: | ||
| − | == | + | ==Problem== |
| − | + | Fourteen white cubes are put together to form the figure on the right. The complete surface of the figure, including the bottom, is painted red. The figure is then separated into individual cubes. How many of the individual cubes have exactly four red faces | |
| + | |||
| + | <asy> | ||
| + | import three; | ||
| + | defaultpen(linewidth(0.8)); | ||
| + | real r=0.5; | ||
| + | currentprojection=orthographic(3/4,8/15,7/15); | ||
| + | draw(unitcube, white, thick(), nolight); | ||
| + | draw(shift(1,0,0)*unitcube, white, thick(), nolight); | ||
| + | draw(shift(2,0,0)*unitcube, white, thick(), nolight); | ||
| + | draw(shift(0,0,1)*unitcube, white, thick(), nolight); | ||
| + | draw(shift(2,0,1)*unitcube, white, thick(), nolight); | ||
| + | draw(shift(0,1,0)*unitcube, white, thick(), nolight); | ||
| + | draw(shift(2,1,0)*unitcube, white, thick(), nolight); | ||
| + | draw(shift(0,2,0)*unitcube, white, thick(), nolight); | ||
| + | draw(shift(2,2,0)*unitcube, white, thick(), nolight); | ||
| + | draw(shift(0,3,0)*unitcube, white, thick(), nolight); | ||
| + | draw(shift(0,3,1)*unitcube, white, thick(), nolight); | ||
| + | draw(shift(1,3,0)*unitcube, white, thick(), nolight); | ||
| + | draw(shift(2,3,0)*unitcube, white, thick(), nolight); | ||
| + | draw(shift(2,3,1)*unitcube, white, thick(), nolight); | ||
| + | </asy> | ||
| + | |||
| + | <math> \textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 12 </math> | ||
| + | |||
| + | ==Solution== | ||
| + | This is the number cubes that are adjacent to another cube on two sides. The bottom corner cubes are connected on three sides, and the top corner cubes are connected on one. The number we are looking for is the number of middle cubes, which is <math>\boxed{\textbf{(B)}\ 6}</math>. | ||
| + | |||
| + | ==See Also== | ||
| + | {{AMC8 box|year=2003|num-b=12|num-a=14}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 20:23, 26 January 2023
Problem
Fourteen white cubes are put together to form the figure on the right. The complete surface of the figure, including the bottom, is painted red. The figure is then separated into individual cubes. How many of the individual cubes have exactly four red faces
![[asy] import three; defaultpen(linewidth(0.8)); real r=0.5; currentprojection=orthographic(3/4,8/15,7/15); draw(unitcube, white, thick(), nolight); draw(shift(1,0,0)*unitcube, white, thick(), nolight); draw(shift(2,0,0)*unitcube, white, thick(), nolight); draw(shift(0,0,1)*unitcube, white, thick(), nolight); draw(shift(2,0,1)*unitcube, white, thick(), nolight); draw(shift(0,1,0)*unitcube, white, thick(), nolight); draw(shift(2,1,0)*unitcube, white, thick(), nolight); draw(shift(0,2,0)*unitcube, white, thick(), nolight); draw(shift(2,2,0)*unitcube, white, thick(), nolight); draw(shift(0,3,0)*unitcube, white, thick(), nolight); draw(shift(0,3,1)*unitcube, white, thick(), nolight); draw(shift(1,3,0)*unitcube, white, thick(), nolight); draw(shift(2,3,0)*unitcube, white, thick(), nolight); draw(shift(2,3,1)*unitcube, white, thick(), nolight); [/asy]](http://latex.artofproblemsolving.com/a/d/8/ad8d99504ec6b48d8afd6b366c02c827e11c198e.png)
Solution
This is the number cubes that are adjacent to another cube on two sides. The bottom corner cubes are connected on three sides, and the top corner cubes are connected on one. The number we are looking for is the number of middle cubes, which is 
.
See Also
| 2003 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 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. 
