Difference between revisions of "1997 AJHSME Problems/Problem 21"
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Latest revision as of 23:27, 4 July 2013
Problem
Each corner cube is removed from this 
cube. The surface area of the remaining figure is
![[asy] draw((2.7,3.99)--(0,3)--(0,0)); draw((3.7,3.99)--(1,3)--(1,0)); draw((4.7,3.99)--(2,3)--(2,0)); draw((5.7,3.99)--(3,3)--(3,0)); draw((0,0)--(3,0)--(5.7,0.99)); draw((0,1)--(3,1)--(5.7,1.99)); draw((0,2)--(3,2)--(5.7,2.99)); draw((0,3)--(3,3)--(5.7,3.99)); draw((0,3)--(3,3)--(3,0)); draw((0.9,3.33)--(3.9,3.33)--(3.9,0.33)); draw((1.8,3.66)--(4.8,3.66)--(4.8,0.66)); draw((2.7,3.99)--(5.7,3.99)--(5.7,0.99)); [/asy]](http://latex.artofproblemsolving.com/a/4/f/a4fe561822f5a96bd789b3402d3a65e14d82fd12.png)
Solution
The original cube has 
square surfaces that each have an area of 
, for a toal surface area of 
.
Since no two corner cubes touch, we can examine the effect of removing each corner cube individually.
Each corner cube contribues 
faces each of surface area 
to the big cube, so the surface area is decreased by when the cube is removed.
However, when the cube is removed, faces on the 3x3x3 cube will be revealed, increasing the surface area by .
Thus, the surface area does not change with the removal of a corner cube, and it remains 
, which is answer 
.
See also
| 1997 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 20 |
Followed by Problem 22 | |
| 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. 
