Difference between revisions of "2003 AMC 8 Problems/Problem 15"
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==See Also== | ==See Also== | ||
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Revision as of 23:47, 4 July 2013
Problem
A figure is constructed from unit cubes. Each cube shares at least one face with another cube. What is the minimum number of cubes needed to build a figure with the front and side views shown?
![[asy] defaultpen(linewidth(0.8)); path p=unitsquare; draw(p^^shift(0,1)*p^^shift(1,0)*p); draw(shift(4,0)*p^^shift(5,0)*p^^shift(5,1)*p); label("FRONT", (1,0), S); label("SIDE", (5,0), S); [/asy]](http://latex.artofproblemsolving.com/e/6/d/e6d30f2f3ae60f8b661b55b5a6167693d0ecc30b.png)
Solution
One example of the solution with 
cubes. Notice the corner cube cannot be removed for a figure of 3 cubes because each face of a cube must be touching another face.
![[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,-1,0)*unitcube, white, thick(), nolight); draw(shift(1,0,0)*unitcube, white, thick(), nolight); draw(shift(1,0,1)*unitcube, white, thick(), nolight);[/asy]](http://latex.artofproblemsolving.com/1/d/c/1dcf40aa4a89c2511ba153f0441f53298e075bef.png)
See Also
| 2003 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. 
