Difference between revisions of "1994 AHSME Problems/Problem 11"
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--Solution by [http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=200685 TheMaskedMagician] | --Solution by [http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=200685 TheMaskedMagician] | ||
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+ | ==See Also== | ||
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+ | {{AHSME box|year=1994|num-b=10|num-a=12}} | ||
+ | {{MAA Notice}} |
Revision as of 16:34, 9 January 2021
Problem
Three cubes of volume and are glued together at their faces. The smallest possible surface area of the resulting configuration is
Solution
We reach the minimum surface area with the configuration above. The cubes from smallest to largest have edge lengths respectively. For the cube with edge , the five faces that are not in contact with another cube have total surface area of . The top face of that cube has surface area . The left face of the cube with edge length has surface area . The four remaining faces of that cube have total surface area . The four faces of the smallest cube that are not in contact with another cube have total surface area . Therefore, our total surface area is
--Solution by TheMaskedMagician
See Also
1994 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.