2016 AMC 12B Problems/Problem 15: Difference between revisions
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We proceed from the factorization in the above solution. By the AM-GM inequality, | We proceed from the factorization in the above solution. By the AM-GM inequality, | ||
<cmath>\frac{a_1+a_2 | <cmath>\frac{a_1+a_2+a_3}{3}\geq\sqrt[3]{a_1a_2a_3}</cmath> | ||
Cubing both sides, | Cubing both sides, | ||
Revision as of 14:07, 20 December 2017
Problem
All the numbers 
are assigned to the six faces of a cube, one number to each face. For each of the eight vertices of the cube, a product of three numbers is computed, where the three numbers are the numbers assigned to the three faces that include that vertex. What is the greatest possible value of the sum of these eight products?
Solution 1
First assign each face the letters 
. The sum of the product of the faces is 
. We can factor this into 
which is the product of the sum of each pair of opposite faces. In order to maximize we use the numbers 
or 
.
Solution 2
We proceed from the factorization in the above solution. By the AM-GM inequality,
![\[\frac{a_1+a_2+a_3}{3}\geq\sqrt[3]{a_1a_2a_3}\]](http://latex.artofproblemsolving.com/a/a/6/aa67eef8049086302c27d7648d8e0d17e704e7dd.png)
Cubing both sides,
![\[\left(\frac{a_1+a_2+a_3}{3}\right)^3\geq{a_1a_2a_3}\]](http://latex.artofproblemsolving.com/2/3/3/233c106eb78fd6060382033db69c868915a0e2ac.png)
Let 
, 
, and 
. Let's substitute in these values.
![\[\left(\frac{a+b+c+d+e+f}{3}\right)^3\geq{(a+f)(b+c)(d+e)}\]](http://latex.artofproblemsolving.com/0/6/d/06d6e468f555fb1ee6e69b3607de98a90f68cee5.png)
is fixed at 27.
![\[\left(\frac{27}{3}\right)^3\geq{(a+f)(b+c)(d+e)}\]](http://latex.artofproblemsolving.com/4/2/e/42e6a5120a4a3947aa06a9123846cb6a117c1564.png)
![\[\boxed{\textbf{(D)}\ 729 }\geq{(a+f)(b+c)(d+e)}\]](http://latex.artofproblemsolving.com/2/8/3/283da114e616eca58fa458e4073c4b6772997789.png)
See Also
| 2016 AMC 12B (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 AMC 12 Problems and Solutions | |
These problems are copyrighted © by the Mathematical Association of America. 
