Difference between revisions of "2010 AIME I Problems/Problem 9"
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Add the three equations to get <math>a^3 + b^3 + c^3 = 28 + 3abc</math>. Now, let <math>abc = p</math>. <math>a = \sqrt [3]{p + 2}</math>, <math>b = \sqrt [3]{p + 6}</math> and <math>c = \sqrt [3]{p + 20}</math>, so <math>p = abc = (\sqrt [3]{p + 2})(\sqrt [3]{p + 6})(\sqrt [3]{p + 20})</math>. Now [[cube]] both sides; the <math>p^3</math> terms cancel out. Solve the remaining [[quadratic]] to get <math>p = - 4, - \frac {15}{7}</math>. To maximize <math>a^3 + b^3 + c^3</math> choose <math>p = - \frac {15}{7}</math> and so the sum is <math>28 - \frac {45}{7} = \frac {196 - 45}{7}</math> giving <math>151 + 7 = \fbox{158}</math>. | Add the three equations to get <math>a^3 + b^3 + c^3 = 28 + 3abc</math>. Now, let <math>abc = p</math>. <math>a = \sqrt [3]{p + 2}</math>, <math>b = \sqrt [3]{p + 6}</math> and <math>c = \sqrt [3]{p + 20}</math>, so <math>p = abc = (\sqrt [3]{p + 2})(\sqrt [3]{p + 6})(\sqrt [3]{p + 20})</math>. Now [[cube]] both sides; the <math>p^3</math> terms cancel out. Solve the remaining [[quadratic]] to get <math>p = - 4, - \frac {15}{7}</math>. To maximize <math>a^3 + b^3 + c^3</math> choose <math>p = - \frac {15}{7}</math> and so the sum is <math>28 - \frac {45}{7} = \frac {196 - 45}{7}</math> giving <math>151 + 7 = \fbox{158}</math>. | ||
| − | == See | + | == See Also == |
{{AIME box|year=2010|num-b=8|num-a=10|n=I}} | {{AIME box|year=2010|num-b=8|num-a=10|n=I}} | ||
[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
Revision as of 16:55, 12 April 2012
Problem
Let 
be the real solution of the system of equations 
, 
, 
. The greatest possible value of 
can be written in the form 
, where 
and 
are relatively prime positive integers. Find 
.
Solution
Add the three equations to get 
. Now, let 
. ![$a = \sqrt [3]{p + 2}$](http://latex.artofproblemsolving.com/e/8/9/e89feedfb60c766267b943720cce4525c8ff02d1.png)
, ![$b = \sqrt [3]{p + 6}$](http://latex.artofproblemsolving.com/c/b/b/cbb6a381df37eb48da97c3256dc67dd15d589ff5.png)
and ![$c = \sqrt [3]{p + 20}$](http://latex.artofproblemsolving.com/0/c/c/0ccbe29741cf95dc3176caff203e96189b0b5d70.png)
, so ![$p = abc = (\sqrt [3]{p + 2})(\sqrt [3]{p + 6})(\sqrt [3]{p + 20})$](http://latex.artofproblemsolving.com/e/a/2/ea2f87da28c91e3043de400384a0037c64267edf.png)
. Now cube both sides; the 
terms cancel out. Solve the remaining quadratic to get 
. To maximize choose 
and so the sum is 
giving 
.
See Also
| 2010 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
