Difference between revisions of "2010 AIME I Problems/Problem 9"
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== Solution == | == Solution == | ||
− | Add the three equations to get <math> | + | Add the three equations to get <math>x^3 + y^3 + z^3 = 28 + 3xyz</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 Also == | == See Also == |
Revision as of 18:16, 18 October 2014
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
.
,
and
, so
. 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 |
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