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Difference between revisions of "2010 AIME I Problems/Problem 9"

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== Solution ==
 
== Solution ==
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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>.
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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>.
  
 
== See Also ==
 
== See Also ==

Revision as of 17:17, 18 October 2014

Problem

Let $(a,b,c)$ be the real solution of the system of equations $x^3 - xyz = 2$, $y^3 - xyz = 6$, $z^3 - xyz = 20$. The greatest possible value of $a^3 + b^3 + c^3$ can be written in the form $\frac {m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.

Solution

Add the three equations to get $a^3 + b^3 + c^3 = 28 + 3abc$. Now, let $abc = p$. $a = \sqrt [3]{p + 2}$, $b = \sqrt [3]{p + 6}$ and $c = \sqrt [3]{p + 20}$, so $p = abc = (\sqrt [3]{p + 2})(\sqrt [3]{p + 6})(\sqrt [3]{p + 20})$. Now cube both sides; the $p^3$ terms cancel out. Solve the remaining quadratic to get $p = - 4, - \frac {15}{7}$. To maximize $a^3 + b^3 + c^3$ choose $p = - \frac {15}{7}$ and so the sum is $28 - \frac {45}{7} = \frac {196 - 45}{7}$ giving $151 + 7 = \fbox{158}$.

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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