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

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===Solution 3===
 
===Solution 3===
 
We have that <math>x^3 = 2 + xyz</math>, <math>y^3 = 6 + xyz</math>, and <math>z^3 = 20 + xyz</math>. Multiplying the three equations, and letting <math>m = xyz</math>, we have that <math>m^3 = (2+m)(6+m)(20+m)</math>, and reducing, that <math>7m^2 + 43m + 60 = 0</math>, which has solutions <math>m = -\frac{15}{7}, -4</math>. Adding the three equations and testing both solutions, we find the answer of <math>\frac{151}{7}</math>, so the desired quantity is <math>151 + 7 = \fbox{158}</math>.
 
We have that <math>x^3 = 2 + xyz</math>, <math>y^3 = 6 + xyz</math>, and <math>z^3 = 20 + xyz</math>. Multiplying the three equations, and letting <math>m = xyz</math>, we have that <math>m^3 = (2+m)(6+m)(20+m)</math>, and reducing, that <math>7m^2 + 43m + 60 = 0</math>, which has solutions <math>m = -\frac{15}{7}, -4</math>. Adding the three equations and testing both solutions, we find the answer of <math>\frac{151}{7}</math>, so the desired quantity is <math>151 + 7 = \fbox{158}</math>.
 +
 +
==Video Solution==
 +
https://youtu.be/LXct4j_rYfw
 +
 +
~Shreyas S
  
 
== See Also ==
 
== See Also ==

Revision as of 16:15, 26 June 2020

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

Solution 1

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 $151 + 7 = \fbox{158}$.

Solution 2

This is almost the same as solution 1. Note $a^3 + b^3 + c^3 = 28 + 3abc$. Next, let $k = a^3$. Note that $b = \sqrt [3]{k + 4}$ and $c = \sqrt [3]{k + 18}$, so we have $28 + 3\sqrt [3]{k(k+4)(k+18)} = 28+3abc=a^3+b^3+c^3=3k+22$. Move 28 over, divide both sides by 3, then cube to get $k^3-6k^2+12k-8 = k^3+22k^2+18k$. The $k^3$ terms cancel out, so solve the quadratic to get $k = -2, -\frac{1}{7}$. We maximize $abc$ by choosing $k = -\frac{1}{7}$, which gives us $a^3+b^3+c^3 = 3k + 22 = \frac{151}{7}$. Thus, our answer is $151+7=\boxed{158}$.

Solution 3

We have that $x^3 = 2 + xyz$, $y^3 = 6 + xyz$, and $z^3 = 20 + xyz$. Multiplying the three equations, and letting $m = xyz$, we have that $m^3 = (2+m)(6+m)(20+m)$, and reducing, that $7m^2 + 43m + 60 = 0$, which has solutions $m = -\frac{15}{7}, -4$. Adding the three equations and testing both solutions, we find the answer of $\frac{151}{7}$, so the desired quantity is $151 + 7 = \fbox{158}$.

Video Solution

https://youtu.be/LXct4j_rYfw

~Shreyas S

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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