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

(4 intermediate revisions by 3 users not shown)
Line 8: Line 8:
  
 
===Solution 2===
 
===Solution 2===
This is almost the same as solution 1. Note <math>a^3 + b^3 + c^3 = 28 + 3abc</math>. Next, let <math>k = a^3</math>. Note that <math>b = \sqrt [3]{k + 4}</math> and  <math>c = \sqrt [3]{k + 18}</math>, so we have <math>28 + \sqrt [3]{k(k+4)(k+18)} = 28+abc=a^3+b^3+c^3=3k+22</math>. Move 28 over, divide both sides by 3, then cube to get <math>k^3-6k^2+12k-8 = k^3+22k^2+18k</math>. The <math>k^3</math> terms cancel out, so solve the quadratic to get <math>k = -2, -\frac{1}{7}</math>. We maximize <math>abc</math> by choosing <math>k = -\frac{1}{7}</math>, which gives us <math>a^3+b^3+c^3 = 3k + 22 = \frac{151}{7}</math>. Thus, our answer is <math>151+7=\boxed{158}</math>.
+
This is almost the same as solution 1. Note <math>a^3 + b^3 + c^3 = 28 + 3abc</math>. Next, let <math>k = a^3</math>. Note that <math>b = \sqrt [3]{k + 4}</math> and  <math>c = \sqrt [3]{k + 18}</math>, so we have <math>28 + 3\sqrt [3]{k(k+4)(k+18)} = 28+3abc=a^3+b^3+c^3=3k+22</math>. Move 28 over, divide both sides by 3, then cube to get <math>k^3-6k^2+12k-8 = k^3+22k^2+18k</math>. The <math>k^3</math> terms cancel out, so solve the quadratic to get <math>k = -2, -\frac{1}{7}</math>. We maximize <math>abc</math> by choosing <math>k = -\frac{1}{7}</math>, which gives us <math>a^3+b^3+c^3 = 3k + 22 = \frac{151}{7}</math>. Thus, our answer is <math>151+7=\boxed{158}</math>.
 +
 
 +
===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>.
 +
 
 +
==Video Solution==
 +
https://youtu.be/LXct4j_rYfw
 +
 
 +
~Shreyas S
  
 
== See Also ==
 
== See Also ==

Revision as of 17:15, 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 (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png