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

Revision as of 21:32, 17 February 2018

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

@@ Line 9: / Line 9: @@
 ===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 + 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>.
 == See Also ==

2010 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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