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 == |
Latest revision as of 16:15, 26 June 2020
Contents
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
Solution 1
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 .
Solution 2
This is almost the same as solution 1. Note . Next, let . Note that and , so we have . Move 28 over, divide both sides by 3, then cube to get . The terms cancel out, so solve the quadratic to get . We maximize by choosing , which gives us . Thus, our answer is .
Solution 3
We have that , , and . Multiplying the three equations, and letting , we have that , and reducing, that , which has solutions . Adding the three equations and testing both solutions, we find the answer of , so the desired quantity is .
Video Solution
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.