2010 AIME I Problems/Problem 9
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 by OmegaLearn
https://youtu.be/SpSuqWY01SE?t=1293
~ pi_is_3.14
Remark
It is tempting to add the equations and then use the well-known factorization . Unfortunately such a factorization is just a red herring: it doesn't give much information on .
Another Remark
The real problem with adding the equations is that are real numbers based on the problem, but the adding trick only works when are integers.
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 |
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