2010 AIME I Problems/Problem 9
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 .
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 .
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 .
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.
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