Difference between revisions of "2011 AIME II Problems/Problem 9"
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Peterungar (talk | contribs) (→Solution: A rational solution) |
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x_1 &= x_2 = \frac{5 - \sqrt{20}}{30} \ | x_1 &= x_2 = \frac{5 - \sqrt{20}}{30} \ | ||
x_5 &= x_4 = \frac{5 + \sqrt{20}}{30} | x_5 &= x_4 = \frac{5 + \sqrt{20}}{30} | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | Another example is | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | x_1 = x_3 = \frac{1}{3} \ | ||
+ | x_2 = \frac{19}{60}, \ x_5 = \frac{1}{60} \ | ||
+ | x_4 &= x_6 = 0 | ||
\end{align*} | \end{align*} | ||
</cmath> | </cmath> |
Revision as of 14:27, 3 June 2011
Problem 9
Let be non-negative real numbers such that , and . Let and be positive relatively prime integers such that is the maximum possible value of . Find .
Solution
Note that neither the constraint nor the expression we need to maximize involves products with . Factoring out say and we see that the constraint is , while the expression we want to maximize is . Adding the left side of the constraint to the expression we get: . This new expression is the product of three non-negative terms whose sum is equal to 1. By AM-GM this product is at most . Since we have added at least the desired maximum is at most . It is easy to see that this upper bound can in fact be achieved by ensuring that the constraint expression is equal to with —for example, by choosing and small enough—so our answer is
An example is:
Another example is