Difference between revisions of "2011 AIME II Problems/Problem 9"
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\end{align*} | \end{align*} | ||
</cmath> | </cmath> | ||
+ | ==Solution 2 (Not legit)== | ||
+ | There's a symmetry between <math>a_1, a_3, a_5</math> and <math>a_2,a_4,a_6</math>. Therefore, a good guess is that <math>x = a_1 = a_3 = a_5</math> and <math>y = a_2 = a_4 = a_6</math>, at which point we know that <math>a+b = 1/3</math>, <math>a^3+b^3 \geq 1/540</math>, and we are trying to maximize <math>3a^2b+3ab^2</math>. Then, | ||
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+ | <cmath>3a^3b+3ab^2 = (a+b)^3-a^3-b^3 \leq 1/27 - 1/540 = 19/540</cmath>, which is the answer. | ||
+ | |||
+ | This solution is extremely lucky; if you attempt to solve for <math>x</math> and <math>y</math> you receive complex answers (which contradict the problem statement), but the final answer is correct. | ||
==See also== | ==See also== |
Revision as of 13:44, 16 February 2020
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
Solution 2 (Not legit)
There's a symmetry between and . Therefore, a good guess is that and , at which point we know that , , and we are trying to maximize . Then,
, which is the answer.
This solution is extremely lucky; if you attempt to solve for and you receive complex answers (which contradict the problem statement), but the final answer is correct.
See also
2011 AIME II (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.