2010 AIME I Problems/Problem 6
Problem
Let be a quadratic polynomial with real coefficients satisfying for all real numbers , and suppose . Find .
Solution
Solution 1
Let , . Completing the square, we have , and , so it follows that for all (by the Trivial Inequality).
Also, , so , and obtains its minimum at the point . Then must be of the form for some constant ; substituting yields . Finally, .
Solution 2
It can be seen that the function must be in the form for some real and . This is because the derivative of is , and a global minimum occurs only at (in addition, because of this derivative, the vertex of any quadratic polynomial occurs at ). Substituting and we obtain two equations:
Solving, we get and , so . Therefore, .
Solution 3
Let ; note that . Setting , we find that and therefore ; this is true iff , so .
Let ; clearly , so we can write , where is some linear function. Plug into the given inequality:
, and thus
For all ; note that the inequality signs are flipped if , and that the division is invalid for . However,
, and thus by the sandwich theorem ; by the definition of a continuous function, . Also, , so ; plugging in and solving, . Thus , and so .
See also
2010 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
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All AIME Problems and Solutions |