2010 AIME I Problems/Problem 6
Let be a quadratic polynomial with real coefficients satisfying for all real numbers , and suppose . Find .
Also, , so , and obtains its minimum at the point . Then must be of the form for some constant ; substituting yields . Finally, .
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, .
Let ; note that . Setting , we find that equality holds when and therefore when ; 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 .
Let , then . Therefore, for some real value A.
|2010 AIME I (Problems • Answer Key • Resources)|
|1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15|
|All AIME Problems and Solutions|