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 (note this is derived from the given inequality chain). Therefore, for some real value A.
Let . Plugging in to the expressions on both sides of the inequality, we see that . We see from the problem statement that . Since we know the vertex of lies at , by symmetry we get as well. Since we now have three equations, we can solve this trivial system and get our answer of .
Similar to Solution 5, let . Note that is a vertex of the polynomial. Additionally, this means that (since is the minimum point). Thus, we have . Therefore . Moreover, . And so our polynomial is . Plug in to get .
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