2010 AIME I Problems/Problem 6
Contents
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 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 .
Solution 4
Let , then (note this is derived from the given inequality chain). Therefore, for some real value A.
.
Solution 5
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 .
Solution 6
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 .
Solution 7
Very similar to Solution 6, start by noticing that is between two polynomials, we try to set them equal to find a point where the two polynomials intersect, meaning that would also have to intersect that point (it must be between the two graphs). Setting , we find that . Note that both of these graphs have the same vertex (at ), and so must also have the same vertex . Setting (this is where we have a vertex at ), we plug in and find that . Evaluating when (our intended goal), we find that .
See Also
2010 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
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