Difference between revisions of "2010 AIME I Problems/Problem 6"
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=== Solution 5 === | === Solution 5 === | ||
− | Let <math>P(x) = ax^2 + bx + c</math>. Plugging in <math>x = 1</math> to the expressions on both sides of the inequality, we see that <math>a + b + c = 1</math>. We see from the problem statement that <math>121a + 11b + c = 181</math>. Since we know the vertex of <math>P(x)</math> lies at <math>x = 1</math>, by symmetry we get <math>81a -9b + c</math>. Since we now have three equations, we can solve this trivial system and get our answer of <math>\boxed{406}</math>. | + | Let <math>P(x) = ax^2 + bx + c</math>. Plugging in <math>x = 1</math> to the expressions on both sides of the inequality, we see that <math>a + b + c = 1</math>. We see from the problem statement that <math>121a + 11b + c = 181</math>. Since we know the vertex of <math>P(x)</math> lies at <math>x = 1</math>, by symmetry we get <math>81a -9b + c = 181</math> as well. Since we now have three equations, we can solve this trivial system and get our answer of <math>\boxed{406}</math>. |
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+ | === Solution 6 === | ||
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+ | Similar to Solution 5, let <math>P(x) = ax^2 + bx + c</math>. Note that <math>(1,1)</math> is a vertex of the polynomial. Additionally, this means that <math>b = -2a</math> (since <math>\frac{-b}{2a}</math> is the minimum <math>x</math> point). Thus, we have <math>P(x) = ax^2 - 2ax + c</math>. Therefore <math>a - 2a + c = 1</math>. Moreover, <math>99a + c = 181</math>. And so our polynomial is <math>\frac{9}{5}x^2 - \frac{18}{5}x + \frac{14}{5}</math>. Plug in <math>x = 16</math> to get <math>\boxed{406}</math>. | ||
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+ | === Solution 7 === | ||
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+ | Very similar to Solution 6, start by noticing that <math>P(x)</math> is between two polynomials, we try to set them equal to find a point where the two polynomials intersect, meaning that <math>P(x)</math> would also have to intersect that point (it must be between the two graphs). Setting <math>x^2 - 2x + 2 = 2x^2 - 4x + 3</math>, we find that <math>x = 1</math>. Note that both of these graphs have the same vertex (at <math>x = 1</math>), and so <math>P(x)</math> must also have the same vertex <math>(1, 1)</math>. Setting <math>P(x) = ax^2 - 2ax + a + 1</math> (this is where we have a vertex at <math>(1, 1)</math>), we plug in <math>11</math> and find that <math>a = 1.8</math>. Evaluating <math>1.8x^2 - 3.6x + 2.8</math> when <math>x = 16</math> (our intended goal), we find that <math>P(16) = \boxed{406}</math>. | ||
== See Also == | == See Also == |
Latest revision as of 19:39, 20 November 2023
Contents
[hide]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 | |
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.