Difference between revisions of "2010 AIME I Problems/Problem 6"
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=== Solution 4 === | === Solution 4 === | ||
− | Let <math>Q(x) = P(x) - (x^2-2x+2)</math>, then <math>0\le Q(x) \le (x-1)^2</math>. Therefore, <math>0\le Q(x+1) \le x^2 \Rightarrow Q(x) = Ax^2</math> for some real value A. | + | Let <math>Q(x) = P(x) - (x^2-2x+2)</math>, then <math>0\le Q(x) \le (x-1)^2</math> (note this is derived from the given inequality chain). Therefore, <math>0\le Q(x+1) \le x^2 \Rightarrow Q(x+1) = Ax^2</math> for some real value A. |
<math>Q(11) = 10^2A \Rightarrow P(11)-(11^2-22+2)=100A \Rightarrow 80=100A \Rightarrow A=\frac{4}{5}</math>. | <math>Q(11) = 10^2A \Rightarrow P(11)-(11^2-22+2)=100A \Rightarrow 80=100A \Rightarrow A=\frac{4}{5}</math>. | ||
<math>Q(16)=15^2A=180 \Rightarrow P(16)-(16^2-32+2) = 180 \Rightarrow P(16)=180+226= \boxed{406}</math> | <math>Q(16)=15^2A=180 \Rightarrow P(16)-(16^2-32+2) = 180 \Rightarrow P(16)=180+226= \boxed{406}</math> | ||
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+ | === Solution 5 === | ||
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+ | 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 === | ||
+ | |||
+ | 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>. | ||
== See Also == | == See Also == |
Latest revision as of 21:20, 17 December 2021
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 .
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.