Difference between revisions of "2010 AIME I Problems/Problem 6"
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Let <math>P(x)</math> be a [[quadratic]] polynomial with real coefficients satisfying <math>x^2 - 2x + 2 \le P(x) \le 2x^2 - 4x + 3</math> for all real numbers <math>x</math>, and suppose <math>P(11) = 181</math>. Find <math>P(16)</math>. | Let <math>P(x)</math> be a [[quadratic]] polynomial with real coefficients satisfying <math>x^2 - 2x + 2 \le P(x) \le 2x^2 - 4x + 3</math> for all real numbers <math>x</math>, and suppose <math>P(11) = 181</math>. Find <math>P(16)</math>. | ||
+ | __TOC__ | ||
== Solution == | == Solution == | ||
+ | === Solution 1 === | ||
<center><asy> | <center><asy> | ||
import graph; real min = -0.5, max = 2.5; pen dark = linewidth(1); | import graph; real min = -0.5, max = 2.5; pen dark = linewidth(1); | ||
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Also, <math>1 = Q(1) \le P(1) \le R(1) = 1</math>, so <math>P(1) = 1</math>, and <math>P</math> obtains its minimum at the point <math>(1,1)</math>. Then <math>P(x)</math> must be of the form <math>c(x-1)^2 + 1</math> for some constant <math>c</math>; substituting <math>P(11) = 181</math> yields <math>c = \frac 95</math>. Finally, <math>P(16) = \frac 95 \cdot (16 - 1)^2 + 1 = \boxed{406}</math>. | Also, <math>1 = Q(1) \le P(1) \le R(1) = 1</math>, so <math>P(1) = 1</math>, and <math>P</math> obtains its minimum at the point <math>(1,1)</math>. Then <math>P(x)</math> must be of the form <math>c(x-1)^2 + 1</math> for some constant <math>c</math>; substituting <math>P(11) = 181</math> yields <math>c = \frac 95</math>. Finally, <math>P(16) = \frac 95 \cdot (16 - 1)^2 + 1 = \boxed{406}</math>. | ||
− | == Solution 2 == | + | === Solution 2 === |
− | It can be seen that the function <math>f(x)</math> must be in the form <math>f(x) = ax^2 - 2ax + c</math> for some real <math>a</math> and <math>c</math>. This is because the [[derivative]] of <math>f(x)</math> is <math>2ax - 2a</math>, and a global minimum occurs only at <math>x = 1</math>. Substituting <math>(1,1)</math> and <math>(11, 181)</math> we obtain two equations: | + | It can be seen that the function <math>f(x)</math> must be in the form <math>f(x) = ax^2 - 2ax + c</math> for some real <math>a</math> and <math>c</math>. This is because the [[derivative]] of <math>f(x)</math> is <math>2ax - 2a</math>, and a global minimum occurs only at <math>x = 1</math> (alternatively, the minimum occurs when <math>x = 1 = -b/[2a]</math>). Substituting <math>(1,1)</math> and <math>(11, 181)</math> we obtain two equations: |
− | <math>f(11) = 99a + c = 181</math>, and <math>f(1) = -a + c = 1</math>. | + | <center><math>f(11) = 99a + c = 181</math>, and <math>f(1) = -a + c = 1</math>.</center> |
Solving, we get <math>a = \frac{9}{5}</math> and <math>c = \frac{14}{5}</math>, so <math>f(x) = \frac{9}{5}x^2 - \frac{18}{5}x + \frac {14}{5}</math>. Therefore, <math>f(16) = \boxed{406}</math>. | Solving, we get <math>a = \frac{9}{5}</math> and <math>c = \frac{14}{5}</math>, so <math>f(x) = \frac{9}{5}x^2 - \frac{18}{5}x + \frac {14}{5}</math>. Therefore, <math>f(16) = \boxed{406}</math>. |
Revision as of 11:08, 18 March 2010
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 (alternatively, the minimum occurs when ). Substituting and we obtain two equations:
Solving, we get and , so . Therefore, .
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 |