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 12:08, 18 March 2010
Problem
Let be a quadratic polynomial with real coefficients satisfying
for all real numbers
, and suppose
. Find
.
Solution
Solution 1
![[asy] import graph; real min = -0.5, max = 2.5; pen dark = linewidth(1); real P(real x) { return 8*(x-1)^2/5+1; } real Q(real x) { return (x-1)^2+1; } real R(real x) { return 2*(x-1)^2+1; } draw(graph(P,min,max),dark); draw(graph(Q,min,max),linetype("6 2")+linewidth(0.7)); draw(graph(R,min,max),linetype("6 2")+linewidth(0.7)); dot((1,1)); label("$P(x)$",(max,P(max)),E,fontsize(10)); label("$Q(x)$",(max,Q(max)),E,fontsize(10)); label("$R(x)$",(max,R(max)),E,fontsize(10)); /* axes */ Label f; f.p=fontsize(8); xaxis(-2, 3, Ticks(f, 5, 1)); yaxis(-1, 5, Ticks(f, 6, 1)); [/asy]](http://latex.artofproblemsolving.com/8/6/4/8642816d5b50f02498e44e455af5a0ae225ce976.png)
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:
![$f(11) = 99a + c = 181$](http://latex.artofproblemsolving.com/f/5/0/f504c12353fd992f58df8f90f81b1b69ee0583bd.png)
![$f(1) = -a + c = 1$](http://latex.artofproblemsolving.com/3/a/9/3a91c23aac72bd1fcaae464d5c4c0fb5cc5470a4.png)
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 |