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Difference between revisions of "2010 AIME I Problems/Problem 6"

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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>.
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== Solution 2 ==
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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:
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<math>f(11) = 99a + c = 181</math>
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<math>f(1) = -a + c = 1</math>.
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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>.
  
 
== See also ==
 
== See also ==

Revision as of 10:45, 18 March 2010

Problem

Let $P(x)$ be a quadratic polynomial with real coefficients satisfying $x^2 - 2x + 2 \le P(x) \le 2x^2 - 4x + 3$ for all real numbers $x$, and suppose $P(11) = 181$. Find $P(16)$.

Solution

[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]

Let $Q(x) = x^2 - 2x + 2$, $R(x) = 2x^2 - 4x + 3$. Completing the square, we have $Q(x) = (x-1)^2 + 1$, and $R(x) = 2(x-1)^2 + 1$, so it follows that $P(x) \ge Q(x) \ge 1$ for all $x$ (by the Trivial Inequality).

Also, $1 = Q(1) \le P(1) \le R(1) = 1$, so $P(1) = 1$, and $P$ obtains its minimum at the point $(1,1)$. Then $P(x)$ must be of the form $c(x-1)^2 + 1$ for some constant $c$; substituting $P(11) = 181$ yields $c = \frac 95$. Finally, $P(16) = \frac 95 \cdot (16 - 1)^2 + 1 = \boxed{406}$.

Solution 2

It can be seen that the function $f(x)$ must be in the form $f(x) = ax^2 - 2ax + c$ for some real $a$ and $c$. This is because the derivative of $f(x)$ is $2ax - 2a$, and a global minimum occurs only at $x = 1$. Substituting $(1,1)$ and $(11, 181)$ we obtain two equations:

$f(11) = 99a + c = 181$ $f(1) = -a + c = 1$.

Solving, we get $a = \frac{9}{5}$ and $c = \frac{14}{5}$, so $f(x) = \frac{9}{5}x^2 - \frac{18}{5}x + \frac {14}{5}$. Therefore, $f(16) = \boxed{406}$.

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
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