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

Revision as of 16:59, 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

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)$",(max,R(max)),E,fontsize(10)); /* axes */ Label f; f.p=fontsize(8); xaxis(-2, 3, Ticks(f, 5, 1)); yaxis(-1, 5, Ticks(f, 6, 1)); [/asy]$

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 $P(x)$ must be in the form $f(x) = ax^2 - 2ax + c$ for some real $a$ and $c$ . This is because the derivative of $P(x)$ is $2ax - 2a$ , and a global minimum occurs only at $x = 1$ (alternatively, the minimum occurs when $x = 1 = -b/[2a]$ ). Substituting $(1,1)$ and $(11, 181)$ we obtain two equations:

$P(11) = 99a + c = 181$ , and $P(1) = -a + c = 1$ .

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

@@ Line 30: / Line 30: @@
 === 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> (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:
+It can be seen that the function <math>P(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>P(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:
-<center><math>f(11) = 99a + c = 181</math>, and <math>f(1) = -a + c = 1</math>.</center>
+<center><math>P(11) = 99a + c = 181</math>, and <math>P(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>P(x) = \frac{9}{5}x^2 - \frac{18}{5}x + \frac {14}{5}</math>. Therefore, <math>P(16) = \boxed{406}</math>.
 == See also ==

2010 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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