2010 AIME I Problems/Problem 6

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 $P(x) = ax^2 - 2ax + c$ for some real $a$ and $c$ . This is because the derivative of $P(x)$ is $2ax - 2a$ , and a global minimum occurs only at $x = 1$ (in addition, because of this derivative, the vertex of any quadratic polynomial occurs at $\frac{-b}{2a}$ ). 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, $P(16) = \boxed{406}$ .

Solution 3

Let $y = x^2 - 2x + 2$ ; note that $2y - 1 = 2x^2 - 4x + 3$ . Setting $y = 2y - 1$ , we find that equality holds when $y = 1$ and therefore when $x^2 - 2x + 2 = 1$ ; this is true iff $x = 1$ , so $P(1) = 1$ .

Let $Q(x) = P(x) - x$ ; clearly $Q(1) = 0$ , so we can write $Q(x) = (x - 1)Q'(x)$ , where $Q'(x)$ is some linear function. Plug $Q(x)$ into the given inequality:

$x^2 - 3x + 2 \le Q(x) \le 2x^2 - 5x + 3$

$(x - 1)(x - 2) \le (x - 1)Q'(x) \le (x - 1)(2x - 3)$ , and thus

$x - 2 \le Q'(x) \le 2x - 3$

For all $x > 1$ ; note that the inequality signs are flipped if $x < 1$ , and that the division is invalid for $x = 1$ . However,

$\lim_{x \to 1} x - 2 = \lim_{x \to 1} 2x - 3 = -1$ ,

and thus by the sandwich theorem $\lim_{x \to 1} Q'(x) = -1$ ; by the definition of a continuous function, $Q'(1) = -1$ . Also, $Q(11) = 170$ , so $Q'(11) = 170/(11-1) = 17$ ; plugging in and solving, $Q'(x) = (9/5)(x - 1) - 1$ . Thus $Q(16) = 390$ , and so $P(16) = \boxed{406}$ .

Solution 4

Let $Q(x) = P(x) - (x^2-2x+2)$ , then $0\le Q(x) \le (x-1)^2$ . Therefore, $0\le Q(x+1) \le x^2 \Rightarrow Q(x) = Ax^2$ for some real value A.

$Q(11) = 10^2A \Rightarrow P(11)-(11^2-22+2)=100A \Rightarrow 80=100A \Rightarrow A=\frac{4}{5}$ .

$Q(16)=15^2A=180 \Rightarrow P(16)-(16^2-32+2) = 180 \Rightarrow P(16)=180+226= \boxed{406}$

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