2010 AIME I Problem 6

Question

Let

P(x)

be a quadratic polynomial with real coefficients satisfying

\begin{aligned} x^2 - 2x + 2 &\le P(x) \\ &\le 2x^2 - 4x + 3 \end{aligned}

for all real numbers

x,

and suppose

P(11) = 181.

Find

P(16).

Answer

Complete the square: both bounding quadratics equal

(x-1)^2 + 1

and

2(x-1)^2 + 1,

with common vertex

(1, 1)

P(x) - (x-1)^2 - 1

is nonnegative everywhere and zero at

x = 1,

so it is

c(x-1)^2;

use

P(11) = 181

to find

c

Completing the square, the condition reads

\begin{aligned} (x-1)^2 + 1 &\le P(x) \\ &\le 2(x-1)^2 + 1. \end{aligned}

At

x = 1

both bounds equal

1,

so

P(1) = 1.

The quadratic

P(x) - \left((x-1)^2 + 1\right)

is nonnegative for all

x

and vanishes at

x = 1,

so

x = 1

is a double root:

P(x) = a(x-1)^2 + 1

for some constant

a.

From

P(11) = 100a + 1 = 181

we get

a = \frac{9}{5}.

Then

\begin{aligned} P(16) &= \frac{9}{5} \cdot 225 + 1 \\ &= 405 + 1 = 406. \end{aligned}

Problem 6 in other years