Problem:
Let P(x) be a quadratic polynomial with real coefficients satisfying
x2−2x+2≤P(x)≤2x2−4x+3
for all real numbers x, and suppose P(11)=181. Find P(16).
Solution:
Completing the square yields
(x−1)2+1≤P(x)≤2(x−1)2+1
The left hand and right hand expressions represent parabolas with a vertex at (1,1), so P(x) must also represent a parabola with vertex at (1,1). Therefore P(x)=a(x−1)2+1,P(11)=100a+1=181, and a=59. Thus P(x)=59(x−1)2+1, and P(16)=406.
The problems on this page are the property of the MAA's American Mathematics Competitions