Problem:
Let f(x)=x2+6x+1, and let R denote the set of points (x,y) in the coordinate plane such that
f(x)+f(y)β€0 and f(x)βf(y)β€0
The area of R is closest to
Answer Choices:
A. 21
B. 22
C. 23
D. 24
E. 25
Solution:
Note that
f(x)+f(y)=x2+6x+y2+6y+2=(x+3)2+(y+3)2β16
and
f(x)βf(y)=x2βy2+6(xβy)=(xβy)(x+y+6).
The given conditions can be written as
(x+3)2+(y+3)2β€16 and (xβy)(x+y+6)β€0
The first inequality describes the region on and inside the circle of radius 4 with center (β3,β3). The second inequality can be rewritten as
(xβyβ₯0 and x+y+6β€0) or (xβyβ€0 and x+y+6β₯0)
Each of these inequalities describes a half-plane bounded by a line that passes through (β3,β3) and has slope 1 or -1 . Thus, the set R is the shaded region in the following diagram, and its area is half the area of the circle, which is 8Οβ25.13β.
The problems on this page are the property of the MAA's American Mathematics Competitions
