Problem:
Consider the region A in the complex plane that consists of all points z such that both z/40 and 40/zΛ have real and imaginary parts between 0 and 1 inclusive. What is the integer that is nearest the area of A? (If z=x+iy with x and y real, then zΛ=xβiy is the conjugate of z.)
Solution:
Let z=x+iy. For z to be in the region in question, we must have 0β€xβ€40 and 0β€yβ€40. Hence the region lies in the square with vertices (0,0),(40,0),(40,40), and (0,40). Next note that
zΛ40β=xβiy40β=x2+y240xβ+ix2+y240yβ
Hence the restrictions on the real and imaginary parts of zΛ40β give
0β€x2+y240xββ€1 and 0β€x2+y240yββ€1
from which
(xβ20)2+y2β₯202 and x2+(yβ20)2β₯202.
Thus the region in question lies outside the circle with center (20,0) and radius 20 and also outside the circle with center (0,20) and radius 20, as indicated by the shaded portion of the diagram. As suggested by the dashed lines in the diagram, the area of the region is 43β the area of the square minus the area of two quarter-circles. Hence
Area(A)=43ββ
402β42β(Οβ
202)=200(6βΟ)β571.7
so that the desired number is 572β.
The problems on this page are the property of the MAA's American Mathematics Competitions
