Problem:
A regular hexagon with center at the origin in the complex plane has opposite pairs of sides one unit apart. One pair of sides is parallel to the imaginary axis. Let be the region outside the hexagon, and let . Then the area of has the form , where and are positive integers. Find .
Solution:
One pair of vertices lies at . Express points on the line segment determined by these two vertices in the form , where is real and . Reciprocals of points on this line segment are then of the form with . Because
the curve traced by the reciprocals of complex numbers on this line segment is an arc of a circle centered at with radius , running from to . The region enclosed by this arc and the lines from the origin to the endpoints can be partitioned into a -degree sector of the disk centered at with radius , together with two triangles, each of base and height . Thus it has area . Multiply by six to find that the total area is . Thus .
Query: The above argument shows that the reciprocals of the points on a line not through the origin fall on a circle through the origin. What happens to a line through the origin? Because pairs of parallel sides are one unit apart, one side of the hexagon lies along the line . If , where is on this line, then
which is equivalent to . Hence , or , which is the equation of a circle of radius centered at . Because is mapped to , it follows that is mapped to the interior of this circle. By symmetry, the five remaining half-planes whose union produces are mapped to corresponding disks. Thus the points that are in the image of the half-plane but none of the other five half-planes belong to the disk and the region bounded by the two rays . The resulting set can be partitioned into a circular sector with radius and central angle and two isosceles triangles with equal sides of length and vertex angle . There are six such nonoverlapping congruent figures forming . It follows that the area of is . Thus .
The problems on this page are the property of the MAA's American Mathematics Competitions