2019 AMC 12B Problems/Problem 24
Let Let denote all points in the complex plane of the form where and What is the area of ?
Notice that , which is one of the cube roots of unity. We wish to find the span of for reals . Observe also that if , then replacing , , and by and leaves the value of unchanged. Therefore, assume that at least one of is equal to . If exactly one of them is , we can form an equilateral triangle of side length using the remaining terms. A similar argument works if exactly two of them are . In total, we get equilateral triangles, whose total area is .
Note: A diagram of the six equilateral triangles is shown below.
Solution 1 but not exactly
WLOG is the smallest of the . Then the expression is equivalent to . To find the area of the region, we need only consider the extremities (), as they will form a polygon which contains all points. So, when we have the origin (diagram omitted). When we have :
When we have :
When we have :
The area of this is . Multiply this by to get .
We can add on each term one at a time. Firstly, the possible values of lie on the following line:
For each point on the line, we can add . This means that we can extend the area to
by "moving" the blue line along the red line. Finally, we can add to every point, giving
by "moving" the previous area along the green line. This leaves us with a regular hexagon with side length , so, as in Solution 1, the total area is .
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