2019 AMC 12B Problems/Problem 24
Contents
Problem
Let Let denote all points in the complex plane of the form where and What is the area of ?
Solution 1
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 2
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 .
See Also
2019 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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