2024 AMC 12A Problems/Problem 24
Problem
There exist exactly four different complex numbers 
that satisfy the equation shown below: ![\[\left|z + \overline{z}\left(\tfrac{3}{5} + \tfrac{4}{5}i\right)\right| = \left|z + \overline{z}\left(\tfrac{4}{5} + \tfrac{3}{5}i\right)\right| = 1.\]](http://latex.artofproblemsolving.com/e/1/a/e1a187572d0c161e36303d4191afedbe4b7bf3f2.png)
What is the area of the convex quadrilateral whose vertices are those four complex numbers in the complex plane?
Solution
Let 
be the rectangle with vertices 
and 
and let 
be its center. Let the reflections of over 
and 
be 
and 
. If the points 
and 
are such that 
and 
are parallelograms, then the given condition is equivalent to and lying on 
. The condition that lies on this circle is equivalent to lying on the union of the perpendicular bisectors of 
and 
and the condition that lies on the circle is equivalent to lying on the union of the perpendicular bisectors of 
and 
. Therefore, the possible are the circumcenters of 
and 
and therefore they form a rhombus. Now, the area of this rhombus is twice the product of the circumradii of 
and 
, and if 
then this is just 
. Now, we can find that 
so 
and 
, giving that the answer is 
.
~Zhaom
See also
| 2024 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 23 |
Followed by Problem 25 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
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