2014 AIME II Problems/Problem 10
Contents
Problem
Let be a complex number with . Let be the polygon in the complex plane whose vertices are and every such that . Then the area enclosed by can be written in the form , where is an integer. Find the remainder when is divided by .
Solution 1 (long but non-bashy)
Note that the given equality reduces to
Now, let and likewise for . Consider circle with the origin as the center and radius 2014 on the complex plane. It is clear that must be one of the points on this circle, as .
By DeMoivre's Theorem, the complex modulus of is cubed when is cubed. Thus must lie on , since its the cube of its modulus, and thus its modulus, must be equal to 's modulus.
Again, by DeMoivre's Theorem, is tripled when is cubed and likewise for . For , , and the origin to lie on the same line, must be some multiple of 360 degrees apart from , so must differ from by some multiple of 120 degrees.
Now, without loss of generality, assume that is on the real axis. (The circle can be rotated to put in any other location.) Then there are precisely two possible distinct locations for ; one is obtained by going 120 degrees clockwise from about the circle and the other by moving the same amount counter-clockwise. Moving along the circle with any other multiple of 120 degrees in any direction will result in these three points.
Let the two possible locations for be and and the location of be point . Note that by symmetry, is equilateral, say, with side length . We know that the circumradius of this equilateral triangle is , so using the formula and that the area of an equilateral triangle with side length is , so we have
Since we're concerned with the non-radical part of this expression and ,
and we are done.
Solution 2 (short)
Assume . Then
Thus is an isosceles triangle with area and
Solution 3
Notice that Hence, , and is an equilateral triangle with circumradius . Then, and the answer is .
Solution 4 (Slick)
I find that generally, the trick to these kinds of AIME problems is to interpret the problem geometrically, and that is just what I did here. Looking at the initial equation, this seems like a difficult task, but rearranging yields a nicer equation: We can interpret the difference of two complex numbers as a vector from one to the other, and we can interpret the quotient as a vector with an angle equal to the angle between the two vectors. Therefore, after labeling the complex numbers with (), (), and (), we can interpret the above equation to mean that the , and hence triangle is isosceles, so length = . Rearranging the equation we find that Taking the magnitude of both sides and using the fact that , we find that so length and triangle is equilateral. There are only two possible 's for which is equilateral, lying on either side of . After drawing these points on the circle of radius 2014 centered at the origin, it is easy to see that and the two 's form an equilateral triangle (this can be verified by simple angle chasing). Drawing a perpendicular bisector of one of the sides and using 30-60-90 triangles shows that the side length of this triangle is and hence its area is for some integer .
~SymbolicPermutation
Notes
This problem is killed by polar coordinates (complex numbers) . Solve using cosine-i-sine.
See also
2014 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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