2014 AIME II Problems/Problem 10
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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 but a little bashy)
Assume . Then
Thus is an isosceles triangle with area and
See also
2014 AIME II (Problems • Answer Key • Resources) | ||
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Followed by Problem 11 | |
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