2013 AIME II Problems/Problem 4
Contents
Problem 4
In the Cartesian plane let and . Equilateral triangle is constructed so that lies in the first quadrant. Let be the center of . Then can be written as , where and are relatively prime positive integers and is an integer that is not divisible by the square of any prime. Find .
Solution 1
The distance from point to point is . The vector that starts at point A and ends at point B is given by . Since the center of an equilateral triangle, , is also the intersection of the perpendicular bisectors of the sides of the triangle, we need first find the equation for the perpendicular bisector to . The line perpendicular to through the midpoint, , can be parameterized by . At this point, it is useful to note that is a 30-60-90 triangle with measuring . This yields the length of to be . Therefore, . Therefore yielding an answer of .
Solution 2
Rather than considering the Cartesian plane, we use complex numbers. Thus A is 1 and B is .
Recall that a rotation of radians counterclockwise is equivalent to multiplying a complex number by , but here we require a clockwise rotation, so we multiply by to obtain C. Upon averaging the coordinates of A, B, and C, we obtain the coordinates of P, viz. .
Therefore is and the answer is .
See Also
2013 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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