2013 AIME II Problems/Problem 4
Contents
Problem
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 .
Solution 3
We can also consider the slopes of the lines. Midpoint of has coordinates . Because line has slope , the slope of line is (Because of perpendicular slopes).
Since is equilateral, and since point is the centroid, we can quickly calculate that . Then, define and to be the differences between points and . Because of the slope, it is clear that .
We can then use the Pythagorean Theorem on line segment : yields and , after substituting . The coordinates of P are thus . Multiplying these together gives us , giving us as our answer.
Solution 4
Since will be segment rotated clockwise , we can use a rotation matrix to find . We first translate the triangle unit to the left, so lies on the origin, and . Rotating clockwise is the same as rotating counter-clockwise, so our rotation matrix is . Then . Thus, . Since the triangle is equilateral, the center of the triangle is the average of the coordinates of the vertices. Then . Our answer is .
Solution 5
We construct point by drawing two circles with radius . One circle will be centered at , while the other is centered at . The equations of the circles are:
Setting the LHS of each of these equations equal to each other and solving for yields after simplification:
Plugging that into the first equation gives the following quadratic in after simplification:
The quadratic formula gives .
Since and , we pick in the hopes that it will give . Plugging into the equation for yields .
Thus, . Averaging the coordinates of the vertices of equilateral triangle will give the center of mass of the triangle.
Thus, , and the product of the coordinates is , so the desired quantity is .
Solution 6
Labeling our points and sketching a graph we get that is to the right of . Of course, we need to find . Note that the transformation from to is , and if we imagine a height dropped to we see that a transformation from the midpoint to is basically the first transformation, with the magnitude and the x and y switched– then multiply the new y by -1. Then, applying this transformation of we get that which means that . Then our answer is .
Solution 7
Transform this into the complex plane and let . We know that 3 complex numbers form an equilateral triangle iff , so plugging in our values of , we get Solving for using Wolfram Alpha, we find that the solutions are . The first one is in the first quadrant, so . The center is the average of the coordinates and we find that it is . Then .
-bobthegod78, krwang, and Simplest14
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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