Difference between revisions of "2013 AIME II Problems/Problem 4"
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Revision as of 13:52, 4 July 2013
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
