Difference between revisions of "2013 AIME II Problems/Problem 4"
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In the Cartesian plane let <math>A = (1,0)</math> and <math>B = \left( 2, 2\sqrt{3} \right)</math>. Equilateral triangle <math>ABC</math> is constructed so that <math>C</math> lies in the first quadrant. Let <math>P=(x,y)</math> be the center of <math>\triangle ABC</math>. Then <math>x \cdot y</math> can be written as <math>\tfrac{p\sqrt{q}}{r}</math>, where <math>p</math> and <math>r</math> are relatively prime positive integers and <math>q</math> is an integer that is not divisible by the square of any prime. Find <math>p+q+r</math>. | In the Cartesian plane let <math>A = (1,0)</math> and <math>B = \left( 2, 2\sqrt{3} \right)</math>. Equilateral triangle <math>ABC</math> is constructed so that <math>C</math> lies in the first quadrant. Let <math>P=(x,y)</math> be the center of <math>\triangle ABC</math>. Then <math>x \cdot y</math> can be written as <math>\tfrac{p\sqrt{q}}{r}</math>, where <math>p</math> and <math>r</math> are relatively prime positive integers and <math>q</math> is an integer that is not divisible by the square of any prime. Find <math>p+q+r</math>. | ||
Revision as of 16:50, 6 April 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 |