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Difference between revisions of "2013 AIME II Problems/Problem 4"

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==Problem 4==
 
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 15:50, 6 April 2013

Problem 4

In the Cartesian plane let $A = (1,0)$ and $B = \left( 2, 2\sqrt{3} \right)$. Equilateral triangle $ABC$ is constructed so that $C$ lies in the first quadrant. Let $P=(x,y)$ be the center of $\triangle ABC$. Then $x \cdot y$ can be written as $\tfrac{p\sqrt{q}}{r}$, where $p$ and $r$ are relatively prime positive integers and $q$ is an integer that is not divisible by the square of any prime. Find $p+q+r$.

Solution 1

The distance from point $A$ to point $B$ is $\sqrt{13}$. The vector that starts at point A and ends at point B is given by $B - A = (1, 2\sqrt{3})$. Since the center of an equilateral triangle, $P$, 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 $\overline{AB}$. The line perpendicular to $\overline{AB}$ through the midpoint, $M = (\dfrac{3}{2},\sqrt{3})$, $\overline{AB}$ can be parameterized by $(\dfrac{2\sqrt{3}}{\sqrt{13}}, \dfrac{-1}{\sqrt{13}}) t + (\dfrac{3}{2},\sqrt{3})$. At this point, it is useful to note that $\Delta BMP$ is a 30-60-90 triangle with $\overline{MB}$ measuring $\dfrac{\sqrt{13}}{2}$. This yields the length of $\overline{MP}$ to be $\dfrac{\sqrt{13}}{2\sqrt{3}}$. Therefore, $P =( \dfrac{2\sqrt{3}}{\sqrt{13}},\dfrac{-1}{\sqrt{13}})(\dfrac{\sqrt{13}}{2\sqrt{3}}) + (\dfrac{3}{2},\sqrt{3}) = (\dfrac{5}{2}, \dfrac{5}{2\sqrt{3}})$. Therefore $xy = \dfrac{25\sqrt{3}}{12}$ yielding an answer of $p + q + r = 25 + 3 + 12 = \boxed{040}$.


Solution 2

Rather than considering the Cartesian plane, we use complex numbers. Thus A is 1 and B is $2 + 2\sqrt{3}i$.

Recall that a rotation of $\theta$ radians counterclockwise is equivalent to multiplying a complex number by $e^{i\theta}$, but here we require a clockwise rotation, so we multiply by $e^{-\frac{i\pi}{3}}$ to obtain C. Upon averaging the coordinates of A, B, and C, we obtain the coordinates of P, viz. $\left(\frac{5}{2}, \frac{5\sqrt{3}}{6}\right)$.

Therefore $xy$ is $\frac{25\sqrt{3}}{12}$ and the answer is $25 + 12 + 3 = \boxed{040}$.

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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