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Difference between revisions of "2013 AIME II Problems/Problem 4"
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==Solution==
 
==Solution==
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The distance from point <math>A</math> to point <math>B</math> is <math> \sqrt{13}</math>. The vector that starts at point A and ends at point B is given by <math>B - A = (1, 2\sqrt{3})</math>. Since the center of an equilateral triangle, <math>P</math>, 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 <math>\overline{AB}</math>. The line perpendicular to <math>\overline{AB}</math> through the midpoint, <math>M =  (\dfrac{3}{2},\sqrt{3})</math>, <math>\overline{AB}</math> can be parameterized by <math> (\dfrac{2\sqrt{3}}{\sqrt{13}}, \dfrac{-1}{\sqrt{13}}) t + (\dfrac{3}{2},\sqrt{3})</math>. At this point, it is useful to note that <math>\Delta BMP</math> is a 30-60-90 triangle with <math>\overline{MB}</math> measuring <math>\dfrac{\sqrt{13}}{2}</math>. This yields the lenght of <math>\overline{MP}</math> to be <math>\dfrac{\sqrt{13}}{2\sqrt{3}}</math>. Therefore, <math>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}})</math>. Therefore <math>xy = \dfrac{25\sqrt{3}}{12}</math> yielding an answer of <math> p + q + r  = 25 + 3 + 12 = 040</math>.
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The distance from point <math>A</math> to point <math>B</math> is <math> \sqrt{13}</math>. The vector that starts at point A and ends at point B is given by <math>B - A = (1, 2\sqrt{3})</math>. Since the center of an equilateral triangle, <math>P</math>, 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 <math>\overline{AB}</math>. The line perpendicular to <math>\overline{AB}</math> through the midpoint, <math>M =  (\dfrac{3}{2},\sqrt{3})</math>, <math>\overline{AB}</math> can be parameterized by <math> (\dfrac{2\sqrt{3}}{\sqrt{13}}, \dfrac{-1}{\sqrt{13}}) t + (\dfrac{3}{2},\sqrt{3})</math>. At this point, it is useful to note that <math>\Delta BMP</math> is a 30-60-90 triangle with <math>\overline{MB}</math> measuring <math>\dfrac{\sqrt{13}}{2}</math>. This yields the lenght of <math>\overline{MP}</math> to be <math>\dfrac{\sqrt{13}}{2\sqrt{3}}</math>. Therefore, <math>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}})</math>. Therefore <math>xy = \dfrac{25\sqrt{3}}{12}</math> yielding an answer of <math> p + q + r  = 25 + 3 + 12 = \boxed{040}</math>.

Revision as of 15:49, 4 April 2013

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

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 lenght 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 $p + q + r = 25 + 3 + 12 = \boxed{040}$.

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