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

Revision as of 11:17, 2 March 2014

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 $25 + 12 + 3 = \boxed{040}$ .

Solution 3

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 $25 + 12 + 3 = \boxed{040}$ .

@@ Line 7: / Line 7: @@
 ==Solution 2==
+Rather than considering the Cartesian plane, we use complex numbers. Thus A is 1 and B is <math>2 + 2\sqrt{3}i</math>.
+Recall that a rotation of <math>\theta</math> radians counterclockwise is equivalent to multiplying a complex number by <math>e^{i\theta}</math>, but here we require a clockwise rotation, so we multiply by <math>e^{-\frac{i\pi}{3}}</math> to obtain C. Upon averaging the coordinates of A, B, and C, we obtain the coordinates of P, viz. <math>\left(\frac{5}{2}, \frac{5\sqrt{3}}{6}\right)</math>.
+Therefore <math>xy</math> is <math>\frac{25\sqrt{3}}{12}</math> and the answer is <math>25 + 12 + 3 = \boxed{040}</math>.
+==Solution 3==
 Rather than considering the Cartesian plane, we use complex numbers. Thus A is 1 and B is <math>2 + 2\sqrt{3}i</math>.

2013 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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