2013 AIME II Problem 4

Question

In the Cartesian plane let

A = (1, 0)

and

B = (2, 2\sqrt{3}).

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

\frac{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.

Answer

C

is the midpoint of

\overline{AB}

shifted perpendicularly by

\frac{\sqrt{3}}{2}

times the length of

\overline{AB},

and

P

is the centroid

A unit vector perpendicular to

\overrightarrow{AB} = (1, 2\sqrt{3})

is

\frac{1}{\sqrt{13}}(2\sqrt{3}, -1);

average the three vertices to get

P

The midpoint of

\overline{AB}

is

M = \left(\frac{3}{2}, \sqrt{3}\right),

and

AB = \sqrt{1 + 12} = \sqrt{13}.

The third vertex lies at distance

\frac{\sqrt{3}}{2}\sqrt{13}

from

M

along a direction perpendicular to

\overrightarrow{AB} = (1, 2\sqrt{3});

a unit perpendicular is

\frac{1}{\sqrt{13}}(2\sqrt{3}, -1).

Taking the sign that lands in the first quadrant,

\begin{aligned} C &= M + \frac{\sqrt{3}}{2}\,(2\sqrt{3}, -1) \\ &= \left(\frac{9}{2}, \frac{\sqrt{3}}{2}\right) \end{aligned}

(the other choice has negative

x

-coordinate).

The center of an equilateral triangle is its centroid, the average of the vertices:

\begin{aligned} P &= \small \left(\frac{1 + 2 + \frac{9}{2}}{3},\ \frac{0 + 2\sqrt{3} + \frac{\sqrt{3}}{2}}{3} \right) \\ &= \left(\frac{5}{2}, \frac{5\sqrt{3}}{6}\right). \end{aligned}

Then

x \cdot y = \frac{5}{2} \cdot \frac{5\sqrt{3}}{6} = \frac{25\sqrt{3}}{12},

so

p + q + r = 25 + 3 + 12 = 40.

Problem 4 in other years