If two of the vertices of an equilateral triangle are $(0,0)$ and $(3, \sqrt{3}),$ find the coordinates of the third vertex.

(A) Let the vertices of the equilateral triangle be $A(0,0)$,$B(3, \sqrt{3})$,and $C(x, y)$.
Since it is an equilateral triangle,$AB = BC = AC$,which implies $AB^2 = BC^2 = AC^2$.
First,calculate $AB^2 = (3-0)^2 + (\sqrt{3}-0)^2 = 9 + 3 = 12$.
Since $AC^2 = AB^2$,we have $x^2 + y^2 = 12$ (Equation $1$).
Since $BC^2 = AB^2$,we have $(x-3)^2 + (y-\sqrt{3})^2 = 12$.
Expanding this: $x^2 - 6x + 9 + y^2 - 2\sqrt{3}y + 3 = 12$.
Substituting $x^2 + y^2 = 12$ into this equation: $12 - 6x - 2\sqrt{3}y + 12 = 12$,which simplifies to $6x + 2\sqrt{3}y = 12$,or $y = \sqrt{3}(2-x)$ (Equation $2$).
Substituting Equation $2$ into Equation $1$: $x^2 + [\sqrt{3}(2-x)]^2 = 12$.
$x^2 + 3(4 - 4x + x^2) = 12$.
$x^2 + 12 - 12x + 3x^2 = 12$.
$4x^2 - 12x = 0$,so $4x(x-3) = 0$.
This gives $x = 0$ or $x = 3$.
If $x = 0$,$y = \sqrt{3}(2-0) = 2\sqrt{3}$.
If $x = 3$,$y = \sqrt{3}(2-3) = -\sqrt{3}$.
Thus,the third vertex is $(0, 2\sqrt{3})$ or $(3, -\sqrt{3})$.