Find The Area Of The Equilateral Triangle

Show that the curve $x^3+3xy+y^3=1$ has only one set of three distinct points, $P$, $Q$, and $R$ which are the vertices of an equilateral triangle, and find its area.

My solution:

The first thing I notice is that there is cyclic symmetry between $x$ and $y$, and so setting $y=x$, we find:

$\displaystyle 2x^3+3x^2-1=(x+1)^2(2x-1)=0$

Thus, we know the points:

$\displaystyle (x,y)=(-1,-1),\,\left(\frac{1}{2},\frac{1}{2} \right)$

are on the given curve. Next, if we begin with the line:

$\displaystyle y=1-x$

and cube both sides, we obtain:

$\displaystyle y^3=1-3x+3x^2-x^3$

We may arrange this as:

$\displaystyle x^3+3x(1-x)+y^3=1$

Since $y=1-x$, we may now write

$\displaystyle x^3+3xy+y^3=1$

And since the point $\displaystyle \left(\frac{1}{2},\frac{1}{2} \right)$ is on the line $y=1-x$, we know the locus of the given curve is the line $y=1-x$ and the point $(-1,-1)$. Hence, there can only be one set of points on the given curve that are the vertices of any triangle, equilateral or otherwise.

Using the formula for the distance between a point and a line, we find the altitude of the equilateral triangle will be:

$\displaystyle h=\frac{|(-1)(-1)+1-(-1)|}{\sqrt{(-1)^2+1}}=\frac{3}{\sqrt{2}}$

Using the Pythagorean theorem, we find that the side lengths of the triangle must be:

$\displaystyle s=\frac{2}{\sqrt{3}}h=\sqrt{6}$

And so the area of the triangle is:

$\displaystyle A=\frac{1}{2}sh=\frac{1}{2}\sqrt{6}\frac{3}{\sqrt{2}}=\frac{3\sqrt{3}}{2}$