Putnam 2006, Problem B-1

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

Solution:
The "curve" $x^3 + 3xy + y^3 - 1 = 0$ is actually reducible, because the left side factors as
$$(x + y - 1)(x^2 - xy + y^2 + x + y + 1).$$
Moreover, the second factor is
$$\frac{1}{2}((x+1)^2 + (y+1)^2 + (x-y)^2),$$
so it only vanishes at $(-1, -1)$. Thus the curve in question consists of the single point $(-1, -1)$ together with the line $x + y = 1$. To form a triangle with three points on this curve, one of its vertices must be $(-1, -1)$. The other two vertices lie on the line $x + y = 1$, so the length of the altitude from $(-1, -1)$ is the distance from $(-1, -1)$ to $(1/2, 1/2)$, or $3\sqrt{2}/2$. The area of an equilateral triangle of height $h$ is $h^2\sqrt{3}/3$, so the desired area is $3\sqrt{3}/2$.

**Remark:** The factorization used above is a special case of the fact that
$$x^3 + y^3 + z^3 - 3xyz$$
$$= (x + y + z)(x + \omega y + \omega^2 z)(x + \omega^2 y + \omega z),$$
where $\omega$ denotes a primitive cube root of unity. That fact in turn follows from the evaluation of the determinant of the circulant matrix
$$\begin{pmatrix} x & y & z \\ z & x & y \\ y & z & x \end{pmatrix}$$
by reading off the eigenvalues of the eigenvectors $(1, \omega^i, \omega^{2i})$ for $i = 0, 1, 2$.