# 2006 B1

#### Problem

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.

#### Discussion

A first idea is to try to visualize what this set should look like. The intuition is that curves of this form generally look something like this:

Obviously for a Putnam B1, they don't expect you to analyze curves this complex, so there is probably some convoluted identity behind this problem. We proceed with this in the back of our minds.

With that in mind, we start trying to manipulate. Seeing the $3xy$ term and also thinking about the similarly-looking equation \[(x+y)^3=x^3+3x^2y+3xy^2+y^3\] we might want to complete the cube. We get \[(x+y)^3=1+3x^2y+3xy^2-3xy=1+3xy(x+y-1)\] where we have grouped similar looking terms on the right-hand side. From here, seeing the lingering $1$ on the right-hand side, we think to subtract the $1$ to the other side because maybe it will factor. It turns out it does! We get \[3xy(x+y-1)=(x+y)^3-1=(x+y-1)((x+y)^2+(x+y)+1)\] From here, we have \[x+y-1=0\quad\text{or}\quad3xy=(x+y)^2+(x+y)+1\] But now what we have on the right is something we're happy about - a quadratic! We employ the quadratic formula to solve for $y$ in terms of $x$: \[3xy=(x+y)^2+(x+y)+1\implies0=y^2+(1-x)y+(x^2+x+1)\] or \[y=\frac{(x-1)\pm\sqrt{(1-x)^2-4(x^2+x+1)}}2=\frac{(x-1)\pm\sqrt{-3x^2-6x-1}}2=\frac{(x-1)\pm\sqrt{-3(x+1)^2}}2\] The only way we can get a real solution is if $x=-1$, because otherwise the thing under the square root would be negative! From $x=-1$, we get $y=-1$. Therefore, this curve consists of two parts: the set where $x+y=1$ or the point $(-1,-1)$:

Now we get to the geometry part of the problem. The three points can't all be part of the line. Therefore, one of the points is $(-1,-1)$. Also, the height of the equilateral triangle is the distance from the point $(-1,-1)$ to the line $x+y=1$. But we already know by symmetry that the altitude from $(-1,-1)$ to that line is the point $\left(\frac12,\frac12\right)$ (and the height of this altitude is \[\sqrt{\left(\frac12-(-1)\right)^2+\left(\frac12-(-1)\right)^2}=\frac3{\sqrt2}\] Using $\tan30^\circ=\frac1{\sqrt3}$, we get that the other two points of the equilateral triangle are the two points at distance $\frac3{\sqrt6}$ away from $\left(\frac12,\frac12\right)$ on either side. Thus, the triangle is unique, and its area is \[\frac12\cdot\text{base}\cdot\text{height}=\frac12\cdot\frac3{\sqrt2}\cdot\frac3{\sqrt6}=\boxed{\frac{3\sqrt3}4}\] With that, we have solved the problem.