Problem: (2006 Putnam – B1) Show that the curve contains only one set of three distinct points, , , and , which are the vertices of an equilateral triangle, and find its area.
Solution: Well, this is a cubic in two variables, but let’s remember the awesome technique of writing it as a polynomial in just one variable, say . Then
Well, testing out a bit (looking at the factorization of ), we get is always a solution, so factor it out to get
Looking at the second term, we’re like let’s hope it has not many solutions, so we take the discriminant and find
But this is always negative unless so that means this is the only case in which this factor can be zero. This gives us the point as a solution. Whoa, that means we have categorized the entire solution set:
(1) the line and (2) the point .
If we have three vertices of an equilateral triangle, they most definitely can’t be collinear, so one point must be . Suppose we choose two points on the line , say and . Since they have to be equidistant from , we know one must be the reflection of the other over the line through perpendicular to , which is . So if the first point is then the other is .
Now setting the squares of the side lengths equal to each other, we know
This gives the -coordinate of both the other two vertices of the triangle, so the only one is the equilateral triangle with vertices
The side length is , so the area is . QED.
Comment: Slightly difficult if your algebraic intuition wasn’t working well, but after you realized the factorization it wasn’t hard to convince yourself that a line and a point can have the vertices of at most one equilateral triangle. A solid B1 problem on the Putnam, quite a bit more difficult than the A1, imo.
Practice Problem: (2006 Putnam – A1) Find the volume of the region of points such that