# Equilateral triangle

1. Jan 13, 2008

### ehrenfest

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

3. The attempt at a solution
I randomly starting plotting points and found that all of them fell on the line y=1-x except (-1,-1). So, I just need to prove that these are the only points that satisfy that equation. It turns out that that equation factors into a very useful form. See B1 http://www.unl.edu/amc/a-activities/a7-problems/putnam/-pdf/2006s.pdf [Broken]
My question is how would you "discover" that really nice form if you were taking the test?

Last edited by a moderator: May 3, 2017
2. Jan 14, 2008

### Shooting Star

The first insticnt would be to try to factorise.

x^3 + y^3 +3xy -1
= (x+y)^3 -3x^2y -3xy^2 +3xy -1
= [(x+y)^3 -1] -3xy(x+y-1)
= [(x+y-1){(x+y)^2 +(x+y) +1}] -3xy(x+y-1)

Now you can take (x+y-1) common and the rest would follow.
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p.s.

If a simpler method occurs to me, I'll immdtly let you know. By inspection, some roots can be found to be (-1,-1), (-1,2) and (2,-1). It's not derivable at (-1,-1). If we plot the three points, we can sense some trouble at (-1,-1).