Thursday, April 23, 2015

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:


Thus, we know the points:

[math](x,y)=(-1,-1),\,\left(\frac{1}{2},\frac{1}{2} \right)[/math]

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


and cube both sides, we obtain:


We may arrange this as:


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


And since the point [math]\left(\frac{1}{2},\frac{1}{2} \right)[/math] 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:


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


And so the area of the triangle is:


1 comment:

  1. Great job on how you did this problem Mark! Thanks for sharing it on your Google+. -M