**Problem**: (2006 Putnam - B1) Show that the curve $x^{3}+3xy+y^{3}=1$ contains only one set of three distinct points, $A$, $B$, and $C$, 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 $ x $. Then

$ x^3+3xy+(y^3-1) = 0 $.

Well, testing out a bit (looking at the factorization of $ y^3-1 $), we get $ x = 1-y $ is always a solution, so factor it out to get

$ (x+y-1)[x^2+x(1-y)+(y^2+y+1)] = 0 $.

Looking at the second term, we're like let's hope it has not many solutions, so we take the discriminant and find

$ (1-y)^2-4(y^2+y+1) = -3y^2-6y-3 = -3(y+1)^2 $.

But this is always negative unless $ y = -1 $ so that means this is the only case in which this factor can be zero. This gives us the point $ (-1,-1) $ as a solution. Whoa, that means we have categorized the entire solution set:

(1) the line $ x+y = 1 $ and (2) the point $ (-1,-1) $.

If we have three vertices of an equilateral triangle, they most definitely can't be collinear, so one point must be $ (-1,-1) $. Suppose we choose two points on the line $ x+y = 1 $, say $ (x_0, 1-x_0) $ and $ (x_1, 1-x_1) $. Since they have to be equidistant from $ (-1, -1) $, we know one must be the reflection of the other over the line through $ (-1, -1) $ perpendicular to $ x+y = 1 $, which is $ x = y $. So if the first point is $ (x_0, 1-x_0) $ then the other is $ (1-x_0, x_0) $.

Now setting the squares of the side lengths equal to each other, we know

$ 2(2x_0-1)^2 = (x_0+1)^2+(2-x_0)^2 $

$ 2x_0^2-2x_0-1 = 0 $ so $ x_0 = \frac{1}{2} \pm \frac{\sqrt{3}}{2} $.

This gives the $ x $-coordinate of both the other two vertices of the triangle, so the only one is the equilateral triangle with vertices

$ (-1, -1); \left(\frac{1}{2}+\frac{\sqrt{3}}{2}, \frac{1}{2}-\frac{\sqrt{3}}{2}\right); \left(\frac{1}{2}-\frac{\sqrt{3}}{2}, \frac{1}{2}+\frac{\sqrt{3}}{2}\right) $.

The side length is $ \sqrt{6} $, so the area is $ \frac{6\sqrt{3}}{4} = \frac{3\sqrt{3}}{2} $. QED.

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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.

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Practice Problem: (2006 Putnam - A1) Find the volume of the region of points $ (x,y,z) $ such that

$ (x^2+y^2+z^2+8)^2 \le 36(x^2+y^2) $.

you dont even need to find the coordinates, once you know one vertex is (-1,-1) and the other two lie on line y=1-x, you are done. since point (1/2,1/2) must be half way between the the unkown vertecies.

ReplyDeletefor the practice problem, substitude w^2 for x^2+y^2 and you get a circle, and its easy from there.

this year's putnam was weired.

funny thing,

ReplyDeletewhen i saw this problem (B1), since i had eaten lots of thai food for lunch, there was not much blood flow to my brain, so i spend like 20 mins and just screwed around and getting no where,

it was until i finished eating an orange, that my brain started working again, i did B2 and B3 first, before i realized how to do B1.

i highly recomend eating Oranges during putnam; who knows, maybe if i had eaten a box of oranges during usamo last year i would have won it....darn....

lol.

Well I figured that finding the points was the best justification as to why only one equilateral triangle exists, though the argument can easily be proven rigorously in other ways as well I suppose.

ReplyDeleteHaha, I think B2 was easier than B1, because it was classic pigeonhole. B1 required a little intuitive factoring. I'll have a couple more Putnam problems here in the next couple of days =).