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Prove that if general equation $$\cos \left(3\alpha\right)\left(x^3 -3xy^2\right) + \sin3\alpha\left(y^3 - 3x^2y\right) + 3a\left(x^2 + y^2\right) - 4a^3 = 0$$ represents three lines then they form a equilateral triangle of area $3a^2\sqrt 3$.

This equation looks like an good candidate to solve in polar coordinates than in cartesian coordinates.

Converting it into polar coordinates by parts \begin{align} \left(x^3 -3xy^2\right) &= r^3 \left(\cos^3 \theta - 3\cos \theta \sin^2 \theta\right) =r^3 \left(4\cos^3 \theta - 3\cos \theta\right) = r^3\cos 3\theta \tag1\label1 \\ \left(y^3 - 3x^2y\right) &= r^3\left(4\sin^3 \theta - 3\sin\theta\right) = r^3\left(\cos\left(3\pi/2 - 3\theta\right)\right) = -r^3\sin 3\theta\tag2\label2 \end{align}

Using $\eqref1$, $\eqref2$ \begin{align} r^3\left(\cos 3\alpha\cos 3\theta - \sin3\alpha\sin 3\theta\right) + 3ar^2 - 4a^3 &= 0 \\ r^3\cos \left(3\alpha+ 3\theta\right) + 3ar^2 - 4a^3 &= 0 \end{align}

On dividing by $a^3$ and substituting $z = r/a$ $$z^3\cos \left(3\alpha+ 3\theta\right) + 3z^2 - 4 = 0.$$

Now this happens very often to me, I am not able to solve this cubic for $z$.I did try to find the factor by putting $\,z = \sin\left(3\alpha + 3\theta\right),\,\tan\left(3\alpha + 3\theta\right),\,\ldots$

What should I do now ? how to solve this cubic ?

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    You might be overinterpreting the goal of the exercise, since it asks if the *assumption* that the equation "represents three lines" implies the formation of an equilateral triangle (of specified area). While the substitution of polar coordinates seems attractive, leading in your work to a polynomial in $r/a$, I don't think you can expect to *prove* the general equation always represents three lines, nor to "solve this cubic for $a$". – 2017-02-21
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    Close vote for what ? Why does ones inability to understand the question should be blamed on the question ? I think my question is very clear, please tell me if you want to rectify anything in the question. – 2017-02-21
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    @hardmath No you are misinterpreting my attempt. What I did was to get the three lines that constitute the given general equation then I will use those line equations to prove that they form a equilateral triangle. – 2017-02-21
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    @hardmath If I am able to solve the cubic, I would be able to get those line equations. If I had already got those then what is the need of this question ? – 2017-02-21
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    Let me try to make the point as simply as possible. Neither the "general equation" nor your rewrite in polar coordinates can be solved "for $a$". Considered as a polynomial in $x,y,a$, the "general equation" is homogeneous of degree $3$, so one naturally expects that the area of any (finite) region bounded by the curve(s) will *scale* with $a$ (which is the gist of the last part of the exercise). – 2017-02-21
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    @hardmath No it is not homogeneous. I maybe wrong but the powers don't add upto to 3 for all terms. If you are correct then what should I answer to this question ? – 2017-02-21
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    Maybe my earlier comment about your polar coordinate version is easier to check by inspection. It can be rewritten as a polynomial in $r/a$. In any case for the Cartesian coordinates we check the total degree in $x,y,a$ of all terms. – 2017-02-21
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    @hardmath Yes I will edit my question now to include this. still it is a 3rd degree polynomial, I still need to solve it. $a$ is a constant, why we include $a$'s power in check of homogenity ? – 2017-02-21

3 Answers 3

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The given equation always represents three lines. When we rotate the axes by an angle $\alpha$, in polar coordinates, $r$ remains unchanged but $\theta$ changes to $\theta - \alpha$. Thus the combined equation becomes $$r^3\cos \left(3\alpha+ 3(\theta-\alpha)\right) + 3ar^2 - 4a^3 = 0$$ This is equivalent to rotating the curve clockwise by $\alpha$ about the origin and thus the shape of the curve does not change. Hence it is enough to prove that the above equation represents three lines. We have \begin{align*} r^3\cos3\theta +3ar^2 - 4a^3 &= 0\\ r^3(4\cos^3 \theta -3\cos\theta) + 3ar^2 - 4a^3 &= 0\\ 4x^3 -3x(x^2+y^2)+3a(x^2+y^2) - 4a^3 &= 0\\ x^3 - 3xy^2 + 3ax^2 + 3ay^2 - 4a^3 &= 0 \end{align*} where in the last two steps, we have converted the equation back to Cartesian. Clearly $x=a$ is a factor and dividing this out we get $$(x-a)(x^2 - 3y^2 + 4ax + 4a^2) = 0$$ and hence $$(x-a)((x+2a)^2 - 3y^2) = 0$$ Thus $$(x-a)(x+2a-\sqrt{3}y)(x+2a +\sqrt{3}y) = 0$$ Now it is easy to see that the circum center of this triangle is the origin and the circum radius is $2a$. Hence the area is $3\sqrt{3}a^2$

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    "Now it is easy to see that the circm center of this triangle is the origin and the circum radius is 2a" how ?? – 2017-02-22
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    The base is $x=a$ and the other vertex is $(-2a,0)$. The altitude through this vertex is the $x$ axis. Since the triangle is equilateral, centroid coincides with the circumcenter. Centroid divides the altitude in the ratio $2:1$ and hence is at the origin. – 2017-02-22
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Real and imaginary parts of $ (x+iy)^3$ can be recognized here.

EDIT1:

Factorization given below tallies fully with the first general equation in OP's post.

$$ (x \cos(\alpha) - y\sin (\alpha) -a)* (x \cos(\alpha + 2 \pi/3) - y\sin (\alpha + 2 \pi/3) -a) * (x \cos(\alpha + 4 \pi/3) - y\sin (\alpha + 4 \pi/3) -a)=0\, \tag1 $$

Area = $ \dfrac { \sqrt3 }{4} \cdot { { (2 \sqrt3 a})^2} = 3 \sqrt3 a^2$

The following is a plot of three sides for $ \alpha = 12.345, a=1. $

BTW, such factorizations are useful in obtaining torsion in non-circular shafts using St. Venant's theory.

3EquilTRIA

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    These are not even lines and I know what went wrong here. 1) : My equation is not homogeneous so it **may or may not** represent a triplet of lines. 2) : It will represent a triplet of lines for only special values of $a, \alpha$. – 2017-02-21
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    Then please delete, correct it and undelete so others do not spend time trying to prove it. – 2017-02-21
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    Done $\phantom{ \ \ \ \ \ \ \ \ \ } $ – 2017-02-21
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    The given equation always represents three lines. I will post a detailed solution. – 2017-02-21
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    @ A---B After a night's sleep :( found and removed a small error in plot. Rest of factorization tallies fully without considering it as any special case $\alpha=0$ by @Muralidharan. – 2017-02-22
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The solutions are $r = a, \theta = -\alpha$, $r = a, \theta = \frac{2\pi}{3} - \alpha$ and $r = a, \theta = \frac{4\pi}{3} - \alpha$

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    How did you get these ? – 2017-02-21
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    It is clear now. @Muralidharan rotated it to make first line parallel to $y$ axis. – 2017-02-22