Why do equilateral triangles relate to cubics

Question

I found this question talking about the relation between an equilateral triangle and cubics with three distinct real roots.

Here's an image from the original post with an example:

What this post says about cubic curves is:

You can always construct an eq. triangle such that:

• the triangle's inscribed circle's center lines up with the point of inflection on the curve
• the function's min. and max. points line up with the inscribed circle's left- and rightmost points (not necessarily tangent to the sides)
• and the triangle's vertices line up with the roots. (as seen on the image)

You can use these facts to solve a cubic of this kind using with the triple angle cosine identity, but my question is not about that.

My question would be about how you can prove this is construction is always possible.

(1.) I started using the fact that cubic curves have point symmetry on the inflection point, thus the min. and max. points are equally distanced from the poi. on the x-axis. This defines a circle.

(2.) Then, if we draw another circle with the same center but twice the radius, we can construct a specific eq. triangle in between the two circles such that the two are the inscribed and circumscribed circles of that triangle.

(3. ??) We have infinitely many ways of rotating this, but for some reason, one of them will line up such that the vertices are in line with the roots. I got stuck here.

Thanks for any help and please explain in a very simple manner.

Answer 1

Let's start with an equilateral triangle, such that one side is parallel to the $x$-axis. The center of the triangle is at $x_c$ along the $x$ axis. The circumscribed circle radius is $R$. The $y_c$ position does not matter, but I will use $y_c=0$ in my example. You can offset it by any value. Now let's rotate the triangle clockwise by an angle $\phi$. The vertex on top, originally at $(x_c,R)$ will have the new $x$ coordinate $x_c+R\sin\phi$. For the vertex on the left, originally at $(x_c-R\cos\frac\pi6,-R\sin\frac\pi6)$, the new $x$ coordinate will be $x_c-R\cos\left(\frac\pi6-\phi\right)$, and for the vertex on the right the new coordinate will be $x_c+ R\cos\left(\frac\pi6+\phi\right)$.

So now let's identify these coordinates with the roots of the cubic. We assume that they are distinct, and in increasing order $a,b,c$. We need to find the $x_c$ coordinate,$R$, and the angle $\phi$. Note that for the equilateral triangle $$x_c=\frac{a+b+c}3$$ The equation of the quadratic (up to a multiplicative factor) will be $$y=(x-a)(x-b)(x-c)=x^3-(a+b+c)x^2+(ab+bc+ca)x-abc$$ Let's calculate the first and second derivatives: $$y'=3x^2-2(a+b+c)x+(ab+bc+ca)\\y''=6x-2(a+b+c)$$ Let's start with the inflection point, $y''=0$ corresponding to $$x_i=\frac{a+b+c}3=x_c$$ Since $y'$ is a quadratic, its roots are symmetric with respect to its vertex (at $x_i$), and the distance between the roots will be $$\frac{\sqrt{2^2(a+b+c)^2-4\cdot3\cdot(ab+bc+ca)}}3=\frac23\sqrt{a^2+b^2+c^2-ab-bc-ca}$$ This is the diameter of the inscribed circle, which also the radius of the circumscribed circle in an equilateral triangle. $$R=\frac23\sqrt{a^2+b^2+c^2-ab-bc-ca}$$ So we need to find $\phi$. We know that the top vertex coordinate is $$b=x_c+R\sin\phi$$ so $$\sin\phi=\frac{b-x_c}R$$ Now let's check if indeed the other coordinates correspond to $a$ and $c$. I will do it for $a$, and $c$ can be done in a similar fashion. $$x_c-R\cos\left(\frac\pi6-\phi\right)=x_c-R\left(\cos\frac\pi6\cos\phi-\sin\frac\pi6\sin\phi\right)\\ =x_c-R\left(\frac{\sqrt{3}}2\sqrt{1-\sin^2\phi}+\frac12\sin\phi\right)\\=x_c-\left(\frac{\sqrt{3}}2\sqrt{R^2-R^2\sin^2\phi}+\frac12R\sin\phi\right)\\=(*)$$ Now we use $$R\sin\phi=b-x_c\\x_c=\frac{a+b+c}3\\R^2=\frac49(a^2+b^2+c^2-ab-bc-ca)$$ to get $$(*)=\frac{a+b+c}3-\left(\frac{\sqrt 3}2\sqrt{\frac49(a^2+b^2+c^2-ab-bc-ca)-\frac{(2b-a-c)^2}9}+\frac12\frac{2b-a-c}3\right)\\=\frac16\left(2a+2b+2c-2b+a+c-\sqrt3\cdot\\\sqrt{4a^2+4b^2+4c^2-4ab-4ac-4bc-4b^2-a^2-c^2+4ab+4bc-2ac}\right)\\=\frac16\left(3a+3c-\sqrt3\sqrt{3a^2+3c^2-6ac}\right)\\=\frac16\left(3a+3c-3\sqrt{(a-c)^2}\right)$$ Since we assumed $c>a$, we then write $$(*)=\frac16(3a+3c-3(c-a))=a$$ Asn mentioned before, you can similarly show that $$x_c+R\cos\left(\frac\pi6+\phi\right)=c$$

Answer 2

Well, the roots of a cubic, with 3 real roots, are determined by the $\cos()$ of an angle, $\psi$ that is divided by 3.

So if one starts with

$\psi = \theta + 2k\pi$

then

$\dfrac{\psi}{3} = \dfrac{\theta}{3} + \dfrac{2\pi}{3} k$

The equilateral triangle comes from the $\dfrac{2\pi}{3}$ spacing between the angles for the solutions.

You might find Figure 2 and Section 2.4 of Nickalls' analytic geometry based solution of the cubic helpful.

If one allows $\psi$ to be a complex hyperbolic angle, then once can use $\cosh()$ to express the roots of the cubic, and go beyond the case of 3 real roots.

See this Geogebra demonstration and vary the slider for the $d$ coefficient of the cubic (or press the play button for that slider and let it vary automatically). The left pane will show the value of the complex hyperbolic angle $\psi$ and the right pane will show its affect on the (real part of the) roots of the cubic.

That Geogebra demonstration is based off of this write up.