Can three points $P(0)$, $P(t_1)$ and $P(t_2)$ form an equilateral triangle where $t_1>0,t_2 >0$?

Question

In a parabola $y^2=4ax$ can three points $P(0)$, $P(t_1)$ and $P(t_2)$ form an equilateral triangle where $t_1>0,t_2 >0$ ? (Given $a>0$)

$P(t)$ stands for the parametric point $(at^2,2at)$.

I know that if one out of $t_1$ and $t_2$ is positive and the other is negative then surely an equilateral triangle can be formed inside the parabola. However, when both are of same sign (positive), how can we check whether such an equilateral triangle is possible or not? Is there any intuitive method?

Answer 1

No, they cannot.

Note that $$|P(0)P(t)|=\sqrt{(at^2)^2+(2at)^2}=a\sqrt{t^4+4t^2}$$ is strictly increasing for $t\gt 0$.

It follows from this that $$t_1\not=t_2\implies |P(0)P(t_1)|\not=|P(0)P(t_2)|$$

Answer 2

Using $$ P_i = P(t_i) \\ P_{ij} = P_j - P_i \\ $$ you could check the lengths (equilateral) via $$ l_{ij} = \lVert P_{ij} \rVert $$ or the angles (equiangular) via $$ P_{ij} \cdot P_{ik} = \lVert P_{ij} \rVert \, \lVert P_{ik} \rVert \cos\angle(P_{ij}, P_{ik}) \iff \\ \cos\angle(P_{ij}, P_{ik}) = \frac{P_{ij} \cdot P_{ik}}{\lVert P_{ij} \rVert \, \lVert P_{ik} \rVert} $$

Answer 3

No, they cannot. Suppose for example $0<t_1<t_2$. The angle with vertex in $t_1$ is always larger than $\pi/2$, so it cannot be equal to $\pi/3$.

Answer 4

Here is a more general solution (I don't assume a priori that one of the vertices is in 0) using complex numbers.

Let us consider the parabola as the set of complex numbers $z=at^2+i2at$ for any real value of $t$ (the parabola has been positionned with a horizontal axis, which is unimportant).

Let $\omega:=e^{2i\pi/3}=-\dfrac12+i\dfrac{\sqrt{3}}{2}$. It is a cubic root of $1$,

with $\omega^2=e^{4i\pi/3}=-\dfrac12-i\dfrac{\sqrt{3}}{2},$ and property

$$\tag{0}1+\omega+\omega^2=0.$$

A necessary and sufficient condition for $z_1,z_2,z_3$ to be an equilateral triangle is:

$$\tag{1}z_1+\omega z_2 + \omega^2 z_3=0.$$

(see proof below).

Setting $z_k:=a(t_k^2+2it_k)$, ($k=1,2,3$) in (1), and separating real and imaginary parts gives:

$$\tag{2}\begin{cases}t_1^2-\frac12t_2^2-\frac12t_3^2-t_2\sqrt{3}+t_3\sqrt{3}=0\\2t_1-t_2+\dfrac{\sqrt{3}}{2}t_2^2-t_3-\dfrac{\sqrt{3}}{2}t_3^2=0\end{cases}$$

Multiplying (2)(a) by $-6$ and (2)(b) by $2\sqrt{3}$, and adding gives:

$$\tag{3}-6t_1^2-4\sqrt{3}t_1-8\sqrt{3}(t_2+t_3)=0.$$

which is impossible to fulfill with $t_1=0$ and $t_2,t_3$ all $>0.$

Relationships (2) give a necessary and sufficient condition for an equilateral triangle to be inscribed in a parabola: as one can extract $t_1$ from the equation (2)(b) and "plug" it into (2)(1), we can get an implicit equation (too long to be reproduced here) $\varphi(t_1,t_2)=0$. See curve below. This equation constitutes a necessary and sufficient condition on $t_2$ and $t_3$ in order to generate an inscribed equilateral triangle. It appears that for a given value of $t_2$, there can be zero, one, two and up to three solutions for $t_3$, the value of $t_1$ being computed from relationship (2)(b).

Connected: (https://www.geogebra.org/material/show/id/130907)).

(note that straight line $t_2=t_3$ represents "point triangles", i.e., triangles reduced to a point).

Proof of property (1):

$z_1,z_2,z_3$ forms an equilateral triangle if and only if

$$\tag{4}(z_1-z_3)=e^{i\pi/3}(z_2-z_3)$$

As $e^{i\pi/3}=-e^{i2\pi/3}=-\omega$, expanding (4) gives:

$$z_1+\omega z_2-(1+\omega)z_3=0$$

yielding (1) (by taking into account (0)).