Consider an equilateral triangle centered at the origin of the 2D Cartesian space. Let the coordinates of its vertices be $v_1=(x_1,y_1)$, $v_2=(x_2,y_2)$ and $v_3=(x_3,y_3)$. All such triangles can be defined by the following conditions:
\begin{gather}
||v_1 – v_2|| = ||v_1 – v_3|| = ||v_2 – v_3|| & \label{eq_dist}\tag{1} \\
v_1 + v_2 + v_3 = (0, 0) & \label{sum_0}\tag{2} \\
\end{gather}
Using these conditions it can be shown that $||v_1|| = ||v_2|| = ||v_3||$ (see proof here). Let
$$r = ||v_1|| = ||v_2|| = ||v_3|| \label{eq_len}\tag{3}$$
My question is whether there is a simple algebraic explanation for the fact that the expression
$$d_x \overset{\text{def}}{=} x_1 x_2 + x_1 x_3 + x_2 x_3$$
does not depend on the angle of rotation of the triangle about the origin. I made an interactive Desmos link where you can observe this behavior. Note that, in that link, I used $s$ as the radius instead of $r$, since $r$ is a special variable in Desmos.
I will proceed by listing and then detailing some of my findings:
- $v_1 \cdot v_2 + v_1 \cdot v_3 + v_2 \cdot v_3$ is also rotation-independent
- I found a trigonometric proof for my question
- This behavior also happens with a tetrahedron in $\mathbb{R}^3$ (see this Desmos link)
1) $v_1 \cdot v_2 + v_1 \cdot v_3 + v_2 \cdot v_3$ is also rotation-independent
This expression, although different than $d_x$, is closely related to it, since each individual dot product can be expressed as:
$$v_k \cdot v_j = x_k x_j + y_k y_j$$
Which implies that
$$\color{blue}{v_1 \cdot v_2} + \color{red}{v_1 \cdot v_3} + \color{green}{v_2 \cdot v_3} = (\color{blue}{x_1 x_2} + \color{red}{x_1 x_3} + \color{green}{x_2 x_3}) + (\color{blue}{y_1 y_2} + \color{red}{y_1 y_3} + \color{green}{y_2 y_3})$$
It is therefore handy to define
\begin{align*}
d_y & \overset{\text{def}}{=} y_1 y_2 + y_1 y_3 + y_2 y_3 \\
d & \overset{\text{def}}{=} v_1 \cdot v_2 + v_1 \cdot v_3 + v_2 \cdot v_3
\end{align*}
Which allows us to write
$$d = d_x + d_y$$
This means that if $d$ and $d_y$ are rotation-independent, so is $d_x$. We can show that $d$ is rotation-independent by noticing that each of the three dot products is also rotation-independent. This happens because the dot product depends only on each vector's norm, and the angle between them – and none of those change when rotating the whole triangle about the origin. Nevertheless, I present a simple algebraic proof for this fact:
\begin{align*}
||v_1||^2 & = r^2 & \text{by (\ref{eq_len})} \\
||- v_2 – v_3||^2 & = r^2 & \text{by (\ref{sum_0})} \\
(- x_2 – x_3)^2 + (- y_2 – y_3)^2 & = r^2 & \text{expand} \\
\color{blue}{x_2^2} + 2 x_2 x_3 + \color{red}{x_3^2} + \color{blue}{y_2^2} + 2 y_2 y_3 + \color{red}{y_3^2} & = r^2 & \text{expand} \\
\color{blue}{r^2} + \color{red}{r^2} + 2 x_2 x_3 + 2 y_2 y_3 & = r^2 & \text{by (\ref{eq_len})} \\
x_2 x_3 + y_2 y_3 & = – \frac{r^2}{2} & \text{simplify} \\
v_2 \cdot v_3 & = – \frac{r^2}{2} & \text{by definition of $\cdot$} \\
\end{align*}
By symmetry, this proof also applies to $v_1 \cdot v_2$ and $v_1 \cdot v_3$, and thus
\begin{gather}
v_1 \cdot v_2 = v_1 \cdot v_3 = v_2 \cdot v_3 = – \frac{r^2}{2} \\
d = v_1 \cdot v_2 + v_1 \cdot v_3 + v_2 \cdot v_3 = – \frac{3}{2} r^2
\end{gather}
Although this is an interesting result, it is not enough, since we would also need to show that $d_y$ is rotation-independent. It could be the case that $d_x$ and $d_y$ both depended on the triangle's rotation but their sum did not – just like the cosine and sine functions depend on their argument but the sum of their squares does not. But if $d_x$ and $d_y$ were indeed rotation-independent (which I know is true, but am trying to find a simple algebraic proof for), since conditions (\ref{eq_dist}) and (\ref{sum_0}) are symmetrical in $x$ and $y$, we would have
$$d_x = d_y = \frac{d}{2} = – \frac{3}{4} r^2 \label{dx_dy}\tag{4}$$
However, I was unable to prove this fact algebraically.
2) I found a trigonometric proof for my question
This proof is based in two observations:
- $v_1$, $v_2$ and $v_3$ lie in a circle centered at $(0, 0)$ (by \ref{sum_0}) and with radius $r$ (by \ref{eq_len})
- The angle between each pair of vertices (seen as vectors) is $\frac{1}{3}$ of a full turn, since the triangle is equilateral (by \ref{eq_dist})
These allow us to write
\begin{align*}
v_1 = (x_1, y_1) & = (r \cos{\theta}, r \sin{\theta}) & \label{trig1}\tag{5.1} \\
v_2 = (x_2, y_2) & = (r \cos{(\theta + \frac{2 \pi}{3})}, r \sin{(\theta + \frac{2 \pi} {3})}) & \label{trig2}\tag{5.2} \\
v_3 = (x_3, y_3) & = (r \cos{(\theta – \frac{2 \pi}{3})}, r \sin{(\theta – \frac{2 \pi}{3})}) & \label{trig3}\tag{5.3}
\end{align*}
where $\theta$ is an arbitrary real number. To simplify a computation below, we will now compute
\begin{align}
\cos(\alpha + \beta) \cos(\alpha – \beta) & = (\cos \alpha \cos \beta – \sin \alpha \sin \beta) (\cos \alpha \cos \beta + \sin \alpha \sin \beta) \\
& = \cos^2 \alpha \cos^2 \beta – \sin^2 \alpha \sin^2 \beta & \label{cos_prod}\tag{6}
\end{align}
So, we have
\begin{align*}
d_x & = x_1 x_2 + x_1 x_3 + x_2 x_3 & \text{by definition} \\
& = x_1(x_2 + x_3) + x_2 x_3 \\
& = x_1(- x_1) + x_2 x_3 & \text{by (\ref{sum_0})} \\
& = x_2 x_3 – x_1^2 \\
& = r^2 (\cos(\theta + \frac{2 \pi}{3}) \cos(\theta – \frac{2 \pi}{3}) – \cos^2 \theta) & \text{by (\ref{trig1}), (\ref{trig2}) and (\ref{trig3})} \\
& = r^2 (\cos^2 \theta \cos^2 \frac{2 \pi}{3} – \sin^2 \theta \sin^2 \frac{2 \pi}{3} – \cos^2 \theta) & \text{by (\ref{cos_prod})} \\
& = r^2 ((\cos^2 \theta) (- \frac{1}{2})^2 – (\sin^2 \theta) (\frac{\sqrt{3}}{2})^2 – \cos^2 \theta) \\
& = r^2 (\frac{1}{4} \cos^2 \theta – \frac{3}{4} \sin^2 \theta – \cos^2 \theta) \\
& = r^2 (- \frac{3}{4} (\cos^2 \theta + \sin^2 \theta)) \\
& = – \frac{3}{4} r^2
\end{align*}
This allows us to conclude that the equality (\ref{dx_dy}) is indeed true. However, we had to resort to trigonometry, whilst I was looking for an algebraic proof. Moreover, this trigonometric proof does not give me a good understanding of how conditions (\ref{eq_dist}) and (\ref{sum_0}) imply that $d_x$ only depends on $r$ and not on $\theta$. It seems that it could have been just a matter of luck that $x_1^2 = r^2 \cos^2 \theta$ and $x_2 x_3 = r^2 (\cos^2 \theta – \frac{3}{4})$. But looking at higher dimensions, the pattern seems to repeat, hinting that there must be something more intricate going on here.
3) This behavior also happens with a tetrahedron in $\mathbb{R}^3$
By considering a tetrahedron centered at the origin of the 3D Cartesian space, defining conditions analagous to (\ref{eq_dist}) and (\ref{sum_0}), and performing a trigonometric change of variables akin to (\ref{trig1}), (\ref{trig2}) and (\ref{trig3}), I was able to show that an expression analogous to $d_x$ (i.e., $\sum_{k \neq j}{x_k x_j}$) does not depend on the rotation of the tetrahedron about the origin. You can observe this behavior by playing around with this Desmos link.
That is it. Any insights on this problem would be appreciated, even if they don't directly answer my question. Thank you in advance.
Best Answer
If we write $v_{1} = r(\cos{\theta},\sin{\theta})$, then, since the triangle is equilateral, we must have $v_{2} = r(\cos{(\theta+\frac{2\pi}{3})},\sin{(\theta+\frac{2\pi}{3})})$ and $v_{3} = r(\cos{(\theta+\frac{4\pi}{3})},\sin{(\theta+\frac{4\pi}{3})})$, so what we want is
\begin{align*} d_{x} &= r\cos{\theta}\cdot r\cos{\left(\theta+\frac{2\pi}{3}\right)} + r\cos{\theta}\cdot r\cos{\left(\theta+\frac{4\pi}{3}\right)} + r\cos{\left(\theta+\frac{2\pi}{3}\right)}\cdot r\cos{\left(\theta+\frac{4\pi}{3}\right)} \\ &= r^2\left(\cos{\theta}\left(-\frac{\cos{\theta}}{2}-\frac{\sqrt{3}\sin{\theta}}{2}\right)+\cos{\theta}\left(-\frac{\cos{\theta}}{2}+\frac{\sqrt{3}\sin{\theta}}{2}\right)+\left(-\frac{\cos{\theta}}{2}-\frac{\sqrt{3}\sin{\theta}}{2}\right)\left(-\frac{\cos{\theta}}{2}+\frac{\sqrt{3}\sin{\theta}}{2}\right)\right) \\ &= r^2\left(-\cos^2{\theta}+\frac{\cos^2{\theta}}{4}-\frac{3\sin^2{\theta}}{4}\right) = -\frac{3r^2}{4} \end{align*}