Note: I am not concerned with the accuracy of my solution so you don't need to redo any of my calculations.
Original question: Suppose $ABC$ is an equilateral triangular lamina of side length unity, resting in two-dimensions. If $A$ and $B$ were constrained to move on the $x$- and $y$-axis respectively, then what is the locus of the centre of the triangle?
My solution: Let the vertices of $\Delta ABC$ be $A=(x_0,0)$, $B=(0,y_0)$ and $C=(x,y)$. Let the centre of the triangle be $D=(X,Y)$. We wish to find the locus of $D$ under the constraints: \begin{align} x_0^2 + y_0^2 &= 0\\ (x-x_0)^2 + y^2 &= 0\\ x^2 + (y-y_0)^2 &= 0 \end{align} Since $D$ is the centre of the triangle, we know $$X = \frac{x+x_0}{3}\quad\text{and}\quad Y = \frac{y+y_0}{3}$$
Parametrizing: Let $\theta$ be the angle that the edge $AB$ makes with the $x$-axis. We use (\theta) as our parameter of choice to write, \begin{align*} x_0 &= \cos\theta\\ y_0 &= \sin\theta \end{align*} for $\theta\in[0,2\pi)$.
This implies, $$(x-\cos\theta)^2 + y^2 = 0$$ We can parametrize this with another parameter $\phi\in[0,2\pi)$ as \begin{align*} x &= \cos\theta + \cos\phi\\ y &= \sin\phi \end{align*} Plugging these back into third equation gives $$(\cos\theta+\cos\phi)^2+(\sin\phi-\sin\theta)^2 = 1$$ After lots of algebraic manipulations... \begin{align*} X &= \frac{1}{2}\cos\theta\pm\frac{1}{2\sqrt{3}}\sin\theta & Y &= \frac{1}{2}\sin\theta\pm\frac{1}{2\sqrt{3}}\cos\theta\\ \end{align*}
Deparametrizing: To obtain equation for the locus of $D$, we eliminate the parameter. Solving for $\sin\theta$ and $\cos\theta$ in terms of $X$ and $Y$, we get $$\cos\theta = 3X\mp\sqrt{3}Y \quad\text{and}\quad \sin\theta = 3Y\mp\sqrt{3}X$$ Therefore, $X$ and $Y$ satisfy $$(3X\mp\sqrt{3}Y)^2+(3Y\mp\sqrt{3}X)^2=1$$ i.e., $$12X^2+12Y^2\mp12\sqrt{3}XY=1$$ These are two oblique ellipses.
My actual questions:
- Is there an easier way to solve the question i.e. was there no need to parametrize, seeing as I am deparametrizing anyway?
- Even before doing any of the calculations, is there some result/theorem that guarantees that a relationship among $X, Y$ and $Z$ exists. (Subquestion: is it possible that making the problem more complicated by say looking at equilateral tetrahedrons in 3D or something might not even be solvable?)
Note: I am not concerned with the accuracy of my solution so you don't need to redo any of my calculations.
It is probably easier to embed the construction in the complex plane and set $$ A=\sin\theta,\qquad B=i\cos\theta $$ to have $$ C = (A-B)e^{\pm i\pi/3}+B = e^{-i\theta\pm i\pi/3}+B$$ so that: $$ 3D = A+B+C = e^{i\theta}+e^{-i\theta\pm i\pi/3}+i\cos\theta. $$ That leads to the same equations for the ellipses.
If three vertices of a tetrahedron lie on the $x-,y-$ and $z-$ axis we may assume they are $A(a,0,0),B(0,b,0),C(0,0,c)$ and $a^2+b^2=a^2+c^2=b^2+c^2$ must hold, so $|a|=|b|=|c|$ and there are just $16$ possible locations for $D$ and the centroid $G$ of $ABCD$.