2022-07-21

Is there an easier way to solve a "Find the locus" problem?
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 $\mathrm{\Delta }ABC$ be $A=\left({x}_{0},0\right)$, $B=\left(0,{y}_{0}\right)$ and $C=\left(x,y\right)$. Let the centre of the triangle be $D=\left(X,Y\right)$. We wish to find the locus of D under the constraints:
$\begin{array}{rl}{x}_{0}^{2}+{y}_{0}^{2}& =0\\ \left(x-{x}_{0}{\right)}^{2}+{y}^{2}& =0\\ {x}^{2}+\left(y-{y}_{0}{\right)}^{2}& =0\end{array}$
Since D is the centre of the triangle, we know
$X=\frac{x+{x}_{0}}{3}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}Y=\frac{y+{y}_{0}}{3}$
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, $\begin{array}{rl}{x}_{0}& =\mathrm{cos}\theta \\ {y}_{0}& =\mathrm{sin}\theta \end{array}$
for $\theta \in \left[0,2\pi \right)$
This implies, $\left(x-\mathrm{cos}\theta {\right)}^{2}+{y}^{2}=0$
We can parametrize this with another parameter $\varphi \in \left[0,2\pi \right)$ as
$\begin{array}{rl}x& =\mathrm{cos}\theta +\mathrm{cos}\varphi \\ y& =\mathrm{sin}\varphi \end{array}$
Plugging these back into third equation gives
$\left(\mathrm{cos}\theta +\mathrm{cos}\varphi {\right)}^{2}+\left(\mathrm{sin}\varphi -\mathrm{sin}\theta {\right)}^{2}=1$
After lots of algebraic manipulations...
$\begin{array}{rlrl}X& =\frac{1}{2}\mathrm{cos}\theta ±\frac{1}{2\sqrt{3}}\mathrm{sin}\theta & Y& =\frac{1}{2}\mathrm{sin}\theta ±\frac{1}{2\sqrt{3}}\mathrm{cos}\theta \end{array}$

Jaylene Tyler

Expert

Step 1
It is probably easier to embed the construction in the complex plane and set $A=\mathrm{sin}\theta ,\phantom{\rule{2em}{0ex}}B=i\mathrm{cos}\theta$ to have $C=\left(A-B\right){e}^{±i\pi /3}+B={e}^{-i\theta ±i\pi /3}+B$
so that: $3D=A+B+C={e}^{i\theta }+{e}^{-i\theta ±i\pi /3}+i\mathrm{cos}\theta .$.
Step 2
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.

Do you have a similar question?