Baladdaa9

Answered

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 $A=({x}_{0},0)$, $B=(0,{y}_{0})$ 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:

$\begin{array}{rl}{x}_{0}^{2}+{y}_{0}^{2}& =0\\ (x-{x}_{0}{)}^{2}+{y}^{2}& =0\\ {x}^{2}+(y-{y}_{0}{)}^{2}& =0\end{array}$

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 $\theta \in [0,2\pi )$

This implies, $(x-\mathrm{cos}\theta {)}^{2}+{y}^{2}=0$

We can parametrize this with another parameter $\varphi \in [0,2\pi )$ as

$\begin{array}{rl}x& =\mathrm{cos}\theta +\mathrm{cos}\varphi \\ y& =\mathrm{sin}\varphi \end{array}$

Plugging these back into third equation gives

$(\mathrm{cos}\theta +\mathrm{cos}\varphi {)}^{2}+(\mathrm{sin}\varphi -\mathrm{sin}\theta {)}^{2}=1$

After lots of algebraic manipulations...

$\begin{array}{rlrl}X& =\frac{1}{2}\mathrm{cos}\theta \pm \frac{1}{2\sqrt{3}}\mathrm{sin}\theta & Y& =\frac{1}{2}\mathrm{sin}\theta \pm \frac{1}{2\sqrt{3}}\mathrm{cos}\theta \end{array}$

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 $A=({x}_{0},0)$, $B=(0,{y}_{0})$ 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:

$\begin{array}{rl}{x}_{0}^{2}+{y}_{0}^{2}& =0\\ (x-{x}_{0}{)}^{2}+{y}^{2}& =0\\ {x}^{2}+(y-{y}_{0}{)}^{2}& =0\end{array}$

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 $\theta \in [0,2\pi )$

This implies, $(x-\mathrm{cos}\theta {)}^{2}+{y}^{2}=0$

We can parametrize this with another parameter $\varphi \in [0,2\pi )$ as

$\begin{array}{rl}x& =\mathrm{cos}\theta +\mathrm{cos}\varphi \\ y& =\mathrm{sin}\varphi \end{array}$

Plugging these back into third equation gives

$(\mathrm{cos}\theta +\mathrm{cos}\varphi {)}^{2}+(\mathrm{sin}\varphi -\mathrm{sin}\theta {)}^{2}=1$

After lots of algebraic manipulations...

$\begin{array}{rlrl}X& =\frac{1}{2}\mathrm{cos}\theta \pm \frac{1}{2\sqrt{3}}\mathrm{sin}\theta & Y& =\frac{1}{2}\mathrm{sin}\theta \pm \frac{1}{2\sqrt{3}}\mathrm{cos}\theta \end{array}$

Answer & Explanation

Jaylene Tyler

Expert

2022-07-22Added 10 answers

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 $C=(A-B){e}^{\pm i\pi /3}+B={e}^{-i\theta \pm i\pi /3}+B$

so that: $3D=A+B+C={e}^{i\theta}+{e}^{-i\theta \pm i\pi /3}+i\mathrm{cos}\theta .$.

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.

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 $C=(A-B){e}^{\pm i\pi /3}+B={e}^{-i\theta \pm i\pi /3}+B$

so that: $3D=A+B+C={e}^{i\theta}+{e}^{-i\theta \pm i\pi /3}+i\mathrm{cos}\theta .$.

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.

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