Equilateral $\triangle ABC$ has $A$ fixed and B moving in a given straight line. Find the locus of $C$.

Question

$\triangle ABC$ is an equilateral triangle with vertex $A$ fixed and $B$ moving in a given straight line. Find the locus of $C$.

I think the locus of $C$ should also be a line.

What I have done: drawn a couple of equilateral triangles, say $\triangle AB'C'$ and $\triangle AB''C''$, and proven that $\triangle AB'B''$ is similar to $\triangle AC'C''$.

Can someone please help me with the next step?

Answer 1

Observe a rotation around $A$ for $60^{\circ}$. Then $B$ goes to $C$ and since $B$ is on a fixed line so is $C$. So $C$ describe a line which closes an angle $60^{\circ}$ with a given line (this new line is a picture of a given line).

Answer 2

Consider that $A$ is the point $(0,1)$ and the line is $y=0$. Then the point $B$ is $(t,0)$. The point $C$ is in the circles $$x^2+(y-1)^2=1+t^2$$ $$(x-t)^2+y^2=1+t^2$$ If you solve the system for $x,y$ you should obtain a pair of parametric equations for the locus.

Probably not the most elegant approach, but safe and effective.

Answer 3

Using the $x$-axis as line on which $B(x,0)$ is moving with $A(0,0)$, you derive using the fact, that $\triangle ABC$ is equilateral:

• $C\left(\frac{x}{2},y \right)$ with $y > 0$ for $x >0$ gives
• $\frac{y}{\frac{x}{2}}=\tan 60^{\circ} \Rightarrow y=\frac{\sqrt{3}}{2}x$

If we consider only $x>0$ we get a ray (half-line) without its starting point.