Three points on the Complex plane

[Hill]

Homework Statement

Find the geometric configuration of the points a, b, and c if (see equation below)

Relevant Equations

(b-a)/(c-a)=(a-c)/(b-c)

$arg((b-a)/(c-a))$ is an angle between $ab$ and $ac$.
$arg((a-c)/(b-c))$ is an angle between $ca$ and $cb$.
For them to be equal, $b$ has to be equidistant from $a$ and $c$, i.e. $|b-a|=|b-c|$.

Then the equation for distances becomes, $|b-a|/|c-a|=|c-a|/|b-a|$.
Thus, $|c-a|=|b-a|=|b-c|$.

My answer, equilateral triangle.

Are there other possibilities or constrains?

 

[pasmith]

Write $$(a- b)(b - c) = (a - c)^2 = (a - b + b - c)^2.$$ If $a - b = A$ and $b - c = B$ then $$\begin{split} AB &= (A + B)^2 \\ 0 &= A^2 + AB + B^2 \end{split}$$ which has solution $$A = e^{\pm i \pi/3} B$$ so that the angle between $A$ and $B$ is $\frac{\pi}3$ and $|A| = |B|$.

However, since $a = b + A$ and $c = b - B$ the angle between $a - b$ and $c - b$ is actually $\frac{2\pi}3$, which is not equilateral. Rather, it is isoceles with angles $\frac{\pi}6$, $\frac{\pi}6$ and $\frac{2\pi}3$ and sides $|A|$, $|A|$ and $\sqrt{3}|A|$.

 

[Hill]

Write $$(a- b)(b - c) = (a - c)^2 = (a - b + b - c)^2.$$ If $a - b = A$ and $b - c = B$ then $$\begin{split} AB &= (A + B)^2 \\ 0 &= A^2 + AB + B^2 \end{split}$$ which has solution $$A = e^{\pm i \pi/3} B$$ so that the angle between $A$ and $B$ is $\frac{\pi}3$ and $|A| = |B|$.

However, since $a = b + A$ and $c = b - B$ the angle between $a - b$ and $c - b$ is actually $\frac{2\pi}3$, which is not equilateral. Rather, it is isoceles with angles $\frac{\pi}6$, $\frac{\pi}6$ and $\frac{2\pi}3$ and sides $|A|$, $|A|$ and $\sqrt{3}|A|$.

If $|a - b| = |b-c| = |A|$ and $|a-c|=\sqrt{3}|A|$, then the first equation, $$(a- b)(b - c) = (a - c)^2$$ makes $$|A||A|=3|A|^2$$ -?

 

[Mark44]

Write
$$(a- b)(b - c) = (a - c)^2 = (a - b + b - c)^2.$$ If $a - b = A$ and $b - c = B$ then $$\begin{split} AB &= (A + B)^2 \\ 0 &= A^2 + AB + B^2 \end{split}$$ which has solution $$A = e^{\pm i \pi/3} B$$ so that the angle between $A$ and $B$ is $\frac{\pi}3$ and $|A| = |B|$.

Isn't the solution to the quadratic $A = B\left(\frac{-1}2 \pm \frac {\sqrt 3} 2 i\right)$? Or in polar form $A = Be^{\pm i \frac{2\pi} 3}$?

Also, I suspect that the triangle is isosceles, of which equilateral is a special case, but I haven't worked out a counterexample.

 

[Hill]

I suspect that the triangle is isosceles, of which equilateral is a special case

As $|A|=|B|$, the first equation becomes $|A|^2=|C|^2$. Thus $|C|=|A|$, i.e., the triangle is equilateral.

 

[pasmith]

Yes, it should be $A = Be^{\pm 2\pi i/3}$, making the triangle equilateral: the interior angle at $b$ is $\frac{\pi}3$ and $|a - b| = |c - b|$.

 

[Hill]

Yes, it should be $A = Be^{\pm 2\pi i/3}$, making the triangle equilateral: the interior angle at $b$ is $\frac{\pi}3$ and $|a - b| = |c - b|$.

Thank you for the clear derivation.