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

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

 

[Hill]

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

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

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

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 [itex]A = Be^{\pm 2\pi i/3}[/itex], making the triangle equilateral: the interior angle at [itex]b[/itex] is [itex]\frac{\pi}3[/itex] and [itex]|a - b| = |c - b|[/itex].

 

[Hill]

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

Thank you for the clear derivation.