There is no Difference Between Equilateral Triangles

Let $M$ be a point in the plane of equilateral triangle $\Delta ABC.$ For a generic point $X$ let $x$ denote the corresponding complex number. Find

$\displaystyle \bigg|\frac{m-a}{b-a}+\frac{m-b}{c-b}+\frac{m-c}{a-c}\bigg|. $


Note interesting identities that hold for the vertices of an equilateral triangle:

  1. $c-b=(b-a)e^{\pm 2i\pi/3}, a-c=(b-a)e^{\mp 2i\pi/3},$ depending on the triangle's orientation.

  2. $\displaystyle\frac{1}{b-a}+\frac{1}{c-b}+\frac{1}{a-c}=0.$

Now define $f(m)=\displaystyle\frac{m-a}{b-a}+\frac{m-b}{c-b}+\frac{m-c}{a-c}.$ Observe that

$\begin{align}\displaystyle f(m)&=\frac{m-a}{b-a}+\frac{m-b}{c-b}+\frac{m-c}{a-c}\\ &=\frac{m}{b-a}+\frac{m}{c-b}+\frac{m}{a-c}+\mbox{const}\\ &=m\bigg(\frac{1}{b-a}+\frac{1}{c-b}+\frac{1}{a-c}\bigg)+\mbox{const}\\ &=\mbox{const}. \end{align}$

Thus $f(m)$ is independent of $M$ (and also of $\Delta ABC)$. To find its constant value substitute, say, $m=a:$

$\displaystyle f(a)=\frac{a-b}{c-b}+\frac{a-c}{a-c}=\frac{a-b}{c-b}+1=e^{\pm i\pi/3}+1,$

so that $f(a)=\frac{1}{2}\pm i\frac{\sqrt{3}}{2}+1,$ implying $f(m)=f(a)=\sqrt{3},$ for all $m.$


The problem and its solution have been posted by Leo Giugiuc (Romania) at CutTheKnotMath facebook page.

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