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Consider a triangle ABC in the plane with angles $A, B, C$ and side lengths $a, b, c$ with the side of length $a$ opposite the angle $A$, etc, as usual.

Suppose also that the angles appear in the order $A, B, C$ as one traverses the vertices in an anticlockwise direction. Regard the three vertices as complex numbers $\alpha, \beta, \gamma$ with $\alpha$ corresponding to the vertex with angle $A$, etc.

So a triangle corresponds to a point $(\alpha, \beta, \gamma)$ in $3$-dimensional complex space. Not all points in $3$-dimensional complex space correspond to triangles since the points defined may be collinear.

I need to show that for the described triangle the following equation holds:

$$(be^{iA} − c)\alpha − be^{iA}\beta + c\gamma = 0$$

I'm pretty stumped! Any help would be appreciated.

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  • $\begingroup$ Have corrected thank you! I did not spot this error when I copied from latex. $\endgroup$
    – Gauss
    Commented Mar 21, 2017 at 14:23

1 Answer 1

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Given a triangle constructed as described in the question, the complex number $\beta - \alpha$ corresponds to the side of the triangle opposite the vertex $\gamma,$ so it has modulus $c.$ The complex number $\gamma - \alpha$ corresponds to the side of the triangle opposite the vertex $\beta,$ so it has modulus $b.$ The complex number $\gamma - \alpha$ also has argument $\arg(\gamma - \alpha) = \arg(\beta - \alpha) + A,$ because the magnitude of the angle between the two sides represented by $\beta - \alpha$ and $\gamma - \alpha$ is $A$ and because $\gamma$ is anticlockwise from $\beta$ when viewed from the position of $\alpha.$

In other words, if $\psi = \arg(\beta - \alpha),$ then \begin{align} \beta - \alpha &= c e^{i\psi}, \\ \gamma - \alpha &= b e^{i(\psi + A)} = be^{iA} e^{i\psi}. \end{align}

Now evaluate $c(\gamma - \alpha) - be^{iA}(\beta - \alpha)$ and compare the result to the fact that is to be proved.

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  • $\begingroup$ This may be a dumb question since I'm not familiar with '3-dimensional complex space'. But is $\arg(\gamma-\alpha)=\arg(\beta-\alpha)+A$ really true? What if $\alpha=[0 \,\,0 \,\,0]^T$, $\beta=[1 \,\,2 \,\,1]^T$, and $\gamma=[1 \,\,2 \,\,-1]^T$? What would $\arg(\beta)$ and $\arg(\gamma)$ be in this case? $\endgroup$
    – T L Davis
    Commented Mar 23, 2017 at 18:56
  • $\begingroup$ The problem statement says that $\alpha,$ $\beta,$ and $\gamma$ are complex numbers ("Regard the three vertices as complex numbers $\alpha, \beta, \gamma$"), not complex vectors. All three of these numbers combined together in a three-tuple $(\alpha, \beta, \gamma)$ produce the coordinates of just one point in the complex space mentioned in the problem statement. $\endgroup$
    – David K
    Commented Mar 23, 2017 at 19:05

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