Geometric proof and extension of |a|=|b|=|c|=a+b+c=1 => a=1 or b=1 or c=1

Question

We have $a,b,c\in\mathbb{C}$ verifying $|a|=|b|=|c|=a+b+c=1$, we have to show that $a=1$ or $b=1$ or $c=1$. That can be rather easily proved using trigonometry formulas.

• Is there a way to prove it by a geometric proof ? (edit : geometry isn't where i'm the best, so please make a detailed answer)

• If we just had $|a|=|b|=|c|=a+b+c=x, x\in\mathbb{R}$, would there still be solutions ? If so, which ones ?

Answer 1

Assume the contrary, i.e. that none of the vectors $a,b,c$ is parallel to the $x$-axes, as in the diagram below:

Since: $$x=\angle OAB = \angle OBC$$ $$y=\angle COA = \angle CBA$$ And: $$2x+2y = 360^\circ$$ we have that $x+y=180^\circ$, or that at least one vector is parallel to the $x$-axes.

Of course, nowhere have we used the actual length, so the claim holds true for all $x$.

Answer 2

1. Seen as vectors, we have $a+b+c-1=0$, i.e. concatenating the vectors $a,b,c,-1$ one after the other will give us a quadrilateral with equal sides, so one could find a permutation of these vectors such that we get a rhombus, and the vector on side parallel to $-1$ will be the $+1$.

2. Yes, for any $x\in\Bbb C$, if $|a|=|b|=|c|=|x|$ and $a+b+c=x$ then one of $a,b,c$ equals to $x$.
   This can be seen either directly by the above argument, or just multiply everything by $x$ in the original result.