I started studying barycentric coordinates and I have some questions.
In my book states, word to word :
Consider $\triangle{ABC}$ and a point $P\in\mathcal{P}$. For any point $M\in\mathcal{P}$ there are uniquely determined 3 real numbers $x,~y,~z$ such that $x+y+z=1$ and
$$\vec{PM}=x\vec{PA}+y\vec{PB}+z\vec{PC}.$$
$x,~y,~z$ are called barycentric coordinates of point $M$ relating to $\triangle{ABC}$.
I tried to prove it in the complex plane:
Consider a cartesian plane with the center in $P$ and $A(a),~B(b),~C(c)$. Then the problem becomes:
Prove $\forall~M(s)\in\mathcal{P}~\exists$ uniquely $x,~y,~z\in\mathbb{R}$, $x+y+z=1$ such that $s=xa+yb+zc$.
Write $a,~b,~c,~s$ in cartesian form:
$a=x_1+y_1i,~b=x_2+y_2i,~c=x_3+y_3i,~s=x_s+y_si$
Then $x_s+y_si=(xx_1+yx_2+zx_3)+(xy_1+yy_2+zy_3)i\Rightarrow$
$
\begin{cases}
xx_1+yx_2+zx_3=x_s&(1) \\
xy_1+yy_2+zy_3=y_s&(2)
\end{cases}
$
Also $x+y+z=1~~~(3)$
Forming a system of equations with $(1),~(2),~(3)$ I got to the conclusion that:
$x=\frac{(y_2-y_3)(x_s-x_3)+(x_3-x_2)(y_s-y_3)}{(y_2-y_3)(x_1-x_3)+(x_3-x_2)(y_1-y_3)}$
Which is exactly the formula on Wikipedia.
$y$ and $z$ will be similar.
My hypothesis: barycentric coordinates are the same $\forall P\in\mathcal{P}$ and constant $A,~B,~C,~M\in\mathcal{P}$.
My proof: Let $P'\in\mathcal{P}$ and $P'=P-\Delta,~\Delta\in\mathbb{C^*}$.
Considering a cartesian plane with the center in $P'$, all the coordinates suffer a simple transformation, that is, they are shifted by $\Delta$:
$x_s'=x_s+\Delta_x,~y_s'=y_s+\Delta_y$
$x_k'=x_k+\Delta_x,~y_k'=y_k+\Delta_y,~k\in\{1,2,3\}$
Finding $x',~y',~z'$ using this substitution $\Delta_x$ and $\Delta_y$ simplify and
$x=x',~y=y',~z=z'$
So for any point $P$ barycentric coordinates will preserve?
If I want to prove, for example, that barycentric coordinates of the circumcenter $O$ of $\triangle{ABC}$ are
$(x,y,z)=(\frac{sin2A}{sin2A+sin2B+sin2C},\frac{sin2B}{sin2A+sin2B+sin2C},\frac{sin2C}{sin2A+sin2B+sin2C})$
then can I consider a cartesian plane with center in O and choose $P=O$? Because, if the logic above is true, if the barycentric coordinates are those for $P=O$, then they must be for any $P$. So I would have to prove
$$\frac{1}{sin2A+sin2B+sin2C}(sin2A\cdot\vec{OA}+sin2B\cdot\vec{OB}+sin2C\cdot\vec{OC})=\vec{0}$$
and since $\frac{1}{sin2A+sin2B+sin2C}\ne0$
$$sin2A\cdot\vec{OA}+sin2B\cdot\vec{OB}+sin2C\cdot\vec{OC}=\vec{0}$$
That is, as a complex equation,
$$a\cdot sin2A+b\cdot sin2B+c\cdot sin2C=0$$
for $|a|=|b|=|c|=R$.
Would that be a correct proof?
EDIT: How I tried to continue the proof:
First attempt
$a=R\cdot e^{i\alpha},~b=R\cdot e^{i\beta},~c=R\cdot e^{i\gamma}$
Divide by R
$(cos\alpha\cdot sin2A + cos\beta\cdot sin2B + cos\gamma\cdot sin2C)+(sin\alpha\cdot sin2A + sin\beta\cdot sin2B + sin\gamma\cdot sin2C)i=0$
It must be proved that both the real and the imaginary part of LHS are $0$, but I have no idea how to do that. The only advantage of this would be working only on angles but I don't know how to continue.
Second attempt
Divide by 2
$a\cdot sinA\cdot cosA+b\cdot sinB\cdot cosB+c\cdot sinC\cdot cosC=0$
Use Law of Sines and Al Kashi's theorem :
$sinA=\frac{BC}{2R},~sinB=\frac{AC}{2R},~sinC=\frac{AB}{2R}$
$cosA=\frac{AB^2+AC^2-BC^2}{2AB\cdot AC},~cosB=\frac{AB^2+BC^2-AC^2}{2AB\cdot BC},~cosC=\frac{AC^2+BC^2-AB^2}{2AC\cdot BC}$
Substitute and multiply by $4R$, then amplify each fraction by $BC,~AC,~AB$ respectively and multiply by the common denominator, $2AB\cdot BC\cdot AC$.
$a\cdot BC^2(AB^2+AC^2-BC^2)+b\cdot AC^2(AB^2+BC^2-AC^2) + c\cdot AB^2(AC^2+BC^2-AB^2)=0.$
The next step would be use $AB^2=|a-b|^2=(a-b)(\overline{a-b})=(2R^2-a\overline{b}-b\overline{a})$ and same for the others but this requires too much computation and might take me up to 1 hour to do and also not make a mistake. Is there any software to handle such expressions and compute the standard expanded form? I tried WolframAlpha but "character limit exceeded" which is kinda disappointing since they say up to 1200 characters when you Google search it (I had to write the Conjugate[] function many times).
Best Answer
That is correct, barycentric coordinates are invariant under translations (and also affine transformations in general, including rotations, reflections, scaling etc).
The center angle $\,2A = \angle BOC = \arg(b)-\arg(c)=\arg\left(\frac{b}{c}\right)\,$, then using that $\,\left|\frac{b}{c}\right|=1\,$, $z - \bar z = 2i\,\text{Im}(z)\,$ and $\,a \bar a = |a|^2 = R^2\,$:
$$ \require{cancel} \sin 2A = \text{Im}\left(\frac{b}{c}\right) = \frac{1}{2i}\left(\frac{b}{c}-\frac{\bar b}{\bar c}\right) = \frac{1}{2i}\left(\frac{b}{c}-\frac{\cancel{R^2} /\, b}{\cancel{R^2} /\,c}\right) = \frac{1}{2i}\left(\frac{b}{c}-\frac{c}{b}\right) $$
It follows that:
$$ 2i\,(a\cdot \sin2A+b\cdot \sin2B+c\cdot \sin2C) \\ \;=\; a\left(\color{blue}{\cancel{\color{black}{\frac{b}{c}}}}-\color{green}{\cancel{\color{black}{\frac{c}{b}}}}\right)+b\left(\color{red}{\cancel{\color{black}{\frac{c}{a}}}}-\color{blue}{\cancel{\color{black}{\frac{a}{c}}}}\right)+c\left(\color{green}{\cancel{\color{black}{\frac{a}{b}}}}-\color{red}{\cancel{\color{black}{\frac{b}{a}}}}\right) \;=\; 0 $$