I have a triangle $ABC$ and a point $P$ with barycentric coordinates ($\alpha, \beta, \gamma)$ that I want to reflect about the sides $a,b$ and $c$.
Calculating the general expression for a displacement vector perpendicular to $c$ and then using $|PB|=|P'B|$, I got
$$P'=\left(\alpha+x, \beta-\frac{S_A}{c^2}x, \gamma-\frac{S_B}{c^2}x\right)$$
for the reflection about $c$, where
$$x=\frac{-a^2\left((\beta-1)(-\frac{S_B}{c^2})+\gamma(-\frac{S_A}{c^2})\right)-b^2\left(\gamma+\alpha(-\frac{S_B}{c^2})\right)-c^2\left(\alpha(-\frac{S_A}{c^2})+(\beta-1)\right)}{\frac{S_A}{c^2}+\frac{S_B}{c^2}-\frac{S_A S_B}{c^4}}$$
and
$$S_A=\frac{-a^2+b^2+c^2}{2}, S_B=\frac{a^2-b^2+c^2}{2}$$
is Conway's Notation.
Can anyone confirm this or provide an easier formula? Any help is appreciated.
Best Answer
Without loss of generaliy let $A=(x_1,y_1), B=(0,0), C(x_3,0)$ be the cartesian coordinates of triangle of reference ABC with $y_1>0$ and $x_3>0$.
If $P=(\alpha, \beta,\gamma)$ are the absolute barycentric coordinates of point P, then its cartesian coordinates are $P=(x_0,y_0)=(\alpha x_1 +\gamma x_3, \alpha y_1)$.
Let $P'$ be the projection of point P on side BC and $(0,\beta',\gamma')$ its absolute barycentric coordinates. Therefore its cartesian coordinates are $(x_0,0)$.
Besides that, by the definition of barycentric coordinates we have $$\gamma'={S_{P'AB}\over S_{ABC}}={(1/2)x_0y_1\over (1/2)x_3y_1}={x_0\over x_3}={\alpha x_1 +\gamma x_3\over x_3}=\gamma +{x_1\over x_3}\alpha$$ $$\gamma'=\gamma +{c\cos B\over a}\alpha$$ $$\gamma'=\gamma +{a^2+c^2-b^2\over 2a^2}\alpha$$
As for $\beta'$, since $\beta'=1-\gamma'$ and $\alpha+\beta+\gamma=1$, we get $$\beta'=\beta+{a^2+b^2-c^2\over 2a^2}\alpha$$
Therefore the barycentric absolute coordinates of the reflexion point of P about BC are
$$(-\alpha,\beta+{a^2+b^2-c^2\over a^2}\alpha,\gamma +{a^2+c^2-b^2\over a^2}\alpha)$$
In like manner we get similar formulas for the reflexion point of P about CA and AB.