# Reflection of point in barycentric coordinates

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.

• I wonder if it might be easier to convert to trilinear coordinates first. – amd Oct 26 '18 at 22:16
• If $P = A$, then you should get $P^\prime = A$. That is, your $x$ should vanish when $\alpha = 1$, $\beta = \gamma = 0$. I don't think it does. – Blue Oct 26 '18 at 23:08

Let's try some brute force with a special case. Assign Cartesian coordinates $$A := (0,0) \qquad B := (c,0) \qquad C := (b\cos A, b\sin A) \tag{1}$$ and define $$P$$ with barycentric coordinates $$(\alpha, \beta, \gamma)$$, so that $$P = \frac{\alpha A + \beta B + \gamma C}{\alpha + \beta + \gamma} = \frac{(\beta c+\gamma b \cos A, \gamma b \sin A )}{\alpha+\beta+\gamma} =: (P_x,P_y) \tag{2}$$ Let $$P^\prime = (\alpha^\prime, \beta^\prime, \gamma^\prime)$$ be the reflection of $$P$$ in $$\overline{AB}$$ (the $$x$$-axis). Then $$P^\prime_x=P_x$$ and $$P^\prime_y=-P_y$$, giving the equations \begin{align} (\beta\,c+\gamma\,b \cos A)(\alpha^\prime+\beta^\prime+\gamma^\prime)&=(\beta^\prime\,c+\gamma^\prime\,b\cos A)(\alpha+\beta+\gamma) \\[4pt] \gamma\,(\alpha^\prime+\beta^\prime+\gamma^\prime)&= - \gamma^\prime\,(\alpha+\beta+\gamma) \end{align}\tag{3} Solving the system for, say, $$\alpha^\prime$$ and $$\beta^\prime$$ gives \begin{align} \alpha^\prime &= -\frac{\gamma^\prime}{\gamma\,c} \left(\;\alpha\,c + 2 \gamma\,(c - b \cos A)\;\right) = -\frac{\gamma^\prime}{\gamma\,c} \left(\;\alpha\,c + 2 \gamma\,a \cos B\;\right) \\[4pt] \beta^\prime &= -\frac{\gamma^\prime}{\gamma\,c}\left(\;\beta\,c + 2 \gamma\,b \cos A\;\right) \end{align}\tag{4} from which we can deduce barycentric coordinates, in a smattering of variants,

\begin{align} \alpha^\prime:\beta^\prime:\gamma^\prime\quad&=\quad \alpha\,c + 2 \gamma\,a \cos B\;:\; \beta\,c + 2 \gamma\,b \cos A\;:\; -\gamma\,c \\[8pt] \quad&=\quad \alpha+\frac{2\gamma\,a\cos B}{c}:\beta+\frac{2\gamma\,b \cos A}{c} : \gamma - 2\gamma \\[8pt] \quad&=\quad \alpha+2\gamma\,\sin A\cos B \csc C:\beta+2\gamma\,\cos A\sin B\csc C : \gamma - 2\gamma \\[8pt] \quad&=\quad \alpha+\frac{2\gamma\,S_B}{c^2}:\beta+\frac{2\gamma\,S_A}{c^2} : \gamma - 2\gamma \end{align} \tag{5}

As a sanity check:

• Any point on the $$x$$-axis has $$\gamma=0$$; reflection fixes the such a point, and we see from $$(5)$$ that, indeed $$\alpha^\prime : \beta^\prime : 0 = \alpha:\beta:0$$.

• The reflection of $$C$$, which has barycentric coordinates $$(0,0,1)$$, should have Cartesian coordinates $$(b\cos A,-b\sin A)$$; from the first form in $$(5)$$, \begin{align} \frac{\alpha^\prime A + \beta^\prime B + \gamma^\prime C}{\alpha^\prime+\beta^\prime+\gamma^\prime} &= \frac{A\cdot 2a\cos B + B\cdot 2b\cos A -C\cdot c}{2a\cos B+2b\cos A-c} \\[8pt] &=\frac{(2bc\cos A-bc \cos A, -bc\sin A)}{2c-c} \\[8pt] &= (b\cos A,-b\sin A) \end{align} \tag{6} as expected.

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.