重心坐标系中的圆

Circle Equation

-- Note: http://www.mit.edu/~evanchen/handouts/bary/bary-full.pdf is really great for learning barycentric coordinates.

It can be proven that for a displacement vector $\vec{PQ} = (x, y, z)$, that $|\vec{PQ}|^{2} = -a^{2}yz - b^{2}xz - c^{2}xy$. For reference, see https://www.expii.com/t/distance-in-barycentric-coordinates-1432?type=explanation&id=1560. From this, we can obtain an equation for the general circle in barycentric coordinates

Let a circle $\omega$ have center $(j, k, l)$ and radius $\sqrt{C}$. Then, since a circle is defined as the set of points equidistant from the center, the circle satisfies $$-a^{2}(y-k)(z-l) - b^{2}(x-j)(z-l) - c^{2}(x-j)(y-k) = C.$$ After some expanding, we have $$-a^{2}yz - b^{2}xz - c^{2}xy +C_{1}x + C_{2}y + C_{3}z = C$$ where the $C$s are arbitrary constants. Then, write $C = C(x+y+z)$ since $x + y + z = 1$. Then subtracting $C(x+y+z)$ and letting $u = C_{1} - C$, $v = C_{2} - C$, and $w = C_{3} - C$, we have $$-a^{2}yz - b^{2}xz - c^{2}xy +ux + vy + wz = 0.$$ To homogenize this equation, we write it as $$-a^{2}yz -b^{2}xz - c^{2}xy + (x+y+z)(ux + vy + wz) = 0$$ for arbitrary constants $u, v, w$. Note that this equation is homogeneous, so we don't need $x + y + z = 1$.

Next are some problems:

Problem 1: Find an equation for the circumcircle of $\triangle ABC$ in barycentric coordinates.

Solution 1: From $A = (1, 0, 0)$, $B = (0, 1, 0)$, and $C = (0, 0, 1)$, we have that $u = 0$, $v = 0$, and $w = 0$ respectively, so the equation for the circumcircle after multiplying both sides by $-1$ is $$a^{2}yz + b^{2}xz + c^{2}xy = 0$$

Problem 2 (Source: Mongolia TST 2000): Let $A_{1}$, $B_{1}$, and $C_{1}$ be the feet of the angle bisectors from vertices $A$, $B$, $C$ respectively. Given that quadrilateral $BA_{1}B_{1}C_{1}$ is cyclic, prove that $$\frac{AC}{AB + BC} = \frac{AB}{AC + BC} +\frac{BC}{BA + AC}$$

Solution 2: We note that $A_{1} = (0 : b : c)$, $B_{1} = (a : 0 : c)$, $C_{1} = (a : b : 0)$, and $B = (0 : 1 : 0)$. Denote $\omega = (BA_{1}B_{1}C_{1})$, with equation $-a^{2}yz - b^{2}xz - c^{2}xy + (x + y + z)(ux + vy + wz) = 0$. Since $B$ is on $\omega$ we get $v = 0$. Plugging in $A_{1}$ and $C_{1}$ can allow us to solve for $u$ and $w$ as well. From plugging in $C_{1}$, we have $$-c^{2}ab + au(a + b) = 0 \implies$$ $$u =\frac{c^{2}ab}{a(a+b)} = \frac{c^{2}b}{a + b}.$$ From plugging in $A_{1}$, we have $$-a^{2}bc + wc(b + c) = 0 \implies$$ $$w = \frac{a^{2}bc}{c(b+c)} = \frac{a^{2}b}{b + c}$$ It just remains to plug in $B_{1}$. Plugging in $B_{1}$ gives $$-b^{2}ac + \left(\frac{c^{2}b}{a+b}a + \frac{a^{2}b}{b+c}c\right)(a+c) = 0 \implies$$ $$\frac{c^{2}ab}{a+b} + \frac{a^{2}bc}{b+c} = \frac{b^{2}ac}{a + c}\implies$$ $$\frac{c}{a+b} + \frac{a}{b+c} = \frac{b}{a + c}$$ which is the desired result.