Second Point Of Intersection Of Circumcircles

OK, this is a horrible problem with a very inelegant solution. Apologies. The original problem asked to show that the second intersection point of the circumcircles of $\triangle BCD, \triangle ADE$ and $\triangle BCE, \triangle ADE$ are equidistant from $A$ (which should be obvious from the following solution). The reader might refer to this helpful article on barycentric coordinates.

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Set the normalized barycentric coordinates $A= (1, 0, 0), B = (0, 1, 0), C = (0, 0, 1).$ Then, the coordinates of $D, E$ are $\left( \dfrac{1}{2}, \dfrac{1}{2}, 0 \right)$ and $\left( \dfrac{1}{2}, 0, \dfrac{1}{2} \right)$ respectively.

Suppose the equation of circle $(ADE)$ is $a^2yz+b^2zx+c^2xy = ux+vy+wz$ (normalized coordinates). Plugging the coordinates of $A,$ we see that $u=0.$ Plugging the coordinates of $D,$ we see that $\dfrac{c^2}{4} = \dfrac{u+v}{2} \implies v = \dfrac{c^2}{2}.$ Similarly, plugging the coordinates of $E$ shows that $w= \dfrac{b^2}{2}.$ To sum it up, the equation of the circumcircle of $\triangle ADE$ is $a^2yz+b^2zx+c^2xy= \dfrac{c^2y + b^2z}{2}.$

Let the equation of $(BCD)$ be $a^2yz + b^2zx + c^2xy = px+qy+rz.$ Plugging the coordinates for $B$ and $C$ show that $q=r=0.$ Plugging the coordinates of $D,$ we see that $\dfrac{c^2}{4} = \dfrac{p}{2} \implies p = \dfrac{c^2}{2}.$ To sum it up, the equation of the circumcircle of $\triangle BCD$ is $a^2yz+b^2zx + c^2xy = \dfrac{c^2x}{2}.$

Let $(x, y, z)$ be the point of intersection of $(ADE)$ and $(BCD)$ apart from $D$ (where $x+y+z=1$ ).

The displacement vector $\vec{AX}$ is $(x-1, y, z),$ so $\begin{aligned} |AX|^2 & = - a^2 yz - b^2 z (x-1) - c^2 (x-1)y \\ & = - (a^2yz + b^2zx + c^2xy) + (b^2z + c^2y) \\ &= - \dfrac{c^2y + b^2z}{2} + b^2z + c^2y \\ &= \dfrac{b^2z + c^2y}{2} \\ & = a^2yz + b^2zx + c^2xy \\ & = \dfrac{c^2x}{2}. \end{aligned}$

Now, we know that $x, y, z$ simultaneously satisfy the equations $\begin{aligned}a^2yz+b^2zx + c^2xy & = \dfrac{c^2x}{2} \\ a^2yz+b^2zx+c^2xy &= \dfrac{c^2y + b^2z}{2} \\ x + y + z & = 1. \end{aligned}$ Equating $a^2yz+b^2zx+c^2xy$ from the first two equations and plugging $z = 1-x-y,$ we obtain $c^2x = c^2y + b^2 (1 - x - y) \implies y = \dfrac{(b^2+c^2)x-b^2}{c^2-b^2}.$
This implies $z = 1 - \dfrac{(b^2+c^2)x-b^2}{c^2-b^2} - x = \dfrac{c^2}{b^2-c^2}(2x-1).$

Plugging this in the second equation, we obtain $2a^2 \dfrac{(b^2+c^2)x-b^2}{c^2-b^2} \dfrac{c^2}{b^2-c^2}(2x-1) + 2b^2 \dfrac{(b^2+c^2)x-b^2}{c^2-b^2} x + 2c^2\dfrac{(b^2+c^2)x-c^2}{c^2-b^2}x = 0.$

We interpret this as a quadratic in $x.$ We know that $x = \dfrac{1}{2}$ is one solution (since $D = \left( \dfrac{1}{2}, \dfrac{1}{2}, 0 \right)$ is a point of intersction of the circles in question). To find the other solution, it suffices to find the ratio of the constant term and coefficient of $x^2.$ Collecting terms, we find out that this ratio is $\dfrac{[x^0]}{[x^2]} = \dfrac{a^2b^2}{2a^2(b^2+c^2) - (b^2-c^2)^2}.$ Hence, the other solution for $x$ is $\dfrac{2a^2b^2}{2a^2(b^2+c^2) - (b^2-c^2)^2}.$ Now, $|AX|^2 = \dfrac{c^2x}{2} = \dfrac{a^2b^2c^2}{2a^2(b^2+c^2) - (b^2-c^2)^2}$ .

Plugging the values shows us that the last three digits of $a+b$ are $\boxed{883}.$

Note that the expression of $AX^2$ is symmetric in $b, c,$ which solves the original problem.