Cartesian plane challenge 2

The coordinates of the centroid are (xA+xB+xC/3 , yA+yB+yC/3). The coordinates of the centroid is the intersection of CK and BE which is (1,1), so xA+xB+xC+yA+yB+yC=3+3 = 6 which is the final answer.

Let the coordinates of triangle $ABC$ be modeled per the following:

$A(x_{A}, 6 - x_{A}); B(x_{B}, \frac{x_{B}+1}{2}); C(1,y_{C})$ (i)

If $CK$ be the median of $AB$ , then we require: $\frac{x_{A} + x_{B}}{2} = 1$ (ii). If $AH$ be the perpendicular altitude to $BC$ , then we require the slope to equal $m_{BC} = 1$ . This translates into:

$\frac{y_{C} - y_{B}}{x_{C}-x_{B}} = \frac{y_{C} - \frac{x_{B}+1}{2}}{1 - x_{B}} = 1 \Rightarrow x_{B} + 2y_{C} = 3$ (iii).

If $E$ is the midpoint of $AC$ and is also contained on the median modeled by the line $y = \frac{x+1}{2}$ , then we require:

$y - [\frac{(6-x_{A}) + y_{C}}{2}] = \frac{1}{2} \cdot (x - \frac{x_{A}+1}{2}) \Rightarrow y = \frac{1}{2}x + [\frac{(6-x_{A}) + y_{C}}{2} - \frac{x_{A}+1}{4}]$ ;

or $\frac{(6-x_{A}) + y_{C}}{2} - \frac{x_{A}+1}{4} = \frac{1}{2} \Rightarrow -3x_{A} + 2y_{C} = -9$ (iv).

Let us now take the equations from (ii), (iii), and (iv) and solve for $x_{A}, x_{B}, y_{C}$ via a 3x3 matrix:

$\begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 2 \\ -3 & 0 & 2 \end{bmatrix} \cdot \begin{bmatrix} x_{A} \\ x_{B} \\y_{C} \end{bmatrix} = \begin{bmatrix} 2 \\ 3 \\ -9 \end{bmatrix}$ (v)

which solving for (v) gives $\begin{bmatrix} x_{A}\\x_{B}\\y_{C} \end{bmatrix} = \begin{bmatrix} 5 \\ -3 \\3 \end{bmatrix}$

and the coordinates $A(5,1); B(-3,-1); C(1,3)$ . Thus the required coordinate sum computes to $\boxed{6}.$