The parabolas with equations $y=x^{2}-2 x-3$ and $y=-x^{2}+4 x+c$ intersect at points $A$ and $B$. Determine the value of $c$ so that the sum of the $x$-coordinate and $y$-coordinate of the midpoint of $A B$ is 2017 .
Let $M$ be the midpoint of $A B$.

Let the coordinates of $A, B$ and $M$ be $\left(x_{A}, y_{A}\right),\left(x_{B}, y_{B}\right)$ and $\left(x_{M}, y_{M}\right)$, respectively.

Since $M$ is the midpoint of $A B$, then $x_{M}=\frac{x_{A}+x_{B}}{2}$ and $y_{M}=\frac{y_{A}+y_{B}}{2}$.

We want $x_{M}+y_{M}=2017$, which is equivalent to $\frac{x_{A}+x_{B}}{2}+\frac{y_{A}+y_{B}}{2}=2017$.

Thus, $x_{A}+x_{B}+y_{A}+y_{B}=4034$.

Since $A$ and $B$ are the points of intersection of the parabolas with equations $y=x^{2}-2 x-3$ and $y=-x^{2}+4 x+c$, then $x_{A}$ and $x_{B}$ are the solutions of the equation $x^{2}-2 x-3=-x^{2}+4 x+c$ or $2 x^{2}-6 x-(3+c)=0$ and so $x^{2}-3 x-\frac{1}{2}(3+c)=0$.

The sum of the roots of this quadratic equation is 3 and the product of the roots is $-\frac{1}{2}(3+c)$. In other words, $x_{A}+x_{B}=3$ and $x_{A} x_{B}=-\frac{1}{2}(3+c)$.

Since points $A\left(x_{A}, y_{A}\right)$ and $B\left(x_{B}, y_{B}\right)$ are on the parabola with equation $y=x^{2}-2 x-3$, then $y_{A}=x_{A}^{2}-2 x_{A}-3$ and $y_{B}=x_{B}^{2}-2 x_{B}-3$.

Therefore, the following equations are equivalent:

$$
\begin{aligned}
x_{A}+x_{B}+y_{A}+y_{B} & =4034 \\
x_{A}+x_{B}+\left(x_{A}^{2}-2 x_{A}-3\right)+\left(x_{B}^{2}-2 x_{B}-3\right) & =4034 \\
x_{A}^{2}+x_{B}^{2}-\left(x_{A}+x_{B}\right)-6 & =4034 \\
\left(x_{A}+x_{B}\right)^{2}-2 x_{A} x_{B}-3-6 & =4034 \\
3^{2}-2\left(-\frac{1}{2}(3+c)\right) & =4043 \\
3+c & =4034
\end{aligned}
$$

and so $c=4031$.

ANSWER: 4031