Flying In The Middle Of Nowhere

See Radical Axis and Nine Point Circle before reading this solution.

This is a hard problem. The goal of this solution is to prove that $K$ is the center of Nine-Point Circle of triangle $ABC$ , which means, $K$ is the midpoint of segment $OH$ . 'Cause it's difficult to prove directly, I let $K'$ be the midpoint of segment $OH$ , then show that $AK'$ is perpendicular to $MN$ .

You may know this result: The circumcenter of triangle $BHC$ reflects that of triangle $BAC$ on line $BC$ , and, $AH = OO'$ . So, $AHO'O$ is a parallelogram. But $K'$ is midpoint of $OH$ then $A, K', O'$ are colinear.

Now, draw the Nine Point Circle of triangle $ABC$ and let it intersects the circumcircle of triangle $BHC$ at $X, Y$ . Line $XY$ is the radical axis of the 2 circles. We'll prove that points $M, N$ also lie on this line.

$HFBD$ is a concylic quadrilateral; so, $NF \times ND = NB \times NH$ . But, $NF \times ND$ is the power of point $N$ with respect to the circle $(K')$ . And, $NB \times NH$ is the power of point $N$ to the circle $(O')$ . Hence, the power of $N$ with respect to these circles are equal; thus, $N \in XY$ . Also, $M \in XY$ .

Radical axis is perpendicular to the line joining the centers; thus, $XY$ is perpendicular to $K'O'$ ; or, $MN$ is perpendicular to $AK'$ (prove complete!). Now $K \equiv K'$ , or $K$ is midpoint of $OH$ .

The last part is easy now. $S_{ABC}=\dfrac12P_{DEF}\times R$ , with $R$ be the radius of $(O)$ . We write the equation in form: $R=2\times \dfrac{S_{ABC}}{P_{DEF}}$ .

$KD$ equals to the radius of the Nine Point Circle; thus, $KD = \dfrac12 R = \dfrac{S_{ABC}}{P_{DEF}} = \dfrac{2016^{2017}}{1344^{2016}} = \left(\dfrac{2016}{1344}\right)^{2016}\times 2016 = \color{#D61F06}{\dfrac{3^{2018} \times 7}{2^{2011}}}$ .

Hence, the answer is $3 \times 2018 + 7 + 2 \times 2011 = \boxed{10083}$ .