help geometry

Connect QR. Then O is the centroid of $\triangle PQR$.

PR = 16 implies NR = 8.

By property of centroid, QO : ON = 2 : 1. Then QN = 10 implies ON = 10/3.

Then,

$OR = \sqrt{8^2 + \left(\dfrac{10}3\right)^2} = \dfrac{26}3$

In triangle PMR  Draw atltitude MS   to  PR

Now  triangles  MPS and  QPN are similar

So    MP  /  QP   =  PS  / PN       

          1/2 =  PS /PN   ⇒   PS   = 4   ⇒  MS = (1/2) QN  =  5

And triangles   RSM  and  RNO   are similar

So    SR   = 12      MS  = 5     RN  =  8

MS / SR   =  ON  / RN

5 / 12  =  ON / 8

40 /12  =  ON   =   10/3

And since   triangle RNO  is right, then

OR   = sqrt   [ RN^2  + ON^2  ]   =   sqrt  [  8^2  + (10/3)^2 ]  =

sqrt [  64 +  100 / 9  ]  =  sqrt  [ (576 + 100) / 9  ]   =   sqrt  [ 676 ] / 3  =   26  / 3

Because $PN = 8$, $QN = 10$, and $\angle N = 90$, $PQ = \sqrt{164} = 2\sqrt{41}$

We know that $M$ is the midpoint, so $PM = \sqrt{41}$.

Now, draw $MA$, so that $MA$ is perpendicular to $PR$.

Because of similar triangles, $MA = 5$ and $AN = 4$

Using the Pythagorean Theorem, we find that $RM = 13$.

Because of similar triangles, $OR = {2 \over 3} \times 13 = \color{brown}\boxed{26\over3}$