Crazy midpoints!

Let $D$ be the origin $(0,0)$ of the $xy$ -plane and $B$ be $(x,0)$ . Then $A(x-8, 0)$ , $M\left(\dfrac x2-4, 0\right)$ , and $N\left(\dfrac x2, 3\right)$ and

$\begin{aligned} MN & = \sqrt{\left(\frac x2 - 4 - \frac x2\right)^2 + (0-3)^2} = \sqrt {16 + 9} = \boxed 5 \end{aligned}$

Extend $BA$ to $E$ such that $BD = AE$

Let $AM= MD = x$

Then $BD = AE= 8-2x$ ( as $AB=8 cm$ )

Then $M$ is midpoint of $BE$ as $BM= ME = 8-x$

In $\triangle{BCE}$ , $M$ and $N$ are midpoints of $BE$ and $BC$ respectively

Therefore, by midpoint theorem 2MN = CE

In $\triangle{CDE}, D=90°, CD = 6, DE= 8$

Therefore, by Pythagoras theorem, $CE = 10$

As $2MN = CE$ , $\boxed{MN = 5 cm }$

This is the 1st problem of the 1993 RMO (Regional Mathematics Olympiad). You can find a link to that here .

Let the position coordinates of A, B and C be (h, k), (0,0) and (a, 0) respectively. Then those of M and N are (ah^2/64, ahk/64) and (a/2,0) respectively. From the given conditions, h^2+k^2=64 and a^2-(ah/8)^2=36. Using these and the distance formula we get the length of MN equal to 5