Day 22: A Forest of Lines

First I shall name the other ends of the trisectors $H, G, E$ and $F$ clockwise from $A$ and call $\angle FAC$ $\alpha$ and $\angle HCA$ $\beta$ :

First notice that $\alpha + \beta = 30^{\circ}$ due to the angle sum in a triangle.

Now $APFB$ is cyclic so $\angle PBF = \alpha$ . We use the same trick to show that $\angle MBH = \beta$ . So $\angle MBP = 90^{\circ} - (\alpha+ \beta) = 60^{\circ}$ .

We can apply a similar method on cyclic quadrilaterals $AOEB$ and $CNGB$ to show that $\angle NBO = 90^{\circ} - 2(\alpha+ \beta) = 30^{\circ}$ .

Therefore $\angle MBN +\angle OBP = 60^{\circ} - 30^{\circ} = \boxed{30^{\circ}}$