2018 AMC 12A Problem 23

Extend $PN$ through $N$ to $Q$ with $PN = NQ.$ Since $N$ is the midpoint of $UG$ and of $PQ,$ the quadrilateral $UPGQ$ is a parallelogram, so $GQ \parallel PT$ and $GQ = PU = 1 = AG.$ Then $\angle QGA$ $= 180^\circ - \angle T$ $= \angle P + \angle A$ $= 36^\circ + 56^\circ = 92^\circ,$ and the isosceles triangle $QGA$ gives $\angle QAG = \tfrac12(180^\circ - 92^\circ) = 44^\circ.$ Because $M, N$ are midpoints, $MN$ is a midline of $\triangle QPA,$ so $MN \parallel AQ$ and $$\begin{aligned} \angle NMP &= \angle QAP \\ &= \angle QAG + \angle GAP \\ &= 44^\circ + 56^\circ = 100^\circ. \end{aligned}$$ The acute angle between line $MN$ and $PA$ is therefore $180^\circ - 100^\circ = 80^\circ.$ Thus, the correct answer is E.