Question

In $\triangle PAT,$ $\angle P = 36^\circ,$ $\angle A = 56^\circ,$ and $PA = 10.$ Points $U$ and $G$ lie on sides $\overline{TP}$ and $\overline{TA},$ respectively, so that $PU = AG = 1.$ Let $M$ and $N$ be the midpoints of segments $\overline{PA}$ and $\overline{UG},$ respectively. What is the degree measure of the acute angle formed by lines $MN$ and $PA?$

$76$

$77$

$78$

$79$

$80$

Solution:

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 $\angle NMP = \angle QAP = \angle QAG + \angle GAP = 44^\circ + 56^\circ = 100^\circ.$ The acute angle between line $MN$ and $PA$ is therefore $180^\circ - 100^\circ = 80^\circ.$

Thus, the correct answer is E.

2018 AMC 12A Problem 23

Solution:

Problem 23 in Other Years