2018 AMC 12A Problem 23

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

Answer: E

Concepts:angle chasing midpoint parallelogram

Difficulty rating: 2370

Small hint:

Extend

PN

to

Q

with

PN = NQ;

then

UPGQ

is a parallelogram, so

GQ \parallel PT

and

GQ = PU = AG

Big hint:

Triangle

QGA

is isosceles, and

MN

is a midline of triangle

QPA,

so

\angle NMA = \angle QAP

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

\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.