2016 AMC 10B Problem 19

Attempt Problem 19 of the 2016 AMC 10B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2016 AMC 10B solutions, or check the answer key.

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19.

Rectangle

ABCD

has

AB=5

and

BC=4.

Point

E

lies on

\overline{AB}

so that

EB=1,

point

G

lies on

\overline{BC}

so that

CG=1,

and point

F

lies on

\overline{CD}

so that

DF=2.

Segments

\overline{AG}

and

\overline{AC}

intersect

\overline{EF}

at

Q

and

P,

respectively. What is the value of

\dfrac{PQ}{EF}?

~\dfrac{\sqrt{3}}{16}

~\dfrac{\sqrt{2}}{13}

~\dfrac{9}{82}

~\dfrac{10}{91}

~\dfrac19

Answer: D

Concepts:similarity parallel lines rectangle

Difficulty rating: 1970

Small hint:

First find where

AC

and

AG

cut

EF

Big hint:

Use similar triangles to express

\frac{PF}{EF}

and

\frac{QF}{EF}

Solution:

We have

AE=AB-EB=4

and

FC=DC-DF=3.

Since

AE\parallel FC,

the triangles

AEP

and

CFP

are similar. Therefore

\frac{PF}{PE}=\frac{FC}{AE}=\frac34,

so

\frac{PF}{EF}=\frac37.

Extend

AG

to meet line

CD

at

X.

Because

BG=3,

similarity gives

\frac{AD}{DX}=\frac{BG}{AB},\qquad \frac4{DX}=\frac35,

and hence

DX=\frac{20}{3}

. Thus

FX=DX-DF=\frac{14}{3}.

Since

AE\parallel FX,

triangles

AEQ

and

XFQ

are similar, so

\frac{QF}{QE}=\frac{FX}{AE}=\frac76.

Consequently

\frac{QF}{EF}=\frac7{13}.

It follows that

\begin{aligned} \frac{PQ}{EF}&=\frac{QF}{EF}-\frac{PF}{EF} \\ &=\frac7{13}-\frac37=\frac{10}{91}. \end{aligned}

Thus, the correct answer is D .