2016 AMC 10A Problem 19

Problem 19 of 25HarderGeometry

In rectangle

ABCD,

AB=6

and

BC=3.

Point

E

between

B

and

C,

and point

F

between

E

and

C

are such that

BE=EF=FC.

Segments

\overline{AE}

and

\overline{AF}

intersect

\overline{BD}

at

P

and

Q,

respectively. The ratio

BP:PQ:QD

can be written as

r:s:t

where the greatest common factor of

r,

s,

and

t

is

1.

What is

r+s+t?

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Solution

Since

BC=3

and

BE=EF=FC

, we have

BE=1

and

BF=2

. Also

AD=3

. From

\triangle APD\sim\triangle EPB

,

\frac{BP}{BD}=\frac{BE}{AD+BE}=\frac14.

From

\triangle AQD\sim\triangle FQB

,

\frac{BQ}{BD}=\frac{BF}{AD+BF}=\frac25.

Therefore

BP=\frac14BD

,

PQ=\left(\frac25-\frac14\right)BD=\frac3{20}BD

, and

QD=\frac35BD

. Thus

\begin{aligned} BP:PQ:QD &= \frac14:\frac3{20}:\frac35 \\ &= 5:3:12. \end{aligned}

The sum is

5+3+12=20

. Thus, the correct answer is E.

More practice

Concepts: similarity · rectangle

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.