1991 AMC 12 Problem 19

Attempt Problem 19 of the 1991 AMC 12 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 1991 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

19.

Triangle

ABC

has a right angle at

C,

AC=3,

and

BC=4.

Triangle

ABD

has a right angle at

A

and

AD=12.

Points

C

and

D

are on opposite sides of

AB.

The line through

D

parallel to

AC

meets

CB

extended at

E.

If

\frac{DE}{DB}=\frac mn,

where

m

and

n

are relatively prime positive integers, then

m+n=

25

128

153

243

256

Answer: B

Concepts:coordinate geometry perpendicular vectors distance

Difficulty rating: 2110

Small hint:

Place

C=(0,0),

A=(0,3),

and

B=(4,0)

Big hint:

Find

D

by using a length-

12

vector perpendicular to

\overrightarrow{AB}

Solution:

Place

C=(0,0),

A=(0,3),

and

B=(4,0).

Since

\overrightarrow{AB}=(4,-3),

the length-

12

vector perpendicular to

\overrightarrow{AB}

and directed away from

C

is

(\frac{36}{5},\frac{48}{5}).

Hence

D=(\frac{36}{5},\frac{63}{5}).

Since

DE\parallel AC,

we have

E=(\frac{36}{5},0),

so

DE=\frac{63}{5}

and

DB=13.

Therefore

\frac{DE}{DB}=\frac{\frac{63}{5}}{13}=\frac{63}{65}.

Thus

m+n=63+65=128.

Thus the correct answer is B.

Problem 19 in other years