2021 AMC 12B Spring Problem 11

Question

Triangle

ABC

has

AB=13,

BC=14,

and

AC=15.

Let

P

be the point on

\overline{AC}

such that

PC=10.

There are exactly two points

D

and

E

on line

BP

such that quadrilaterals

ABCD

and

ABCE

are trapezoids. What is the distance

DE?

\dfrac{42}{5}

6\sqrt2

\dfrac{84}{5}

12\sqrt2

18

Answer: D

Concepts:coordinate geometry parallel lines trapezoid

Difficulty rating: 1690

Small hint:

Place

A=(0,0),

C=(15,0);

then

B=\left(\tfrac{33}{5},\tfrac{56}{5}\right)

and

P=(5,0)

Big hint:

One trapezoid needs

CD\parallel AB,

the other needs

AE\parallel BC;

find where each parallel line meets line

BP

Solution:

Place

A=(0,0)

and

C=(15,0).

Then

B=\left(\tfrac{33}{5},\tfrac{56}{5}\right),

and since

PC=10,

P=(5,0).

Line

BP

has slope

7,

so it is

y=7(x-5).

For

ABCD

to be a trapezoid with

D

on line

BP,

take

CD\parallel AB.

The line through

C

parallel to

AB

meets line

BP

at

(\tfrac95,-\tfrac{112}{5}).

For

ABCE

with

E

on line

BP,

take

AE\parallel BC.

The line through

A

parallel to

BC

meets line

BP

at

(\tfrac{21}{5},-\tfrac{28}{5}).

Their coordinate differences are

\tfrac{12}{5}

and

\tfrac{84}{5},

so

DE=\tfrac15\sqrt{12^2+84^2}=12\sqrt2.

Thus, the correct answer is D.

2021 AMC 12B Spring Problem 11

Answer: D

Difficulty rating: 1690

Small hint:

Big hint:

Solution:

Problem 11 in other years