2021 AMC 12B Fall Problem 24

Question

Triangle

ABC

has side lengths

AB = 11,

BC = 24,

and

CA = 20.

The bisector of

\angle BAC

intersects

\overline{BC}

in point

D,

and intersects the circumcircle of

\triangle ABC

in point

E \neq A.

The circumcircle of

\triangle BED

intersects the line

AB

in points

B

and

F \neq B.

What is

CF?

28

20\sqrt{2}

30

32

20\sqrt{3}

Answer: C

Concepts:power of a point angle bisector similarity

Difficulty rating: 2650

Small hint:

Apply power of the point

A

to the circle through

B,

E,

D

:

AB \cdot AF = AD \cdot AE

Big hint:

Because

\triangle ABE \sim \triangle ADC,

we get

AD \cdot AE = AB \cdot AC,

so

AF = AC

Solution:

Points

A,

D,

E

are collinear on the bisector, and

A,

B,

F

are collinear on line

AB.

The power of

A

with respect to the circle through

B,

E,

D

gives

AB \cdot AF = AD \cdot AE.

Since

\angle BAE = \angle DAC

and

\angle AEB = \angle ACB

(subtending

AB

), triangles

ABE

and

ADC

are similar, so

AD \cdot AE = AB \cdot AC.

Therefore

AF = AC = 20.

Place

A = (0, 0),

B = (11, 0).

From

CA = 20,

CB = 24,

point

C = \left(-\tfrac52, \tfrac{\sqrt{1575}}{2}\right).

Point

F

lies on ray

AB

with

AF = 20,

so

F = (20, 0).

Then

CF^2 = \left(20 + \tfrac52\right)^2 + \tfrac{1575}{4}

= \tfrac{2025}{4} + \tfrac{1575}{4}

= 900,

so

CF = 30.

Thus, the correct answer is C.