2001 AMC 12 Problem 24

Attempt Problem 24 of the 2001 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 2001 AMC 12 solutions, or check the answer key.

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

24.

In triangle

ABC,

\angle ABC = 45^\circ.

Point

D

is on

\overline{BC}

so that

2 \cdot BD = CD

and

\angle DAB = 15^\circ.

Find

\angle ACB.

54^\circ

60^\circ

72^\circ

75^\circ

90^\circ

Answer: D

Concepts:angle chasing isosceles triangle special right triangle

Difficulty rating: 2110

Small hint:

The exterior angle gives

\angle ADC

= \angle DAB + \angle ABD

= 15^\circ + 45^\circ

= 60^\circ

Big hint:

Drop a perpendicular from

C

to line

AD

at

E;

then

\triangle CDE

is

30

-

60

-

90

with

DE = \tfrac{1}{2}CD = BD

Solution:

Let

E

be the foot of the perpendicular from

C

to line

AD.

The exterior angle of

\triangle ADB

gives

\angle ADC = 15^\circ + 45^\circ = 60^\circ,

so

\triangle CDE

is a

30

-

60

-

90

triangle with

DE = \tfrac{1}{2}CD = BD.

Then

\triangle BDE

is isosceles with

\angle EBD = \angle BED = 30^\circ,

and since

\angle ECB = 30^\circ

too,

\triangle BEC

is isosceles with

BE = EC.

Also

\angle ABE

= 45^\circ - 30^\circ

= 15^\circ

= \angle EAB,

so

\triangle ABE

is isosceles with

AE = BE.

Hence

AE = BE = EC,

making right triangle

AEC

isosceles with

\angle ECA = 45^\circ.

Therefore

\angle ACB

= \angle ECA + \angle ECD

= 45^\circ + 30^\circ

= 75^\circ.

Thus, the correct answer is D.

Problem 24 in other years