2010 AMC 12A Problem 8

Attempt Problem 8 of the 2010 AMC 12A 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 2010 AMC 12A solutions, or check the answer key.

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8.

Triangle

ABC

has

AB=2 \cdot AC.

Let

D

and

E

be on

\overline{AB}

and

\overline{BC},

respectively, such that

\angle BAE = \angle ACD.

Let

F

be the intersection of segments

AE

and

CD,

and suppose that

\triangle CFE

is equilateral. What is

\angle ACB?

60^\circ

75^\circ

90^\circ

105^\circ

120^\circ

Answer: C

Concepts:angle chasing equilateral triangle special right triangle

Difficulty rating: 1660

Small hint:

Let

\angle BAE=\angle ACD=x

Big hint:

Use

\triangle CFE

to find

\angle AFC,

then angle-chase

\angle BAC

Solution:

Let

\angle BAE=\angle ACD=x.

Because

\triangle CFE

is equilateral,

\angle CFE=60^\circ.

Rays

FA

and

FE

are opposite, so

\angle AFC=120^\circ.

In

\triangle AFC,

we get

\angle FAC=180^\circ-120^\circ-x

=60^\circ-x.

Since

F

lies on

AE,

this is

\angle EAC.

Hence

\angle BAC=x+(60^\circ-x)=60^\circ.

Put

AC=s,

so

AB=2s.

The Law of Cosines gives

\begin{aligned} BC^2 &= s^2+(2s)^2 \\ &\quad-2(s)(2s)\cos60^\circ \\ &=3s^2. \end{aligned}

Thus the side lengths are in the ratio

1:\sqrt3:2,

with

AB

the hypotenuse, so

\angle ACB=90^\circ.

Thus, C is the correct answer.

Problem 8 in other years