Question

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

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.

2010 AMC 12A Problem 8

Answer: C

Difficulty rating: 1660

Solution:

Problem 8 in Other Years