2010 AMC 10A Problem 14

Let $\theta = \angle BAE = \angle ACD$. Since $\triangle CFE$ is equilateral, $\angle CFE = 60^\circ$, and $F$ lies on segment $AE$, so the supplementary angle $\angle AFC = 120^\circ$.

In $\triangle AFC$ the angles are $\angle FAC$, $\angle FCA = \theta$, and $120^\circ$, hence $\angle FAC = 60^\circ - \theta$. Therefore
$$\angle BAC = \angle BAE + \angle EAC = \theta + (60^\circ - \theta) = 60^\circ.$$

Now use $AB = 2\,AC$. Let $M$ be the midpoint of $AB$, so $AM = AC$. Triangle $AMC$ has two equal sides enclosing a $60^\circ$ angle, so it is equilateral and $MC = MA = MB$. Thus $M$ is equidistant from $A$, $B$, $C$: it is the circumcenter of $\triangle ABC$, and $AB$ is a diameter. The angle at $C$ subtends that diameter, so $\angle ACB = 90^\circ$.

The answer is $\boxed{\textbf{(C)}\ 90^\circ}$.

The equilateral triangle is a source of known angles; chasing them through the intersection point $F$ pins down $\angle A$ without ever locating $D$ or $E$. The second half is a standard fact worth memorizing: a triangle with a $60^\circ$ angle whose adjacent sides are in ratio $2 : 1$ is a $30$-$60$-$90$ triangle.

With $\angle A = 60^\circ$, $AC = 1$, $AB = 2$, the Law of Cosines gives $BC^2 = 1 + 4 - 2 \cdot 2 \cdot 1 \cdot \tfrac12 = 3$. Then $AC^2 + BC^2 = 1 + 3 = 4 = AB^2$, so the angle at $C$ is right by the converse of the Pythagorean theorem.

• Assuming the 60-degree angles of triangle CFE transfer directly to angle ACB (answer 60) without using AB = 2AC.
• Trying to find $\theta$ itself; it cancels, and only $\angle BAC = 60^\circ$ matters.