2010 AMC 10A Problem 14

The equilateral triangle $CFE$ gives us $60^\circ$ angles for free. Use them to find angle $A$, then use $AB = 2 \cdot AC$ to finish.

Part 1: angle $A$ is $60^\circ$.

1. Call the two equal angles $x$: $\angle BAE = \angle ACD = x$.
2. $\triangle CFE$ is equilateral, so $\angle CFE = 60^\circ$. Points $A$, $F$, $E$ lie on one straight line, so the angle on the other side at $F$ is $\angle AFC = 180^\circ - 60^\circ = 120^\circ$.
3. Look at $\triangle AFC$. Its angles are $\angle FCA = x$ (this is the same angle as $\angle ACD$), $\angle AFC = 120^\circ$, and $\angle FAC$. They add to $180^\circ$, so
$$\angle FAC = 180^\circ - 120^\circ - x = 60^\circ - x.$$
4. Angle $A$ of the big triangle is made of two pieces, $\angle BAE$ and $\angle EAC$:
$$\angle BAC = x + (60^\circ - x) = 60^\circ.$$
The unknown $x$ cancels.

Part 2: angle $C$ is $90^\circ$.

1. Now we know $\angle A = 60^\circ$ and $AB = 2 \cdot AC$. Let $M$ be the midpoint of $AB$. Then $AM = MB = AC$.
2. $\triangle AMC$ has $AM = AC$ and a $60^\circ$ angle between them. An isosceles triangle with a $60^\circ$ top angle has base angles $(180^\circ - 60^\circ)/2 = 60^\circ$ too, so it is equilateral. That gives $MC = AC = MB$ and $\angle ACM = 60^\circ$, $\angle AMC = 60^\circ$.
3. $\triangle MBC$ has $MB = MC$, so it is isosceles. Its angle at $M$ is $180^\circ - 60^\circ = 120^\circ$, so its other two angles are each $(180^\circ - 120^\circ)/2 = 30^\circ$. In particular $\angle MCB = 30^\circ$.
4. So $\angle ACB = \angle ACM + \angle MCB = 60^\circ + 30^\circ = 90^\circ$.

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

An equilateral triangle hands you known angles, and following them around the point $F$ pins down angle $A$ without knowing where $D$ and $E$ are. The second part is a fact worth remembering: if a triangle has a $60^\circ$ angle and the two sides next to it are in the ratio $1 : 2$, it is a $30$-$60$-$90$ triangle, with the right angle opposite the longer side.

Once angle $A$ is $60^\circ$, use lengths instead. Say $AC = 1$ and $AB = 2$.

1. Drop a line from $C$ straight down to $AB$, meeting it at $H$ at a right angle.
2. Triangle $AHC$ has angles $60^\circ$, $90^\circ$, $30^\circ$, so its sides are in the ratio $1 : \sqrt3 : 2$. With longest side $AC = 1$: $AH = \tfrac12$ and $CH = \tfrac{\sqrt3}{2}$.
3. Then $HB = 2 - \tfrac12 = \tfrac32$, and in right triangle $CHB$:
$$BC^2 = \left(\tfrac32\right)^2 + \left(\tfrac{\sqrt3}{2}\right)^2 = \tfrac94 + \tfrac34 = 3.$$
4. Now $AC^2 + BC^2 = 1 + 3 = 4 = AB^2$. The sides fit the Pythagorean theorem, so the angle at $C$, across from $AB$, is $90^\circ$.

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