Let $S$ be the set of points on the rays forming the sides of a $120^{\circ}$ angle, and let $P$ be a fixed point inside the angle 
on the angle bisector. Consider all distinct equilateral triangles $PQR$ with $Q$ and $R$ in $S$. 
(Points $Q$ and $R$ may be on the same ray, and switching the names of $Q$ and $R$ does not create a distinct triangle.) 
There are
$\text{(A) exactly 2 such triangles} \quad\\ \text{(B) exactly 3 such triangles} \quad\\ \text{(C) exactly 7 such triangles} \quad\\ \text{(D) exactly 15 such triangles} \quad\\ \text{(E) more than 15 such triangles}$
$\fbox{E}$
Take the "obvious" equilateral triangle $OAP$, where $O$ is the vertex, $A$ is on the upper ray, and $P$ is our central point.  Slide $A$ down on the top ray to point $A'$, and slide $O$ down an equal distance on the bottom ray to point $O'$.
Now observe $\triangle AA'P$ and $\triangle OO'P$.  We have $m\angle A = 60^\circ$ and $m \angle O = 60^\circ$, therefore $\angle A \cong \angle O$.  By our construction of moving the points the same distance, we have $AA' = OO'$.  Also, $AP = OP$ by the original equilateral triangle.  Therefore, by SAS congruence, $\triangle AA'P \cong \triangle OO'P$.  
Now, look at $\triangle A'PO'$.  We have $PA' = PO'$ from the above congruence.  We also have the included angle $\angle A'PO'$ is $60^\circ$.  To prove that, start with the $60^\circ$ angle $APO$, subtract the angle $APA'$, and add the congruent angle $OPO'$.
Since $\triangle A'PO'$ is an isosceles triangle with vertex of $60^\circ$, it is equilateral.