smc

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 $A^{\prime }$, and slide $O$ down an equal distance on the bottom ray to point $O^{\prime }$. Now observe $△A{A}^{\prime }P$ and $△O{O}^{\prime }P$. 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 $A{A}^{\prime }=O{O}^{\prime }$. Also, $AP=OP$ by the original equilateral triangle. Therefore, by SAS congruence, $△A{A}^{\prime }P\cong △O{O}^{\prime }P$. Now, look at $△A^{\prime }P{O}^{\prime }$. We have $P{A}^{\prime }=P{O}^{\prime }$ from the above congruence. We also have the included angle $\angle A^{\prime }P{O}^{\prime }$ is $60^{\circ }$. To prove that, start with the $60^{\circ }$ angle $APO$, subtract the angle $AP{A}^{\prime }$, and add the congruent angle $OP{O}^{\prime }$. Since $△A^{\prime }P{O}^{\prime }$ is an isosceles triangle with vertex of $60^{\circ }$, it is equilateral. $\boxed{\text{more than 15 such triangles}}$