imc

Let the brackets denote areas. We are given that $$[PQR]=\frac{1}{2}\cdot PQ\cdot h_R=12.$$ Since $PQ=8,$ it follows that $h_R=3.$ We construct a circle with diameter $\overline{PQ}.$ All such locations for $R$ are shown below: We apply casework to the right angle of $△PQR:$ 1. If $\angle P=90^{\circ },$ then $R\in \{R_1,R_5\}$ by the tangent. 2. If $\angle Q=90^{\circ },$ then $R\in \{R_4,R_8\}$ by the tangent. 3. If $\angle R=90^{\circ },$ then $R\in \{R_2,R_3,R_6,R_7\}$ by the Inscribed Angle Theorem. Together, there are $\boxed{8}$ such locations for $R.$ Remarks 1. The reflections of $R_1,R_2,R_3,R_4$ about $\overleftrightarrow{PQ}$ are $R_5,R_6,R_7,R_8,$ respectively. 2. The reflections of $R_1,R_2,R_5,R_6$ about the perpendicular bisector of $\overline{PQ}$ are $R_4,R_3,R_8,R_7,$ respectively.