Let $M$ be the intersection of diagonals of the convex quadrilateral $ABCD$, where $m(\widehat{AMB})=60^\circ$. Let the points $O_1$, $O_2$, $O_3$, $O_4$ be the circumcenters of the triangles $ABM$, $BCM$, $CDM$, $DAM$, respectively. What is $Area(ABCD)/Area(O_1O_2O_3O_4)$?

$ 
\textbf{(A)}\ \dfrac 12
\qquad\textbf{(B)}\ \dfrac 32
\qquad\textbf{(C)}\ \dfrac {\sqrt 3}2
\qquad\textbf{(D)}\ \dfrac {1+2\sqrt 3}2
\qquad\textbf{(E)}\ \dfrac {1+\sqrt 3}2
$
1. **Calculate the area of quadrilateral \(ABCD\):**
   - The area of \(ABCD\) can be divided into the sum of the areas of triangles \(ABM\), \(BCM\), \(CDM\), and \(DAM\).
   - The area of each triangle can be calculated using the formula for the area of a triangle: 
     \[
     \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \times \sin(\text{angle})
     \]
   - Given \(m(\widehat{AMB}) = 60^\circ\), we have:
     \[
     [ABM] = \frac{1}{2} \times AM \times BM \times \sin(60^\circ) = \frac{1}{2} \times AM \times BM \times \frac{\sqrt{3}}{2} = \frac{AM \times BM \times \sqrt{3}}{4}
     \]
   - Similarly, we can calculate the areas of the other triangles:
     \[
     [BCM] = \frac{1}{2} \times BM \times CM \times \sin(60^\circ) = \frac{BM \times CM \times \sqrt{3}}{4}
     \]
     \[
     [CDM] = \frac{1}{2} \times CM \times DM \times \sin(60^\circ) = \frac{CM \times DM \times \sqrt{3}}{4}
     \]
     \[
     [DAM] = \frac{1}{2} \times DM \times AM \times \sin(60^\circ) = \frac{DM \times AM \times \sqrt{3}}{4}
     \]
   - Summing these areas, we get:
     \[
     [ABCD] = [ABM] + [BCM] + [CDM] + [DAM] = \frac{AM \times BM \times \sqrt{3}}{4} + \frac{BM \times CM \times \sqrt{3}}{4} + \frac{CM \times DM \times \sqrt{3}}{4} + \frac{DM \times AM \times \sqrt{3}}{4}
     \]
   - Since \(M\) is the intersection of the diagonals, we can express the area in terms of the diagonals \(AC\) and \(BD\):
     \[
     [ABCD] = \frac{AC \times BD \times \sqrt{3}}{4}
     \]

2. **Calculate the area of quadrilateral \(O_1O_2O_3O_4\):**
   - The points \(O_1, O_2, O_3, O_4\) are the circumcenters of triangles \(ABM\), \(BCM\), \(CDM\), and \(DAM\), respectively.
   - The quadrilateral \(O_1O_2O_3O_4\) is a parallelogram because the perpendicular bisectors of the sides of the triangles intersect at right angles.
   - The area of the parallelogram \(O_1O_2O_3O_4\) can be calculated using the formula for the area of a parallelogram:
     \[
     \text{Area} = \text{base} \times \text{height}
     \]
   - The length of the sides of the parallelogram can be related to the diagonals \(AC\) and \(BD\) of the original quadrilateral \(ABCD\):
     \[
     O_1O_2 = O_3O_4 = \frac{BD}{2 \sin(60^\circ)} = \frac{BD}{\sqrt{3}}
     \]
     \[
     O_1O_4 = O_2O_3 = \frac{AC}{2 \sin(60^\circ)} = \frac{AC}{\sqrt{3}}
     \]
   - Therefore, the area of \(O_1O_2O_3O_4\) is:
     \[
     [O_1O_2O_3O_4] = \frac{AC}{\sqrt{3}} \times \frac{BD}{\sqrt{3}} = \frac{AC \times BD}{3}
     \]

3. **Calculate the ratio of the areas:**
   - The ratio of the area of \(ABCD\) to the area of \(O_1O_2O_3O_4\) is:
     \[
     \frac{[ABCD]}{[O_1O_2O_3O_4]} = \frac{\frac{AC \times BD \times \sqrt{3}}{4}}{\frac{AC \times BD}{3}} = \frac{\frac{\sqrt{3}}{4}}{\frac{1}{3}} = \frac{\sqrt{3}}{4} \times 3 = \frac{3\sqrt{3}}{4} = \frac{3}{2}
     \]

The final answer is \(\boxed{\frac{3}{2}}\)