Medians $AM$ and $BE$ of a triangle $ABC$  intersect at $O$. The points $O, M, E, C$ lie on one circle. Find the length of $AB$ if $BE = AM =3$.
1. **Claim 1: If two medians of a triangle are equal, then it is an isosceles triangle.**

   Given that \(AM\) and \(BE\) are medians of \(\triangle ABC\) and they intersect at \(O\), the centroid. Since \(AM = BE = 3\), we can use the properties of medians and centroids to deduce that \(\triangle ABC\) is isosceles.

2. **Centroid Properties:**

   The centroid \(O\) divides each median into a ratio of \(2:1\). Therefore, \(AO = \frac{2}{3}AM\) and \(BO = \frac{2}{3}BE\). Since \(AM = BE\), it follows that \(AO = BO\).

3. **Congruence of Triangles:**

   By the Side-Angle-Side (SAS) Congruence Theorem, \(\triangle AOM \cong \triangle BOM\) because:
   - \(AO = BO\)
   - \(OM\) is common
   - \(\angle AOM = \angle BOM\) (since \(O\) is the centroid and divides the medians equally)

   Hence, \(\triangle AOM \cong \triangle BOM\).

4. **Isosceles Triangle:**

   Since \(\triangle AOM \cong \triangle BOM\), it follows that \(AM = BM\). Therefore, \(\triangle ABC\) is isosceles with \(AC = BC\).

5. **Cyclic Quadrilateral:**

   Given that points \(O, M, E, C\) lie on one circle, we can use the properties of cyclic quadrilaterals. Specifically, opposite angles of a cyclic quadrilateral sum to \(180^\circ\).

6. **Angle Chasing:**

   Let \(\angle A = \angle B = \angle CEM = \angle CME\). By angle chasing and using the fact that \(CC_1 \perp EM\) and \(\angle COE = \angle COM = x\), we have:
   \[
   \angle MEO = \angle EMO = 90^\circ - x \implies \angle BEC = \angle AMC = 90^\circ
   \]

7. **Equilateral Triangle:**

   Since the medians are also altitudes, \(\triangle ABC\) must be equilateral. Therefore, all sides of \(\triangle ABC\) are equal.

8. **Calculating the Side Length:**

   In an equilateral triangle, the length of the median is related to the side length by the formula:
   \[
   \text{Median} = \frac{\sqrt{3}}{2} \times \text{Side}
   \]
   Given \(AM = 3\), we have:
   \[
   3 = \frac{\sqrt{3}}{2} \times AB \implies AB = \frac{3 \times 2}{\sqrt{3}} = 2\sqrt{3}
   \]

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