A triangle ABC with $\angle A = 60^o$ is given. Points $M$ and $N$ on $AB$ and $AC$ respectively are such that the circumcenter of $ABC$ bisects segment $MN$. Find the ratio $AN:MB$.


by E.Bakaev
To solve the problem, we need to find the ratio \( \frac{AN}{MB} \) given that the circumcenter \( O \) of triangle \( \triangle ABC \) bisects segment \( MN \). Here is a step-by-step solution:

1. **Identify Key Elements:**
   - Given \( \angle A = 60^\circ \).
   - Points \( M \) and \( N \) are on \( AB \) and \( AC \) respectively.
   - The circumcenter \( O \) of \( \triangle ABC \) bisects \( MN \).

2. **Use Properties of the Circumcenter:**
   - The circumcenter \( O \) is equidistant from all vertices of the triangle \( \triangle ABC \).
   - Since \( \angle A = 60^\circ \), \( \triangle ABC \) is an equilateral triangle. This is because in any triangle, if one angle is \( 60^\circ \) and the circumcenter bisects a segment, the triangle must be equilateral.

3. **Equilateral Triangle Properties:**
   - In an equilateral triangle, all sides are equal, and all angles are \( 60^\circ \).
   - The circumcenter \( O \) coincides with the centroid and the orthocenter of the triangle.

4. **Midpoint Condition:**
   - Since \( O \) bisects \( MN \), \( M \) and \( N \) must be such that \( O \) is the midpoint of \( MN \).

5. **Symmetry in Equilateral Triangle:**
   - In an equilateral triangle, the medians, angle bisectors, and altitudes all coincide.
   - Therefore, \( M \) and \( N \) must be symmetric with respect to the centroid \( O \).

6. **Ratio Calculation:**
   - Since \( O \) is the midpoint of \( MN \) and \( \triangle ABC \) is equilateral, the segments \( AN \) and \( MB \) are equal in length.
   - Thus, the ratio \( \frac{AN}{MB} = 1 \).

The final answer is \( \boxed{1} \).