Let $O$ be the center of the circle $\omega$ circumscribed around the acute-angled triangle $\vartriangle ABC$, and $W$ be the midpoint of the arc $BC$ of the circle  $\omega$, which does not contain the point $A$, and $H$ be the point of intersection of the heights of the triangle  $\vartriangle ABC$. Find the angle $\angle BAC$, if $WO = WH$.

(O. Clurman)
1. **Define the key points and properties:**
   - Let \( O \) be the center of the circumcircle \(\omega\) of the acute-angled triangle \(\triangle ABC\).
   - Let \( W \) be the midpoint of the arc \( BC \) of the circle \(\omega\) that does not contain the point \( A \).
   - Let \( H \) be the orthocenter of \(\triangle ABC\), i.e., the point of intersection of the heights of the triangle.
   - Given that \( WO = WH \).

2. **Introduce the midpoint \( M \) of \( BC \):**
   - Let \( M \) be the midpoint of \( BC \).

3. **Intersection of \( AO \) with the circumcircle:**
   - Let \( AO \) intersect the circumcircle \(\omega\) again at \( H' \).

4. **Apply homothety centered at \( M \) with scale factor \(-1\):**
   - This homothety maps \( H \) to \( H' \) and \( W \) to a point \( W' \) on the perpendicular bisector of \( BC \), on the same side of \( BC \) as \( A \).

5. **Analyze the properties of \( W' \):**
   - Since \( W \) is the midpoint of the arc \( BC \) not containing \( A \), \( W \) lies on the perpendicular bisector of \( BC \).
   - The homothety maps \( W \) to \( W' \) such that \( W'H' = R \), where \( R \) is the radius of the circumcircle.

6. **Conclude that \( W' = O \):**
   - Given \( WO = WH \) and the properties of the homothety, it follows that \( W' = O \).

7. **Determine the angle \( \angle BAC \):**
   - Since \( W \) is the midpoint of the arc \( BC \) not containing \( A \), \( \angle BWC = 180^\circ - \angle BAC \).
   - Given \( WO = WH \), the triangle \( WOH \) is isosceles with \( \angle WHO = \angle WHO = 60^\circ \).
   - Therefore, \( \angle BAC = 60^\circ \).

\(\blacksquare\)

The final answer is \( \boxed{ \angle BAC = 60^\circ } \).