Let $ABC$ be an acute-angled, nonisosceles triangle. Altitudes $AA'$ and $BB' $meet at point $H$, and the medians of triangle $AHB$ meet at point $M$. Line $CM$ bisects segment $A'B'$. Find angle $C$.

(D. Krekov) 
1. **Inscribing Triangle in the Unit Circle:**
   Let \(ABC\) be an acute-angled, non-isosceles triangle inscribed in the unit circle. This implies that the complex numbers representing the vertices \(A\), \(B\), and \(C\) have magnitudes equal to 1:
   \[
   |a| = |b| = |c| = 1
   \]

2. **Finding Coordinates of Altitudes' Feet:**
   The feet of the altitudes from \(A\) and \(B\) to the opposite sides are given by:
   \[
   a' = \frac{1}{2} \left( a + b + c - \frac{bc}{a} \right)
   \]
   \[
   b' = \frac{1}{2} \left( a + b + c - \frac{ac}{b} \right)
   \]

3. **Finding the Orthocenter \(H\):**
   The orthocenter \(H\) of triangle \(ABC\) is given by:
   \[
   h = a + b + c
   \]

4. **Finding the Centroid \(M\) of Triangle \(AHB\):**
   The centroid \(M\) of triangle \(AHB\) is the average of the coordinates of \(A\), \(H\), and \(B\):
   \[
   m = \frac{a + h + b}{3} = \frac{a + (a + b + c) + b}{3} = \frac{2a + 2b + c}{3}
   \]

5. **Finding the Midpoint \(N\) of Segment \(A'B'\):**
   The midpoint \(N\) of segment \(A'B'\) is given by:
   \[
   n = \frac{a' + b'}{2} = \frac{1}{4} \left( 2a + 2b + 2c - \frac{bc}{a} - \frac{ac}{b} \right)
   \]

6. **Collinearity Condition:**
   Since \(CMN\) is collinear, the following ratio must be real:
   \[
   \frac{n - c}{m - c} \in \mathbb{R}
   \]
   Substituting the expressions for \(n\) and \(m\):
   \[
   \frac{\frac{1}{4} \left( 2a + 2b + 2c - \frac{bc}{a} - \frac{ac}{b} \right) - c}{\frac{2a + 2b + c}{3} - c} \in \mathbb{R}
   \]
   Simplifying the numerator and the denominator:
   \[
   \frac{2a + 2b + 2c - \frac{bc}{a} - \frac{ac}{b} - 4c}{2a + 2b - 2c} \in \mathbb{R}
   \]

7. **Simplifying the Expression:**
   \[
   \frac{(a + b)(ac + bc - 2ab)}{ab(a + b - c)} = \overline{\left( \frac{(a + b)(ac + bc - 2ab)}{ab(a + b - c)} \right)}
   \]
   Since the expression must be real, equate it to its conjugate:
   \[
   \frac{(a + b)(ac + bc - 2ab)}{ab(a + b - c)} = \frac{(a + b)(a + b - 2c)}{ac + bc - ab}
   \]

8. **Cross-Multiplying and Simplifying:**
   \[
   (ac + bc - 2ab)(ac + bc - ab) = ab(a + b - 2c)(a + b - c)
   \]
   Expanding both sides:
   \[
   a^2c^2 + b^2c^2 + 2a^2b^2 + 2abc^2 - 3a^2bc - 3ab^2c = a^3b + ab^3 + 2abc^2 + 2a^2b^2 - 3a^2bc - 3ab^2c
   \]
   Simplifying further:
   \[
   a^2c^2 + b^2c^2 = a^3b + ab^3
   \]
   Factoring out common terms:
   \[
   (a^2 + b^2)(c^2 - ab) = 0
   \]

9. **Analyzing the Factors:**
   Since \(ABC\) is scalene, \(c^2 \neq ab\). Therefore:
   \[
   a^2 + b^2 = 0
   \]
   This implies:
   \[
   \frac{b}{a} = \pm i
   \]
   Hence, the arc \(AB\) measures \(90^\circ\). The inscribed angle \(\angle ACB\) is half of that:
   \[
   \angle ACB = 45^\circ
   \]

The final answer is \(\boxed{45^\circ}\)