Let $ABC$ be an obtuse triangle, and let $H$ denote its orthocenter. Let $\omega_A$ denote the circle with center $A$ and radius $AH$. Let $\omega_B$ and $\omega_C$ be defined in a similar way. For all points $X$ in the plane of triangle $ABC$ let circle $\Omega(X)$ be defined in the following way (if possible): take the polars of point $X$ with respect to circles $\omega_A$, $\omega_B$ and $\omega_C$, and let $\Omega(X)$ be the circumcircle of the triangle defined by these three lines.
With a possible exception of finitely many points find the locus of points $X$ for which point $X$ lies on circle $\Omega(X)$.

[i]Proposed by Vilmos Molnár-Szabó, Budapest[/i]
1. **Understanding the Problem:**
   We are given an obtuse triangle \(ABC\) with orthocenter \(H\). We define circles \(\omega_A\), \(\omega_B\), and \(\omega_C\) centered at \(A\), \(B\), and \(C\) respectively, each with radius equal to the distance from the center to \(H\). For any point \(X\) in the plane, we define \(\Omega(X)\) as the circumcircle of the triangle formed by the polars of \(X\) with respect to \(\omega_A\), \(\omega_B\), and \(\omega_C\). We need to find the locus of points \(X\) such that \(X\) lies on \(\Omega(X)\).

2. **Using Simson Line:**
   By the Simson line theorem, the condition that \(X\) lies on \(\Omega(X)\) is equivalent to the inverses of \(X\) with respect to the three circles being collinear.

3. **Complex Numbers and Inversion:**
   Using complex numbers, let \(a\), \(b\), and \(c\) be the complex coordinates of \(A\), \(B\), and \(C\) respectively, and let \(h\) be the complex coordinate of \(H\). The inverse of \(X\) with respect to \(\omega_A\) is given by:
   \[
   a + \frac{|a-h|^2}{\overline{x-a}}
   \]
   Simplifying this expression, we get:
   \[
   a + \frac{a(b+c)^2}{bc(a\overline{x}-1)}
   \]
   Subtracting \(a + b + c\) from this, we obtain:
   \[
   (b+c)\frac{ab+bc+ca-abc\overline{x}}{bc(a\overline{x}-1)}
   \]
   We need this expression and its cyclic permutations to be collinear.

4. **Collinearity Condition:**
   Dividing by \(ab + bc + ca - abc\overline{x}\) and multiplying by \(abc\), we get:
   \[
   \frac{a(b+c)}{a\overline{x}-1}
   \]
   Taking the conjugate and multiplying by \(abc\), we have:
   \[
   \frac{a(b+c)}{x-a}
   \]
   We need to check the collinearity of:
   \[
   \frac{a(b+c)}{x-a} - \frac{b(a+c)}{x-b}
   \]
   Simplifying this, we get:
   \[
   \frac{a(x-b)(b+c) - b(x-a)(a+c)}{(x-a)(x-b)} = \frac{(a-b)(xc+ab)}{(x-a)(x-b)}
   \]
   The three points lying on a line is equivalent to:
   \[
   \frac{(a-b)(xc+ab)}{(a-c)(xb+ac)} \frac{x-c}{x-b}
   \]
   being real.

5. **Final Condition:**
   Simplifying further, we get:
   \[
   \frac{(xc+ab)(x-c)}{(xb+ac)(x-b)} = \frac{(ab\overline{x}+c)(c\overline{x}-1)}{(ac\overline{x}+b)(b\overline{x}-1)}
   \]
   Expanding and simplifying, we obtain:
   \[
   (x^2c + x(ab-c^2) - abc)(\overline{x}^2abc + \overline{x}(b^2-ac) - b) = (x^2b + x(ac-b^2) - abc)(\overline{x}^2abc + \overline{x}(c^2-ab) - c)
   \]
   This simplifies to:
   \[
   x^2\overline{x}^2abc - x^2\overline{x}(ab+bc+ca) - x\overline{x}^2abc(a+b+c) + x(ab+bc+ca) + \overline{x}abc(a+b+c) - abc = 0
   \]
   Finally, we get:
   \[
   (x\overline{x} - 1)(abc(x\overline{x} + 1) - x(ab+bc+ca) - \overline{x}abc(a+b+c)) = 0
   \]
   This implies:
   \[
   abc(x\overline{x} - 1)|x - a - b - c|^2 = 0
   \]

6. **Conclusion:**
   Therefore, \(X\) lies on \(\Omega(X)\) if and only if \(X\) is on the circumcircle of \(\triangle ABC\) or \(X = H\).

The final answer is \( \boxed{ X } \) lies on the circumcircle of \(\triangle ABC\) or \(X = H\).