Let $I$ be the incenter of triangle $ABC$. $K_1$ and $K_2$ are the points on $BC$ and $AC$ respectively, at which the inscribed circle is tangent. Using a ruler and a compass, find the center of the inscribed circle for triangle $CK_1K_2$ in the minimal possible number of steps (each step is to draw a circle or a line).

(Hryhorii Filippovskyi)
1. **Draw the incircle of triangle \(ABC\)**:
   - Let \(I\) be the incenter of triangle \(ABC\).
   - Draw the incircle of triangle \(ABC\) with center \(I\) and radius \(IK_1\), where \(K_1\) is the point of tangency on \(BC\).

2. **Identify the points of tangency**:
   - Let \(K_1\) be the point where the incircle touches \(BC\).
   - Let \(K_2\) be the point where the incircle touches \(AC\).

3. **Draw line \(CI\)**:
   - Draw the line segment \(CI\) from vertex \(C\) to the incenter \(I\).

4. **Intersection of \(CI\) with the incircle**:
   - The line \(CI\) intersects the incircle at two points. One of these points is \(I\) itself, and the other point is the incenter of triangle \(CK_1K_2\).

5. **Identify the incenter of triangle \(CK_1K_2\)**:
   - The intersection point of \(CI\) with the incircle, other than \(I\), is the incenter of triangle \(CK_1K_2\).

Thus, the minimal steps to find the incenter of triangle \(CK_1K_2\) are:
1. Draw the incircle of triangle \(ABC\).
2. Draw the line segment \(CI\).
3. Identify the intersection point of \(CI\) with the incircle, other than \(I\).

The final answer is the intersection point of \( \boxed{ CI } \) with the incircle, other than \(I\).