Let $\omega$ be the unit circle centered at the origin of $R^2$. Determine the largest possible value for the radius of the circle inscribed to the triangle $OAP$ where $ P$ lies the circle and $A$ is the projection of $P$ on the axis $OX$.
1. **Identify the points and their coordinates:**
   - The unit circle $\omega$ is centered at the origin $O(0,0)$.
   - Let $P$ be a point on the unit circle, so $P(\cos \alpha, \sin \alpha)$.
   - Let $A$ be the projection of $P$ on the $x$-axis, so $A(\cos \alpha, 0)$.

2. **Determine the coordinates of the incenter:**
   - The incenter of triangle $OAP$ is the point where the angle bisectors of the triangle intersect.
   - The coordinates of the incenter $I$ can be found using the angle bisector theorem. For simplicity, we use the fact that the incenter lies on the angle bisector of $\angle OAP$.
   - The angle bisector of $\angle OAP$ divides the angle $\alpha$ into two equal parts, so the coordinates of the incenter $I$ are:
     \[
     I\left(\left(\cos \frac{\alpha}{2} - \sin \frac{\alpha}{2}\right) \cos \frac{\alpha}{2}, \left(\cos \frac{\alpha}{2} - \sin \frac{\alpha}{2}\right) \sin \frac{\alpha}{2}\right)
     \]

3. **Calculate the radius of the incircle:**
   - The radius $r$ of the incircle is the perpendicular distance from the incenter $I$ to any side of the triangle.
   - For simplicity, we calculate the distance from $I$ to the $x$-axis (which is the side $OA$).
   - The $y$-coordinate of $I$ gives this distance:
     \[
     r = \left(\cos \frac{\alpha}{2} - \sin \frac{\alpha}{2}\right) \sin \frac{\alpha}{2}
     \]

4. **Maximize the radius $r$:**
   - We need to find the maximum value of the function:
     \[
     f(\alpha) = \left(\cos \frac{\alpha}{2} - \sin \frac{\alpha}{2}\right) \sin \frac{\alpha}{2}
     \]
   - To find the critical points, we take the derivative of $f(\alpha)$ with respect to $\alpha$ and set it to zero:
     \[
     f'(\alpha) = \frac{d}{d\alpha} \left[ \left(\cos \frac{\alpha}{2} - \sin \frac{\alpha}{2}\right) \sin \frac{\alpha}{2} \right]
     \]
   - Using the product rule and chain rule:
     \[
     f'(\alpha) = \left(\cos \frac{\alpha}{2} - \sin \frac{\alpha}{2}\right) \cos \frac{\alpha}{2} \cdot \frac{1}{2} - \left(\sin \frac{\alpha}{2} + \cos \frac{\alpha}{2}\right) \sin \frac{\alpha}{2} \cdot \frac{1}{2}
     \]
     \[
     f'(\alpha) = \frac{1}{2} \left[ \left(\cos \frac{\alpha}{2} - \sin \frac{\alpha}{2}\right) \cos \frac{\alpha}{2} - \left(\sin \frac{\alpha}{2} + \cos \frac{\alpha}{2}\right) \sin \frac{\alpha}{2} \right]
     \]
   - Setting $f'(\alpha) = 0$:
     \[
     \left(\cos \frac{\alpha}{2} - \sin \frac{\alpha}{2}\right) \cos \frac{\alpha}{2} = \left(\sin \frac{\alpha}{2} + \cos \frac{\alpha}{2}\right) \sin \frac{\alpha}{2}
     \]
   - Simplifying:
     \[
     \cos^2 \frac{\alpha}{2} - \cos \frac{\alpha}{2} \sin \frac{\alpha}{2} = \sin^2 \frac{\alpha}{2} + \cos \frac{\alpha}{2} \sin \frac{\alpha}{2}
     \]
     \[
     \cos^2 \frac{\alpha}{2} - \sin^2 \frac{\alpha}{2} = 2 \cos \frac{\alpha}{2} \sin \frac{\alpha}{2}
     \]
     \[
     \cos \alpha = \sin \alpha
     \]
   - This implies:
     \[
     \alpha = \frac{\pi}{4}
     \]

5. **Calculate the maximum radius:**
   - Substitute $\alpha = \frac{\pi}{4}$ into the expression for $r$:
     \[
     r = \left(\cos \frac{\pi}{8} - \sin \frac{\pi}{8}\right) \sin \frac{\pi}{8}
     \]
   - Using trigonometric identities:
     \[
     \cos \frac{\pi}{8} = \frac{\sqrt{2 + \sqrt{2}}}{2}, \quad \sin \frac{\pi}{8} = \frac{\sqrt{2 - \sqrt{2}}}{2}
     \]
     \[
     r = \left(\frac{\sqrt{2 + \sqrt{2}}}{2} - \frac{\sqrt{2 - \sqrt{2}}}{2}\right) \cdot \frac{\sqrt{2 - \sqrt{2}}}{2}
     \]
     \[
     r = \frac{\sqrt{2 - \sqrt{2}}}{2} \cdot \frac{\sqrt{2 - \sqrt{2}}}{2} = \frac{2 - \sqrt{2}}{4} = \frac{\sqrt{2} - 1}{2}
     \]

The final answer is $\boxed{\frac{\sqrt{2} - 1}{2}}$