Let $ABC$ be a triangle with incenter $I$. The line through $I$, perpendicular to $AI$, intersects the circumcircle of $ABC$ at points $P$ and $Q$. It turns out there exists a point $T$ on the side $BC$ such that $AB + BT = AC + CT$ and $AT^2 =  AB \cdot AC$. Determine all possible values of the ratio $IP/IQ$.
1. **Assume WLOG \( AB < AC \)**:
   - We start by assuming \( AB < AC \) without loss of generality. This assumption helps simplify the problem by focusing on one specific case.

2. **Identify point \( T \)**:
   - The point \( T \) on side \( BC \) such that \( AB + BT = AC + CT \) and \( AT^2 = AB \cdot AC \) is the touchpoint of the \( A \)-excircle with \( BC \).

3. **Define \( M_A \), \( E \), and \( D \)**:
   - Let \( M_A \) be the touchpoint of the \( A \)-mixtilinear incircle with the circumcircle \( \odot(ABC) \).
   - Let \( E \) be the midpoint of the minor arc \( \widehat{BC} \) of \( \odot(ABC) \).
   - Let \( D \) be the touchpoint of the incircle of \( \triangle ABC \) with \( BC \).

4. **Isogonal Conjugates and Lengths**:
   - It is known that \( AT \) is isogonal to \( AM_A \) with respect to \( \angle BAC \).
   - Also, \( AT \times AM_A = AB \times AC \) (by considering the \( \sqrt{AB \times AC} \) inversion). Therefore, \( AT = AM_A \).

5. **Perpendicular Bisector**:
   - Since \( AE \) is the bisector of \( \angle M_AAT \), we conclude \( AE \) is the perpendicular bisector of \( M_A T \). Therefore, \( ED = ET = EM_A \).

6. **Radical Axes and Concurrency**:
   - By considering the pairwise radical axes of the circles \( \odot(BIC) \), \( \odot(IE) \), and \( \odot(ABC) \), we find that the three lines \( PQ \), \( M_AE \), and \( BC \) are concurrent at a point \( Y \).

7. **Intersection and Perpendicularity**:
   - Let \( YA \) meet \( \odot(ABC) \) for the second time at \( X \). Since \( YX \times YA = YI^2 \) and \( AI \perp IY \), we have \( AX \perp XI \). Therefore, \( X \) is the \( A \)-sharkydevil point, so \( BX/XC = BD/DC \implies X, D, E \) are collinear.

8. **Cyclic Points**:
   - The five points \( I, X, Y, M_A, D \) are located on a circle with diameter \( IY \).

9. **Intersection and Equal Lengths**:
   - Let \( M_AD \) cut \( \odot(ABC) \) for the second time at \( A' \). Since \( ED = EM_A \), we have \( A'X = A'D \).

10. **Angle Bisectors and Perpendicular Bisectors**:
    - The angle bisectors of \( \angle XA'D \) and \( \angle DEM_A \) meet at the midpoint of arc \( \widehat{XM_A} \) of \( \odot(ABC) \). These lines are the perpendicular bisectors of \( XD \) and \( DM_A \), respectively, which meet at the center of \( \odot(XIDM_AY) \), located on \( IY \).

11. **Intersection and Midpoint**:
    - The center of \( \odot(XIDM_AY) \) is the intersection of \( IY \) and \( \odot(ABC) \), namely \( P \). Therefore, \( P \) is the midpoint of \( IY \).

12. **Final Calculation**:
    - To finish off:
    \[
    YP \times YQ = YI^2 \implies IP \times (2IP + IQ) = 4IP^2 \implies IQ = 2IP,
    \]
    so \( IP/IQ = 1/2 \). For the other case (\( AB > AC \)), we will have \( IP/IQ = 2 \).

The final answer is \( \boxed{\{1/2, 2\}} \)