Acute triangle $ABP$, where $AB > BP$, has altitudes $BH$, $PQ$, and $AS$. Let $C$ denote the intersection of lines $QS$ and $AP$, and let $L$ denote the intersection of lines $HS$ and $BC$. If $HS = SL$ and $HL$ is perpendicular to $BC$, find the value of $\frac{SL}{SC}$.
1. **Identify the given information and setup the problem:**
   - We have an acute triangle \( ABP \) with \( AB > BP \).
   - The altitudes are \( BH \), \( PQ \), and \( AS \).
   - \( C \) is the intersection of lines \( QS \) and \( AP \).
   - \( L \) is the intersection of lines \( HS \) and \( BC \).
   - Given \( HS = SL \) and \( HL \) is perpendicular to \( BC \).

2. **Analyze the harmonic division:**
   - Since \( AHPC \) is harmonic, the points \( A, H, P, C \) form a harmonic division. This implies that the cross-ratio \((A, H; P, C)\) is \(-1\).

3. **Use the harmonic property:**
   - Given \( HS = SL \), the segment \( HS \) is equal to \( SL \). This implies that \( S \) is the midpoint of \( HL \).

4. **Parallel lines and equal segments:**
   - Since \( HS = SL \) and \( HL \) is perpendicular to \( BC \), it follows that \( HS \parallel AB \). This implies that \( PA = PB \).

5. **Relationship between segments:**
   - From \( HL \perp BC \), we have \( BC \parallel PQ \). This implies that \( AP = PC \).

6. **Using the harmonic property again:**
   - Since \( \frac{AH}{HP} = \frac{AC}{CP} \), and given \( AP = PC \), we have \( AH = 2HP \).

7. **Calculate the ratio \( \frac{SL}{SC} \):**
   - Since \( HS = SL \) and \( HL \) is perpendicular to \( BC \), we have \( \frac{SL}{SC} = \frac{HS}{2SQ} = \frac{AB}{6SQ} = \frac{1}{3} \).

The final answer is \( \boxed{ \frac{1}{3} } \).