Mathematical Olympiad Rioplatense

Since $SH = SL$, we compute the ratio $\frac{SH}{SC}$. Note that $AB > BP$ implies that $P$ is between $A$ and $C$. Denote $\angle BAP = \alpha$ and observe that $\angle PSC = \angle PSH = \angle BSL = \angle BSQ = \alpha$. Indeed we have $\angle HSB = \angle QSP = 180^\circ - \alpha$ from the cyclic quadrilaterals $AHSB$ and $APSQ$. On the other hand each of the four angles in the above equality completes $\angle HSB$ or $\angle QSP$ to $180^\circ$. In particular $\angle PSC = \angle PSH$ means that $SP$ is the internal bisector of $\angle CSH$. Since $AS \perp SP$, it follows that $SA$ is the external bisector of $\angle CSH$. Hence $\frac{SH}{SC} = \frac{AH}{AC}$ by the external angle theorem.

To find $\frac{AH}{AC}$ consider the midpoint $M$ of $HC$. The right triangles $BHL$ and $BCH$ are similar as they share an acute angle at vertex $B$. Since $BS$ and $BM$ are respective medians, $\angle BMH = \angle BSL$. We proved above that $\angle BSL = \alpha$, hence $\angle BMH = \alpha = \angle BAH$. Therefore triangle $ABM$ is isosceles with base $AM$. Its altitude $BH$ is also a median, so $AH = HM$. In addition $HM = MC$, and we obtain $AH = \frac{1}{3}AC$. In conclusion $\frac{SL}{SC} = \frac{SH}{SC} = \frac{AH}{AC} = \frac{1}{3}$.