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 $\frac{SL}{SC}=\frac{SH}{SC}=\frac{AH}{AC}=\frac{1}{3}$.