12th Czech-Polish-Slovak Mathematics Competition

Throughout the solution, we shall denote by $S_{XYZ}$ the area of the triangle $XYZ$.

Fig. 3

Let $PR$ be the diameter of the circumcircle of $ABC$ (Fig. 3). Then $ARBP$ is a rectangle. Since $BR$ is parallel to $PA$ we have $S_{PBK}=S_{PRK}$, which implies $$S_{BKL}=S_{LPR}.$$ Since $PA=BR$ we have $\angle PCA=\angle BPR=\angle LPR$. Also, because $CPAR$ is cyclic we have $\angle CAP=\angle CRP$. Hence $LPR$ and $PCA$ are similar triangles. Finally, we have $$S_{BKL}:S_{ACP}=S_{LPR}:S_{ACP}=P{R}^2:A{C}^2=A{B}^2:A{C}^2$$ which does not depend on the point $P$.

Second solution.

Denote the lengths of the sides and the sizes of the angles of the triangle $ABC$ by $a$, $b$, $c$ and $\alpha$, $\beta$, $\gamma$, respectively (Fig. 4). Let $\angle PAC=\phi$. Simple angle chasing gives $\angle CPB=\alpha$, $\angle PBC=\phi$, $\angle LCB=\angle ACP=\angle ABP=\beta -\phi$, $\angle BLC=\angle APC=90^{\circ }+\alpha$, $\angle PKL=\alpha$. Also, $AP=c\cos (\alpha +\phi )$, $BP=c\sin (\alpha +\phi )$.

Fig. 4

Using the sine law in $BCP$ and $BCL$ we get $$CP=\frac{BC\sin \phi }{\sin \alpha }=\frac{a\sin \phi }{\sin \alpha },$$ $$BL=\frac{BC\sin (\beta -\phi )}{\sin (90^{\circ }+\alpha )}=\frac{a\cos (\alpha +\phi )}{\cos \alpha }.$$ Then $$\begin{gathered} PL=BP-BL=c\sin (\alpha +\phi )-\frac{a\cos (\alpha +\phi )}{\cos \alpha }= \\ =\frac{c\sin (\alpha +\phi )\cos \alpha -c\sin \alpha \cos (\alpha +\phi )}{\cos \alpha }=\frac{c\sin \phi }{\cos \alpha }. \end{gathered}$$ $$KL=\frac{PL}{\sin \alpha }=\frac{c\sin \phi }{\sin \alpha \cos \alpha }.$$

$$\frac{S_{KLB}}{S_{APC}}=\frac{\frac{1}{2}BL\cdot KL\cdot \sin (90^{\circ }+\alpha )}{\frac{1}{2}AP\cdot CP\cdot \sin (90^{\circ }+\alpha )}=\frac{BL\cdot KL}{AP\cdot CP}=\frac{\frac{a\cos (\alpha +\phi )}{\cos \alpha }\cdot \frac{c\sin \phi }{\sin \alpha \cos \alpha }}{c\cos (\alpha +\phi )\cdot \frac{a\sin \phi }{\sin \alpha }}=\frac{1}{{\cos }^2\alpha },$$ and $\alpha$ is clearly independent on the choice of the point $P$.

Remark. We can shorten the previous solution by noticing that from the angles we have a pair of similar triangles $BLC$, $APC$ (Fig. 4). Then $BL/AP=LC/PC$ and the desired ratio can be expressed as $$ \frac{S_{KLB}}{S_{APC}} = \frac{BL \cdot KL}{AP \cdot CP} = \frac{LC \cdot KL}{PC^2}.