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
$$S_{BKL} : S_{ACP} = S_{LPR} : S_{ACP} = PR^2 : AC^2 = AB^2 : AC^2$$
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 $\angle PAC = \varphi$. Simple angle chasing gives $\angle CPB = \alpha$, $\angle PBC = \varphi$, $\angle LCB = \angle ACP = \angle ABP = \beta - \varphi$, $\angle BLC = \angle APC = 90^\circ + \alpha$, $\angle PKL = \alpha$. Also, $AP = c \cos(\alpha + \varphi)$, $BP = c \sin(\alpha + \varphi)$.

Fig. 4

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

$$\frac{S_{KLB}}{S_{APC}} = \frac{\frac{1}{2}BL \cdot KL \cdot \sin(90^\circ + \alpha)}{\frac{1}{2}AP \cdot CP \cdot \sin(90^\circ + \alpha)} = \frac{BL \cdot KL}{AP \cdot CP} = \frac{\frac{a \cos(\alpha+\varphi)}{\cos \alpha} \cdot \frac{c \sin \varphi}{\sin \alpha \cos \alpha}}{c \cos(\alpha + \varphi) \cdot \frac{a \sin \varphi}{\sin \alpha}} = \frac{1}{\cos^2 \alpha},$$
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
$$$$
$\text{S KLB S APC = BL}$ KL}{AP $\cdot$ CP} = $\text{LC}$ KL}{PC^2}.