Macedonian Mathematical Olympiad

Without loss of generality we can assume that $p_B$ is between $p_A$ and $p_C$. The first case is if $q_A$ is between $q_B$ and $q_C$ as shown in the picture, obviously $KL$ intersects $PN$. Analogously, if $q_C$ is between $q_A$ and $q_B$ the case is symmetrical to the one we are considering. The second case, if $q_B$ is between $q_A$ and $q_C$, then $MQ$ and $PN$ are not parallel, so $KL$ intersects at least one of them and the two cases are equivalent. According to this we can assume that $KL$ intersects $PN$. The line $MN$ cannot be parallel to $p_B$, since in that case $p$ is perpendicular to $AC$ and analogously $PQ$ is not parallel to $q_A$.

Let $X$, $Y$ and $Z$ be the points of intersection of the lines $PN$, $q_A$ and $p_B$ with $KL$, $PQ$ and $MN$ respectively. From the similarity of the triangles $LNZ$ and $PMZ$ we get $$\frac{LZ}{PZ}=\frac{LN}{PM}(1)$$ Similarly from the similarity of $LPY$ and $NQY$ we get $$\frac{NQ}{LY}=\frac{NQ}{LP}(2)$$ If $KL$ does not pass through $C$, let it intersect $NC$ and $MC$ in $U$ and $V$ respectively. From the triangle $CPN$ and Menelaus' theorem for the line $KL$ we get $$\frac{PX}{XN}=\frac{NU}{UC}=\frac{CV}{VP}=-1,\text{ i.e.}$$ $$\frac{PX}{NX}=\frac{UC}{NU}=\frac{VP}{CV}(3).$$ From the similarity of the triangles $KQU$ and $LNU$ we get $\frac{UQ}{UN}=\frac{KQ}{LN}$, from where $$1+\frac{NQ}{UN}=1+\frac{KB}{LN},\text{ so }\overline{UN}=\frac{\overline{LNNQ}}{KB}\text{ and }\overline{UC}=\overline{UN}+\overline{NC}=\frac{\overline{LNNQ}+\overline{NCKB}}{KB}\text{ and from here}$$ $$\frac{\overline{LNNQ}}{\overline{NU}}=\frac{\overline{KB}}{\overline{LNNQ}+\overline{NCKB}}=\frac{-\overline{LNNQ}}{\overline{LNNQ}+\overline{NCKB}}(4)$$ Analogously, for the similar triangles $KMV$ and $LPV$ we get $$\frac{CV}{VP}=\frac{LPPM+PCKA}{-LPPM}(5)$$ If we substitute (4) and (5) in (3) we get: $$\frac{PX}{NX}=\frac{UC}{NU}=\frac{VP}{CV}=\frac{LNNQ+NCKB}{-LNNQ}=\frac{-LPPM}{LPPM+PCKA}=\frac{LNNQ+NCKB}{LPPM+PCKA}=\frac{LPPM}{LNNQ}=\frac{LPPM}{LNNQ}(6)$$

If we now substitute (1), (2) and (6) in Ceva's equality for the triangle $LNP$ and the lines $LK$, $NM$ and $PQ$ we get: $$\frac{\overline{LZ}}{\overline{PZ}}\frac{\overline{PX}}{\overline{NX}}\frac{\overline{NY}}{\overline{LY}}=\frac{\overline{LN}}{\overline{PM}}\frac{\overline{NQ}}{\overline{LP}}-\frac{\overline{LP}}{\overline{NM}}\frac{\overline{PM}}{\overline{NQ}}=-1$$ If $KL$ passes through $C$, then $X$ is the midpoint of $PN$ and $$\frac{\overline{LN}}{\overline{BK}}=\frac{\overline{NC}}{\overline{LB}}$$ (7) If we now substitute (1), (2) and (7) in Ceva's equality for the triangle $LNP$ and the lines $LK$, $NM$ and $PQ$ we get: $$\frac{\overline{LZ}}{\overline{PZ}}\frac{\overline{PX}}{\overline{NX}}\frac{\overline{NY}}{\overline{LY}}=-\frac{\overline{LN}}{\overline{PM}}\frac{\overline{NQ}}{\overline{LP}}=-\frac{\overline{NC}}{\overline{LB}}\frac{\overline{NQ}}{\overline{LP}}=-1$$ From the converse of Ceva's theorem, the lines $KL$, $MN$ and $PQ$ intersect in one point or are parallel. Without loss of generality we can assume that $p_B$ is between $p_A$ and $p_C$. If $q_B$ is not between $q_A$ and $q_C$ as shown in the picture it is obvious that the lines cannot be parallel. If $q_B$ is between $q_A$ and $q_C$, then in order for the lines to be parallel it is required that $\frac{\overline{AK}}{\overline{AM}}=\frac{\overline{AL}}{\overline{AN}}$, but then $\frac{\overline{AK}}{\overline{AM}}=\frac{\overline{AL}}{\overline{AN}}=\frac{\overline{KB}}{\overline{MC}}$, therefore $AB$ is parallel to $AC$, which is impossible since $ABC$ is a triangle.