Problem 2102

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
$$\frac{LZ}{PZ} = \frac{LN}{PM} \quad (1)$$
Similarly from the similarity of $LPY$ and $NQY$ we get
$$\frac{NQ}{LY} = \frac{NQ}{LP} \quad (2)$$
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
$$\frac{PX}{XN} = \frac{NU}{UC} = \frac{CV}{VP} = -1, \text{ i.e.}$$
$$\frac{PX}{NX} = \frac{UC}{NU} = \frac{VP}{CV} \quad (3).$$
From the similarity of the triangles $KQU$ and $LNU$ we get $\frac{UQ}{UN} = \frac{KQ}{LN}$, from where
$$1 + \frac{NQ}{UN} = 1 + \frac{KB}{LN}, \text{ so } \overline{UN} = \frac{\overline{LN NQ}}{KB} \text{ and } \overline{UC} = \overline{UN} + \overline{NC} = \frac{\overline{LN NQ} + \overline{NC KB}}{KB} \text{ and from here}$$
$$\frac{\overline{LN NQ}}{\overline{NU}} = \frac{\overline{KB}}{\overline{LN NQ} + \overline{NC KB}} = \frac{-\overline{LN NQ}}{\overline{LN NQ} + \overline{NC KB}} \quad (4)$$
Analogously, for the similar triangles $KMV$ and $LPV$ we get
$$\frac{CV}{VP} = \frac{LPPM + PCKA}{-LPPM} \quad (5)$$
If we substitute (4) and (5) in (3) we get:
$$\frac{PX}{NX} = \frac{UC}{NU} = \frac{VP}{CV} = \frac{LN NQ + NC KB}{-LN NQ} = \frac{-LPPM}{LPPM + PCKA} = \frac{LN NQ + NC KB}{LPPM + PCKA} = \frac{LPPM}{LN NQ} = \frac{LPPM}{LN NQ} \quad (6)$$

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.