66th Belarusian Mathematical Olympiad

Let $P$ and $Q$ be the projections of $L$ and $N$ onto the line $BC$, respectively. Let $a=BC$, $b=CA$, $c=AB$, let $p$ and $S$ be the half-perimeter and the area of the triangle $ABC$, respectively. Let $X$, $Y$ be the intersection points of the lines $LN$, $MK$ and $BC$, respectively. Without loss of generality we assume that $X$, $Y$ lie on the segment $BC$ (all other positions of these points can be considered in similar way).



It is well known that the radius of the excircle with the center lying opposite to the vertex $C$ is equal to $S/(p-c)$, so $LP=S/(p-c)$. By condition, $A{C}_1=B{C}_1$ and since $N$ is symmetric to $C$ about $C_1$, we have $N{C}_1=C_{1}C$, so $ANBC$ is a parallelogram. If $AH$ is the altitude of $ABC$, then $NQ=AH=2S/a$. Since $CP=p$, $CB=a$, we have $BP=p-a$. From the equality of the triangles $CHA$ and $BQN$ it follows that $BQ=CH=b\cos \angle C$. Since $LP\parallel NQ$, we see that $△XLP\sim △XNQ$. Then $\frac{XP}{XQ}=\frac{LP}{NQ}$. Let $BX=x$, then

$$\frac{x+p-a}{x+b\cos \angle C}=\frac{S/(p-c)}{2S/a}\Rightarrow x=\frac{2(p-c)(p-a)-ab\cos \angle C}{c-b}$$

Similarly, if $y=CY$, then $y=\frac{2(p-b)(p-a)-ac\cos \angle B}{b-c}$. To prove the required statement, it suffices to show that $x+y=a$. Indeed,

$$\begin{array}{rl}x+y& =\frac{2(p-c)(p-a)-ab\cos \angle C}{c-b}+\frac{2(p-b)(p-a)-ac\cos \angle B}{b-c}=\\ & =\frac{2(p-c)(p-a)-2(p-b)(p-a)+ac\cos \angle B-ab\cos \angle C}{c-b}=\\ & =\frac{2(b-c)(p-a)+ac\cdot \frac{a^2+c^2-b^2}{2ac}-ab\cdot \frac{a^2+b^2-c^2}{2ab}}{c-b}=\\ & =\frac{(b-c)(b+c-a)+c^2-b^2}{c-b}=\frac{a(c-b)}{c-b}=a.\end{array}$$