Sharygin 2019 P23 wrote: In the plane, let $a, b$ be two closed broken lines (possibly self-intersecting), and $K, L, M, N$ be four points. The vertices of $a, b$ and the points $K L, M, N$ are in general position (i.e. no three of these points are collinear, and no three segments between them concur at an interior point). Each of segments $KL$ and $MN$ meets $a$ at an even number of points, and each of segments $LM$ and $NK$ meets a at an odd number of points. Conversely, each of segments $KL$ and $MN$ meets $b$ at an odd number of points, and each of segments $LM$ and $NK$ meets $b$ at an even number of points. Prove that $a$ and $b$ intersect. Solution . When we refer to broken lines we mean closed broken lines (possibly self-intersecting). For a closed curve or broken line $\mathfrak{c}$ we define $\text{Interior}(\mathfrak{c})$ as the set of points strictly in the interior of it and define $\text{Exterior}(\mathfrak{c})$ as the set of points strictly outside $\mathfrak c$.Consider the following key claim, Claim 1. Let $k$ be a given broken line. $\overline{XY}$ be a line segment such that the line segment and the broken line $k$ are in general position and $X,Y\notin k$. Then the following holds: $X\in\text{Interior}(k),~Y\in\text{Interior}(k)\implies |k\cap\overline{XY}|$ is even. $X\in\text{Exterior}(k),~Y\in\text{Exterior}(k)\implies |k\cap\overline{XY}|$ is even. $X\in\text{Interior}(k),~Y\in\text{Exterior}(k)\implies |k\cap\overline{XY}|$ is odd. Proof. Consider an arbitrary line $l$ and broken line $k$ in general position. Let the points of intersection of $l$ with $k$ be $P_1,P_2,\ldots,P_n$ in serial order i.e. there are no points of intersection between $P_i$ and $P_{i+1}$ for all $1\le n\le n-1$ also $P_1$ and $P_n$ are the first and last intersections. [asy][asy] import graph; size(6cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -8.463648657878554, xmax = 5.892052331249103, ymin = -4.981027003868339, ymax = 5.003387870922874; /* image dimensions */ draw((-4.08,3.76)--(-7,1.36)--(-4.04,0.04)--(-7.72,-3.28)--(3.4,-0.14)--cycle, linewidth(0.8) + blue); /* draw figures */ draw((-4.08,3.76)--(-7,1.36), linewidth(0.8) + blue); draw((-7,1.36)--(-4.04,0.04), linewidth(0.8) + blue); draw((-4.04,0.04)--(-7.72,-3.28), linewidth(0.8) + blue); draw((-7.72,-3.28)--(3.4,-0.14), linewidth(0.8) + blue); draw((3.4,-0.14)--(-4.08,3.76), linewidth(0.8) + blue); draw((-5.787347065521166,4.321242107442294)--(-5.773296757127461,2.368249240717156), linewidth(0.8) + gray); draw((-5.773296757127461,2.368249240717156)--(-5.762071588803278,0.8079508436555154), linewidth(0.8)); draw((-5.762071588803278,0.8079508436555154)--(-5.745477429881128,-1.4986372465231916), linewidth(0.8) + gray); draw((-5.745477429881128,-1.4986372465231916)--(-5.736690900169162,-2.7199648764866153), linewidth(0.8)); draw((-5.736690900169162,-2.7199648764866153)--(-5.725467342925167,-4.280039333402341), linewidth(0.8) + gray); /* dots and labels */ dot((-4.08,3.76),dotstyle); dot((-7,1.36),dotstyle); dot((-4.04,0.04),dotstyle); dot((-7.72,-3.28),dotstyle); dot((3.4,-0.14),dotstyle); label("$k$", (-0.14330292888588061,2.06022079962248), NE * labelscalefactor,blue); label("$l$", (-6.2,-0.33653550738585847), NE * labelscalefactor,black); dot((-5.7732967571274605,2.3682492407171574),linewidth(4pt) + dotstyle); label("$P_1$", (-5.7191764099570905,2.4700288728415223), NE * labelscalefactor); dot((-5.762071588803279,0.807950843655516),linewidth(4pt) + dotstyle); label("$P_2$", (-5.7067579834959075,0.9053071387324515), NE * labelscalefactor); dot((-5.745477429881128,-1.4986372465231923),linewidth(4pt) + dotstyle); label("$P_3$", (-5.694339557034724,-1.404520183047605), NE * labelscalefactor); dot((-5.736690900169162,-2.719964876486616),linewidth(4pt) + dotstyle); label("$P_4$", (-5.681921130573541,-2.6215259762435488), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] In total, the line $l$ is divided into $n+1$ intercepts. Colour an intercept as black if the intercept is in the interior of $k$ and white otherwise. Observe that two consecutive intercepts cannot be of the same colour because otherwise, $k$ and $l$ wouldn't be in general position. Hence we conclude that the intercepts will get coloured black and white alternatingly. Now, obviously the first and last intercepts will be coloured both white since $k$ is closed and the first and last intercepts are both unbounded at one end. Consider two points $X,Y$ on $l$ such that $X,Y\notin k.$ We divide the possibilities into the following three cases: $X\in\text{Interior}(k),~Y\in\text{Interior}(k)$ : Note that the intercepts in which $X$ and $Y$ belong to are coloured black. Now since the intercepts are alternatingly coloured, it is easy to see that there must be an even number of intersections between $X$ and $Y.$ $X\in\text{Exterior}(k),~Y\in\text{Exterior}(k)$ : Note that the intercepts in which $X$ and $Y$ belong to are coloured white. Now since the intercepts are alternatingly coloured, it is easy to see that there must be an even number of intersections between $X$ and $Y.$ $X\in\text{Interior}(k),~Y\in\text{Exterior}(k)$ : Note that the intercepts in which $X$ and $Y$ belong to are coloured black and white respectively. Now since the intercepts are alternatingly coloured, it is easy to see that there must be an odd number of intersections between $X$ and $Y.$ Thus, our claim is proved. $~~\square$ Coming to the original problem, Assume the contrary that $a$ and $b$ don't intersect. Using Claim 1, the following statements are true: $K,L$ both belong to $\text{Interior}(a)$ or both belong to $\text{Exterior}(a).$ $N,M$ both belong to $\text{Interior}(a)$ or both belong to $\text{Exterior}(a).$ All of $K,L,M,N$ cannot be in $\text{Interior}(a).$ All of $K,L,M,N$ cannot be in $\text{Exterior}(a).$ From the above we can conclude that exactly one pair of points $(K,L)$ or $(N,M)$ is in $\text{Interior}(a).$ Without loss of generality assume $K,L\in\text{Interior}(a)$ and $N,M\in \text{Exterior}(a).$ Again doing the same thing for $b,$ we obtain exactly one pair of points $(K,N)$ or $(L,M)$ is in $\text{Interior}(b).$ Without loss of generality assume $K,N\in\text{Interior}(b)$ and $L,M\in \text{Exterior}(b).$ Therefore it follows that $K\in\text{Interior}(a)\cap \text{Interior}(b)$ but since $a$ and $b$ don't intersect, we have either $\text{Interior}(a)\subset \text{Interior}(b)$ or $\text{Interior}(b)\subset \text{Interior}(a),$ without loss generality assume $\text{Interior}(a)\subset \text{Interior}(b)$. Therefore using this we get that $K,L\in\text{Interior}(a)\subset\text{Interior}(b)\implies K,L\in\text{Interior}(b).$ But we already obtained $L\in\text{Exterior}(b).$ Which means $\text{Interior}(b)\cap\text{Exterior}(b)$ is non-empty. Contradiction! $~~\blacksquare$

I got a somewhat similar solution

ayan.nmath wrote: When we refer to broken lines we mean closed broken lines (possibly self-intersecting). OH, I didn't realize they were closed until I read this sentence.