Verified: no
Verdicts: FAIL + FAIL
Solve time: 37m28s
Source on kvant.digital

Problem

Is it possible to draw on the plane a closed 20-segment polygonal chain and number its segments 1, 2, 3, ..., 20 in the order of traversal so that for each natural $i=1$, 2, 3, ..., 10, the segments numbered $i$ and $10+i$ intersect each other and do not intersect any other segments?

I. Efremov

Ural Young Mathematicians Tournament (62nd)

Solution

We construct a closed 20-segment polygonal chain satisfying the required intersection properties explicitly. The construction is combinatorial and geometric, with a precise embedding of the segments in the plane.

Construction

Place ten points $P_1, P_2, \dots, P_{10}$ evenly on a circle of radius $R$ in the plane, labeled in clockwise order. Around each point $P_i$ place a small disk $D_i$ of radius $r \ll R$, such that the disks are pairwise disjoint. Inside each disk $D_i$ draw two crossing segments forming an $X$. Let the two diagonals of the $X$ inside $D_i$ be denoted $s_i$ and $t_i$, with the convention that $s_i$ corresponds to segment $i$ of the polygonal chain and $t_i$ corresponds to segment $10+i$.

Each segment $s_i$ and $t_i$ is entirely contained in $D_i$, so they intersect exactly once at the center of the $X$ and do not meet any other segment_