Bariz - Problem

Source: Ukrainian Geometry Olympiad 2020, X p5 , XI p4

Tags: Coloring, combinatorics, combinatorial geometry, geometry, points

parmenides51

27.04.2020 04:19

On the plane painted $101$ points in brown and another $101$ points in green so that there are no three lying on one line. It turns out that the sum of the lengths of all $5050$ segments with brown ends equals the length of all $5050$ segments with green ends equal to $1$, and the sum of the lengths of all $10201$ segments with multicolored equals $400$. Prove that it is possible to draw a straight line so that all brown points are on one side relative to it and all green points are on the other.

Woow, nice

Lets take a look on two green points (call them P and Q). All other green points are $N_1, N_2,..., N_{99}$ Using triangle inequality $\sum_{i=1}(PN_i + QN_i) > 99PQ$, since that $PQ + \sum_{i=1}(PN_i + QN_i)<1 \implies PQ<\frac{1}{100}$ Lets prove that convex hull of all green points does not intersect the convex hull of brown This claim is equivalent proving that green segments do not intersect brown segments Let there are two such segments $G_1G_2$ and $B_1B_2$ with intersection at $O$ For arbitrary green point $G$ and brown $B$ we have $GB < BO + GO < max(GG_1; GG_2) + max(BB_1;BB_2) < \frac{1}{50}$, but $\frac{10201}{50} < 400$, contradiction.