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Part of 2020 Ukrainian Geometry Olympiad - April

Problems(4)

line a detaches line b, but b does not detach a, in a set of 2020 lines

Source: Ukrainian Geometry Olympiad 2020, VIII p5 , IX p4

4/27/2020
The plane shows 20202020 straight lines in general position, that is, there are none three intersecting at one point but no two parallel. Let's say, that the drawn line aa detaches the drawn line bb if all intersection points of line bb with the other drawn lines lie in one half plane wrt to line aa (given the most straightforward aa). Prove that you can be guaranteed find two drawn lines aa and bb that aa detaches bb, but bb does not detach aa.
combinatorial geometrycombinatorics
101 brown points, 101 green points, equal sums of 1-coloured segments

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

4/27/2020
On the plane painted 101101 points in brown and another 101101 points in green so that there are no three lying on one line. It turns out that the sum of the lengths of all 50505050 segments with brown ends equals the length of all 50505050 segments with green ends equal to 11, and the sum of the lengths of all 1020110201 segments with multicolored equals 400400. 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.
Coloringcombinatoricscombinatorial geometrygeometrypoints
BCDE parallelogram wanted insides a convex pentagon ABCDE

Source: Ukrainian Geometry Olympiad 2020, IX p5

4/27/2020
Given a convex pentagon ABCDEABCDE, with ∠BAC=∠ABE=∠DEA−90o\angle BAC = \angle ABE = \angle DEA - 90^o, ∠BCA=∠ADE\angle BCA = \angle ADE and also BC=EDBC = ED. Prove that BCDEBCDE is parallelogram.
geometryparallelogrampentagonequal anglesequal segments
AMxCM + BMxDM >= \sqrt{ABxBCxCDxDA} if <AMB =<ADM+<BCM ...

Source: Ukrainian Geometry Olympiad 2020, XI p5

4/27/2020
Inside the convex quadrilateral ABCDABCD there is a point MM such that ∠AMB=∠ADM+∠BCM\angle AMB = \angle ADM + \angle BCM and ∠AMD=∠ABM+∠DCM\angle AMD = \angle ABM + \angle DCM. Prove that AM⋅CM+BM⋅DM≥AB⋅BC⋅CD⋅DAAM \cdot CM + BM \cdot DM \ge \sqrt{AB \cdot BC\cdot CD \cdot DA}
geometryanglesgeometric inequality