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Belarusian Mathematical Olympiad

Geometry Difficulty 5.1 AIME, harder Prove it Belarus

Points $M$ and $N$ are the midpoints of the sides $BC$ and $AD$ , respectively, of a convex quadrilateral $ABCD$ . Is it possible that
$AB + CD > \max(AM + DM, BN + CN)$

Solution

Answer: no.
Since $(\angle ABC + \angle BCD) + (\angle BAD + \angle CDA) = 360^\circ$ , one of these summands is not less than $180^\circ$ . Without loss of generality, assume that $\angle ABC + \angle BCD \ge 180^\circ$ . Denote the reflection of the triangle $MCD$ about $M$ by $MBD_1$ . The inequality
$\angle ABM + \angle MBE = \angle ABC + \angle BCD \ge 180^\circ$
implies that the point $B$ lies either on the segment $AE$ or inside the triangle $AME$ . In both cases $AB + BE < AM + ME$ , which implies $AB + CD < AM + DM$ .

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