AoPS - Problem

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Problem

Source: Polish Junior Math Olympiad 2nd Round 2020

Tags: geometry, parallelogram, perpendicular bisector

pggp

22.10.2020 20:29

Let $ABCD$ be the parallelogram, such that angle at vertex $A$ is acute. Perpendicular bisector of the segment $AB$ intersects the segment $CD$ in the point $X$ . Let $E$ be the intersection point of the diagonals of the parallelogram $ABCD$ . Prove that $XE = \frac{1}{2}AD$ .

zuss77

22.10.2020 20:52

super easy, probably P1?

$M,N$ are midpoints of

$AB,CD$ then

$E$ - circumcenter of

$\triangle MXN$ and

$EX=EN=AD/2$

pggp

22.10.2020 21:01

It is 2nd stage of olympiad for 8th grade in Poland, so won't be hard.

franzliszt

22.10.2020 23:05

Fast sketch: Let the midpoint of $AD$ be $M$ . $MDFE$ is cyclic and $DF \parallel ME$ so $MDFE$ is an isosceles trapezoid.

Attachments:

parmenides51

08.04.2021 20:25

posted and solved also here

ZJ42

08.04.2021 20:26

are we allowed to post answers to math olympiad problems?

franzliszt

08.04.2021 20:31

Yes you are lol

hakN

08.04.2021 21:05

Let $F$ be the midpoint of $AB$ and let $G$ be the midpoint of $CD$ . Obviously $F,E,G$ are collinear. We have $\angle FXD = \angle FXA + \angle AXD = \angle FXA + \angle XAF = 90$ and combined with the fact that $E$ is the midpoint of $FG$ , we get that $\frac{AD}{2} = EF = EG = EX$ .