Let $ABCDE$ be a cyclic pentagon such that $BC = DE$ and $AB$ is parallel to $DE$. Let $X, Y,$ and $Z$ be the midpoints of $BD, CE,$ and $AE$ respectively. Show that $AE$ is tangent to the circumcircle of the triangle $XYZ$. Proposed by Nikola Velov, Macedonia
Problem
Source: JBMO Shortlist 2022
Tags: geometry, pentagon, Cyclic, Junior, Balkan, shortlist, tangent
on_gale
27.06.2023 05:37
By looking at our isosceles trapezoids, we get $XY \parallel BE, XZ \parallel AB$ and by midline we get $ZY \parallel AC$. We're using directed angles mod $180^{\circ}$. Now $\angle{XYZ} = \angle{DCA} = \angle{DBA} = \angle{BDE} = \angle{BXZ}$. This implies $BD$ tangent to $(XYZ)$ at $X$. By symmetry, $AE$ is also tangent and the conclusion follows.
Orestis_Lignos
27.06.2023 11:05
Very nice problem Note that $XY \parallel BE$, as $BCDE$ is an isosceles trapezoid. Hence, $\angle YZE=\angle CAE=180^\circ-\angle CDE=\angle BED=\angle(BE,ED)=\angle(XY,XZ)=\angle ZXY,$ as desired.
mathmax12
21.09.2023 23:45
Note, that $XY \parallel BE,$ and $XZ \parallel AB,$ also $YZ \parallel AC.$ Now $\angle EZY=180-\angle CDE=\angle BED=\angle YXZ \blacksquare$
vsamc
14.05.2024 21:56
Solved with Jack_w.
Let $O$ be the center of $(XYZ)$. First of all, since $BC = DE$, $BCDE$ is an isosceles trapezoid. Therefore, $XY\parallel CD \parallel BE$, and they all share the same perpendicular bisector as well. Note that $OX=OY$, so $O$ is on the perpendicular bisector of $XY$, which is the perpendicular bisector of $BE$. Now, note that since $AB$ is parallel to $DE$, $ABDE$ is an isosceles trapezoid as well, so it follows that $XZ$ is parallel to $DE$, and furthermore, the perpendicular bisector of $XZ$ is the perpendicular bisector of $DE$ and the perpendicular bisector of $AB$. Therefore, since $OX=OZ$, $O$ lies on the perpendicular bisector of $XZ$, which is the perpendicular bisector of $DE$. Thus, since $O$ lies on the perpendicular bisectors of $BE$ and $DE$, it follows that $O$ is the center of $(ABDE)$, so $O$ is the center of $(ABCDE)$. Therefore, $OZ\perp AE$, i.e. $(XYZ)$ is tangent to $AE$, as desired. $\blacksquare$
Math_.only.
24.05.2024 11:47
Great solutions
MathLuis
18.08.2024 07:06
Let $E'$ a point such that $EDE'B$ is a parallelogram, we get then $EX \cap CD=E'$, so now $\angle XZY=\angle E'AC=\angle BDC=\angle YXD=\angle XYC$ so we have $(XYZ)$ tangent to $BD,CE$, now to finish note that $AE=BD=CE$ by arc lenghts so $EZ=EY$ which means $AB$ is tangent to $(XYZ)$ as desired, thus done .