Let $ABCD$ a convex quadrilateral such that $AD$ and $BC$ are not parallel. Let $M$ and $N$ the midpoints of $AD$ and $BC$ respectively. The segment $MN$ intersects $AC$ and $BD$ in $K$ and $L$ respectively, Show that at least one point of the intersections of the circumcircles of $AKM$ and $BNL$ is in the line $AB$.
Problem
Source: Peru IMO Shortlist
Tags: geometry, circumcircle, IMO Shortlist
AnArtist
14.01.2018 07:28
Please check your question BN intersects AC at C.
mathisreal
18.01.2018 05:51
Solutions or ideas??
Davrbek
18.01.2018 11:13
AnArtist wrote: Please check your question BN intersects AC at C. I thaink $MN$
hectorleo123
27.05.2023 04:08
Menelao i think
MathLuis
27.05.2023 04:25
Looks so complicated yet the sol its pretty clean, i liked this somehow Let $P$ the midpoint of $AB$ and let $(BNL) \cap AB=E$, by Reim's we get $MPEN$ cyclic, but now by Reim's again $MAEK$ is cyclic so $(BNL), (AKM)$ meet at $E$ which lies on $ABC$ so we are done (notice for the Reim's i only used midbases).
hectorleo123
27.05.2023 05:53
MathLuis wrote: Looks so complicated yet the sol its pretty clean, i liked this somehow Let $P$ the midpoint of $AB$ and let $(BNL) \cap AB=E$, by Reim's we get $MPEN$ cyclic, but now by Reim's again $MAEK$ is cyclic so $(BNL), (AKM)$ meet at $E$ which lies on $ABC$ so we are done (notice for the Reim's i only used midbases). Excuse my ignorance, but what is Reim's?