parmenides51 wrote: In the non-isosceles triangle $ABC$ consider the points $X$ on $[AB]$ and $Y$ on $[AC]$ such that $[BX]=[CY]$, $M$ and $N$ are the midpoints of the segments $[BC]$, respectively $[AT]$, and the straight lines $XY$ and $BC$ meet in $K$. Prove that the circumcircle of triangle $KMN$ contains a point, different from $M$ , which is independent of the position of the points $X$ and $Y$. Can you please fix the problem statement? $M$ and $N$ are midpoints of which segments? and what is $T$?

indeed there was a typo, instead of $[AT]$, the correct is $[XY]$

Let $D$ is intersection of perpendicular bisector of $BC$ and circumcircle of $KMN$. $ \angle DNK=\angle DMK=90$, so DN is perpendicular bisector of $XY$. So $BD=CD,XD=YD \to \triangle XBD=\triangle YCD \to \angle ABD=\angle ACD$ So point $D$ lies on circumcenter of $ABC$ It means that circumcircle of triangle $KMN$ contains point of intersection of circumcenter $ABC$ and perpendicular bisector of $BC$

Shouldn't the shortlist problems be ascending depending on their difficulty? This can't be the difficulty of G7..

at that time, the bmo shortlist was not on ascending order, there were randomly labeled as g1,g2, etc