Bariz - Problem

Source: 2016 Saudi Arabia GMO TST level 4+, I p3

Tags: geometry, collinear, midpoint, perpendicular bisector

parmenides51

01.08.2020 23:21

Let $ABC$ be an acute, non-isosceles triangle with the circumcircle $(O)$. Denote $D, E$ as the midpoints of $AB,AC$ respectively. Two circles $(ABE)$ and $(ACD)$ intersect at $K$ differs from $A$. Suppose that the ray $AK$ intersects $(O)$ at $L$. The line $LB$ meets $(ABE)$ at the second point $M$ and the line $LC$ meets $(ACD)$ at the second point $N$. a) Prove that $M, K, N$ collinear and $MN$ perpendicular to $OL$. b) Prove that $K$ is the midpoint of $MN$

See that $\angle ABL +\angle ACL=180^{\circ}=\angle LKM+\angle LKN$ and $M, K, N$ collinear. By Reim's notice that $EM\|CL$ and $DN \| BL$, this means $\angle DEM=\angle AEM-\angle AED=\angle BCL=\angle LMN=\angle DNM$, so $DENM$ cyclic. If $F=DN\cap EM$ then $F\in AK$ with this a) and b) are almost immediate, for a) see that $F$ is midpoint of $AL$ so $OF\perp AL$ but $\angle FOL=\angle ACL=\angle LKN$ so $O, F, K, OL\cap MN$ lie in a circle with diameter $OK$. For b) just see $FMLN$ is a parallelogram.