Problem A. 758. (October 2019)

A. 758. In quadrilateral \(\displaystyle ABCD\), \(\displaystyle AB=BC=\frac1{\sqrt2}DA\), and \(\displaystyle \angle ABC\) is a right angle. The midpoint of side \(\displaystyle BC\) is \(\displaystyle E\), the orthogonal projection of \(\displaystyle C\) on \(\displaystyle AD\) is \(\displaystyle F\), and the orthogonal projection of \(\displaystyle B\) on \(\displaystyle CD\) is \(\displaystyle G\). The second intersection point of circle \(\displaystyle BCF\) (with center \(\displaystyle H\)) and line \(\displaystyle BG\) is \(\displaystyle K\), and the second intersection point of circle \(\displaystyle BHC\) and line \(\displaystyle HK\) is \(\displaystyle L\). The intersection of lines \(\displaystyle BL\) and \(\displaystyle CF\) is \(\displaystyle M\). The center of the Feurbach circle of triangle \(\displaystyle BFM\) is \(\displaystyle N\). Prove that \(\displaystyle \angle BNE\) is a right angle.

Proposed by Zsombor Fehér, Cambridge

(7 pont)

Deadline expired on November 11, 2019.

Statistics:

20 students sent a solution.
7 points:	Al-Hag Máté Amin, Bán-Szabó Áron, Csaplár Viktor, Füredi Erik Benjámin, Győrffi Ádám György, Hegedűs Dániel, Pooya Esmaeil Akhoondy, Somogyi Dalma, Szente Péter, Tiderenczl Dániel, Tóth 827 Balázs.
6 points:	Beke Csongor, Bursics András, Hervay Bence, Nguyen Bich Diep, Osztényi József, Seres-Szabó Márton, Szendrei Botond, Várkonyi Zsombor, Weisz Máté.

Problems in Mathematics of KöMaL, October 2019