 Mathematical and Physical Journal
for High Schools
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# 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