Olympiad problems

1979 Austrian-Polish Competition, 1

On sides $AB$ and $BC$ of a square $ABCD$ the respective points $E$ and $F$ have been chosen so that $BE = BF$. Let $BN$ be the altitude in triangle $BCE$. Prove that $\angle DNF = 90$.

2021 Yasinsky Geometry Olympiad, 6

Tags: geometry , incenter , arc midpoint , right angle

In an acute-angled triangle $ABC$, point $I$ is the center of the inscribed circle, point $T$ is the midpoint of the arc $ABC$ of the circumcircle of triangle $ABC$. It turned out that $\angle AIT = 90^o$ . Prove that $AB + AC = 3BC$. (Matthew of Kursk)

2018 NZMOC Camp Selection Problems, 4

Tags: geometry , right angle , equal angles

Let $P$ be a point inside triangle $ABC$ such that $\angle CPA = 90^o$ and $\angle CBP = \angle CAP$. Prove that $\angle P XY = 90^o$, where $X$ and $Y$ are the midpoints of $AB$ and $AC$ respectively.

2016 Thailand TSTST, 3

Tags: geometry , parallel , orthocenter , right angle

Let $H$ be the orthocenter of acute-angled $\vartriangle ABC$, and $X, Y$ points on the ray $AB, AC$. ($B$ lies between $X, A$, and $C$ lies between $Y, A$.) Lines $HX, HY$ intersect $BC$ at $D, E$ respectively. Let the line through $D$ parallel to $AC$ intersect $XY$ at $Z$. Prove that $\angle XHY = 90^o$ if and only if $ZE \parallel AB$.

2004 Germany Team Selection Test, 2

Tags: geometry , angle , combinatorial geometry , polygon , extremal combinatorics , right angle

Let $n \geq 5$ be a given integer. Determine the greatest integer $k$ for which there exists a polygon with $n$ vertices (convex or not, with non-selfintersecting boundary) having $k$ internal right angles. [i]Proposed by Juozas Juvencijus Macys, Lithuania[/i]

2020 Balkan MO Shortlist, G2

Tags: right angle , euler line , angle chasing , geometry

Let $G, H$ be the centroid and orthocentre of $\vartriangle ABC$ which has an obtuse angle at $\angle B$. Let $\omega$ be the circle with diameter $AG$. $\omega$ intersects $\odot(ABC)$ again at $L \ne A$. The tangent to $\omega$ at $L$ intersects $\odot(ABC)$ at $K \ne L$. Given that $AG = GH$, prove $\angle HKG = 90^o$ . [i]Sam Bealing, United Kingdom[/i]