Olympiad problems

2001 Austria Beginners' Competition, 4

Let $ABC$ be a triangle whose angles $\alpha=\angle CAB$ and $\beta=\angle CBA$ are greater than $45^{\circ}$. Above the side $AB$ a right isosceles triangle $ABR$ is constructed with $AB$ as the hypotenuse, such that $R$ is inside the triangle $ABC$. Analogously we construct above the sides $BC$ and $AC$ the right isosceles triangles $CBP$ and $ACQ$, right at $P$ and in $Q$, but with these outside the triangle $ABC$. Prove that $CQRP$ is a parallelogram.

1961 Czech and Slovak Olympiad III A, 2

Tags: construction , geometry , isosceles right triangle , geometric construction , square

Let a right isosceles triangle $APQ$ with the hypotenuse $AP$ be given in plane. Construct such a square $ABCD$ that the lines $BC, CD$ contain points $P, Q,$ respectively. Compute the length of side $AB = b$ in terms of $AQ=a$.

1999 German National Olympiad, 6a

Tags: combinatorics , combinatorial geometry , isosceles right triangle

Suppose that an isosceles right-angled triangle is divided into $m$ acute-angled triangles. Find the smallest possible $m$ for which this is possible.

2022 China Northern MO, 1

Tags: isosceles right triangle , equal angles , geometry

As shown in the figure, given $\vartriangle ABC$ with $AB \perp AC$, $AB=BC$, $D$ is the midpoint of the side $AB$, $DF\perp DE$, $DE=DF$ and $BE \perp EC$. Prove that $\angle AFD= \angle CEF$. [img]https://cdn.artofproblemsolving.com/attachments/9/2/f16a8c8c463874f3ccb333d91cdef913c34189.png[/img]

2022 Yasinsky Geometry Olympiad, 3

Tags: isosceles right triangle , angle , geometry

In an isosceles right triangle $ABC$ with a right angle $C$, point $M$ is the midpoint of leg $AC$. At the perpendicular bisector of $AC$, point $D$ was chosen such that $\angle CDM = 30^o$, and $D$ and $B$ lie on different sides of $AC$. Find the angle $\angle ABD$. (Volodymyr Petruk)