Olympiad problems

1996 Argentina National Olympiad, 4

Let $ABCD$ be a parallelogram with center $O$ such that $\angle BAD <90^o$ and $\angle AOB> 90^o$. Consider points $A_1$ and $B_1$ on the rays $OA$ and $OB$ respectively, such that $A_1B_1$ is parallel to $AB$ and $\angle A_1B_1C = \frac12 \angle ABC$. Prove that $A_1D$ is perpendicular to $B_1C$.

2019 Saudi Arabia Pre-TST + Training Tests, 5.2

Tags: angle bisector , geometry , equal angles

Let the bisector of the outside angle of $A$ of triangle $ABC$ and the circumcircle of triangle $ABC$ meet at point $P$. The circle passing through points $A$ and $P$ intersects segments $BP$ and $CP$ at points $E$ and $F$ respectively. Let $AD$ is the angle bisector of triangle $ABC$. Prove that $\angle PED = \angle PFD$. [img]https://cdn.artofproblemsolving.com/attachments/0/3/0638429a220f07227703a682479ed150302aae.png[/img]

Novosibirsk Oral Geo Oly VIII, 2019.4

Tags: equal angles , geometry

Given a triangle $ABC$, in which the angle $B$ is three times the angle $C$. On the side $AC$, point $D$ is chosen such that the angle $BDC$ is twice the angle $C$. Prove that $BD + BA = AC$.

2015 Caucasus Mathematical Olympiad, 3

Tags: equal segments , angle bisector , circumcircle , geometry , equal angles

Let $AL$ be the angle bisector of the acute-angled triangle $ABC$. and $\omega$ be the circle circumscribed about it. Denote by $P$ the intersection point of the extension of the altitude $BH$ of the triangle $ABC$ with the circle $\omega$ . Prove that if $\angle BLA= \angle BAC$, then $BP = CP$.

2007 Postal Coaching, 1

Tags: angle , equal angles , isosceles , geometry

Let $ABC$ be an isosceles triangle with $AC = BC$, and let $M$ be the midpoint of $AB$. Let $P$ be a point inside the triangle such that $\angle PAB =\angle PBC$. Prove that $\angle APM + \angle BPC = 180^o$.

Estonia Open Senior - geometry, 2018.2.5

Tags: geometry , angle , equal angles , reflection

Let $A'$ be the result of reflection of vertex $A$ of triangle ABC through line $BC$ and let $B'$ be the result of reflection of vertex $B$ through line $AC$. Given that $\angle BA' C = \angle BB'C$, can the largest angle of triangle $ABC$ be located: a) At vertex $A$, b) At vertex $B$, c) At vertex $C$?

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.

2017 Junior Balkan Team Selection Tests - Romania, 4

Tags: geometry , angle , equilateral , equal angles

Let $ABC$ be a right triangle, with the right angle at $A$. The altitude from $A$ meets $BC$ at $H$ and $M$ is the midpoint of the hypotenuse $[BC]$. On the legs, in the exterior of the triangle, equilateral triangles $BAP$ and $ACQ$ are constructed. If $N$ is the intersection point of the lines $AM$ and $PQ$, prove that the angles $\angle NHP$ and $\angle AHQ$ are equal. Miguel Ochoa Sanchez and Leonard Giugiuc

2011 Ukraine Team Selection Test, 9

Tags: geometry , cyclic quadrilateral , tangent , circumcircle , tangent circle , equal angles

Inside the inscribed quadrilateral $ ABCD $, a point $ P $ is marked such that $ \angle PBC = \angle PDA $, $ \angle PCB = \angle PAD $. Prove that there exists a circle that touches the straight lines $ AB $ and $ CD $, as well as the circles circumscribed by the triangles $ ABP $ and $ CDP $.

2019 Istmo Centroamericano MO, 3

Tags: geometry , tangent circle , equal angles

Let $ABC$ be an acute triangle, with $AB <AC$. Let $M$ be the midpoint of $AB$, $H$ the foot of the altitude from $A$, and $Q$ be point on side $AC$ such that $\angle ABQ = \angle BCA$. Show that the circumcircles of the triangles $ABQ$ and $BHM$ are tangent.

2020 Novosibirsk Oral Olympiad in Geometry, 6

Tags: geometry , equal angles , tangent

In triangle $ABC$, point $M$ is the midpoint of $BC$, $P$ the point of intersection of the tangents at points $B$ and $C$ of the circumscribed circle of $ABC$, $N$ is the midpoint of the segment $MP$. The segment $AN$ meets the circumcircle $ABC$ at the point $Q$. Prove that $\angle PMQ = \angle MAQ$.