Olympiad problems

2017 Peru IMO TST, 9

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

Let $ABCD$ be a cyclie quadrilateral, $\omega$ be it's circumcircle and $M$ be the midpoint of the arc $AB$ of $\omega$ which does not contain the vertices $C$ and $D$. The line that passes through $M$ and the intersection point of segments $AC$ and $BD$, intersects again $\omega$ in $N$. Let $P$ and $Q$ be points in the $CD$ segment such that $\angle AQD = \angle DAP$ and $\angle BPC = \angle CBQ$. Prove that the circumcircle of $NPQ$ and $\omega$ are tangent to each other.

1996 Argentina National Olympiad, 4

Tags: perpendicular , geometry , parallelogram , equal angles , parallel

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 $B_1C$.

2022 Regional Olympiad of Mexico West, 3

Tags: equal angles , isosceles , geometry

In my isosceles triangle $\vartriangle ABC$ with $AB = CA$, we draw $D$ the midpoint of $BC$. Let $E$ be a point on $AC$ such that $\angle CDE = 60^o$ and $M$ the midpoint of $DE$. Prove that $\angle AME = \angle BMD$.

2018 India PRMO, 17

Tags: geometry , equal angles

Triangles $ABC$ and $DEF$ are such that $\angle A = \angle D, AB = DE = 17, BC = EF = 10$ and $AC - DF = 12$. What is $AC + DF$?

2022 China Team Selection Test, 1

Tags: geometry , circles , equal angles

Given two circles $\omega_1$ and $\omega_2$ where $\omega_2$ is inside $\omega_1$. Show that there exists a point $P$ such that for any line $\ell$ not passing through $P$, if $\ell$ intersects circle $\omega_1$ at $A,B$ and $\ell$ intersects circle $\omega_2$ at $C,D$, where $A,C,D,B$ lie on $\ell$ in this order, then $\angle APC=\angle BPD$.

2013 BMT Spring, 15

Tags: geometry , equal angles

Let $ABCD$ be a convex quadrilateral with $\angle ABD = \angle BCD$, $AD = 1000$, $BD = 2000$, $BC = 2001$, and $DC = 1999$. Point $E$ is chosen on segment $DB$ such that $\angle ABD = \angle ECD$. Find $AE$.

2010 Mathcenter Contest, 3

Tags: equal angles , geometry

$ABCD$ is a convex quadrilateral, and the point $K$ is a point on side $AB$, where $\angle KDA=\angle BCD$, let $L$ be a point on the diagonal $AC$, where $KL$ is parallel to $BC$. Prove that $$\angle KDB=\angle LDC.$$ [i](tatari/nightmare)[/i]

2010 Junior Balkan Team Selection Tests - Romania, 2

Tags: equal angles , geometry , convex quadrilateral , angle

Let $ABCD$ be a convex quadrilateral with $\angle BCD= 120^o, \angle {CBA} = 45^o, \angle {CBD} = 15^o$ and $\angle {CAB} = 90^o$. Show that $AB = AD$.

2020 Dutch IMO TST, 1

Tags: geometry , incenter , equal angles

In acute-angled triangle $ABC, I$ is the center of the inscribed circle and holds $| AC | + | AI | = | BC |$. Prove that $\angle BAC = 2 \angle ABC$.

2019 Yasinsky Geometry Olympiad, p4

Tags: geometry , perimeter , equal angles , incenter

In the triangle $ABC$, the side $BC$ is equal to $a$. Point $F$ is midpoint of $AB$, $I$ is the point of intersection of the bisectors of triangle $ABC$. It turned out that $\angle AIF = \angle ACB$. Find the perimeter of the triangle $ABC$. (Grigory Filippovsky)

2001 239 Open Mathematical Olympiad, 2

Tags: geometry , circumcircle , diagonal , equal angles

In a convex quadrangle $ ABCD $, the rays $ DA $ and $ CB $ intersect at point $ Q $, and the rays $ BA $ and $ CD $ at the point $ P $. It turned out that $ \angle AQB = \angle APD $. The bisectors of the angles $ \angle AQB $ and $ \angle APD $ intersect the sides quadrangle at points $ X $, $ Y $ and $ Z $, $ T $ respectively. Circumscribed circles of triangles $ ZQT $ and $ XPY $ intersect at $ K $ inside quadrangle. Prove that $ K $ lies on the diagonal $ AC $.