Olympiad problems

2002 India IMO Training Camp, 13

Tags: asymptote , geometry , circumcircle , geometric transformation , reflection , trapezoid , perpendicular bisector

Let $ABC$ and $PQR$ be two triangles such that [list] [b](a)[/b] $P$ is the mid-point of $BC$ and $A$ is the midpoint of $QR$. [b](b)[/b] $QR$ bisects $\angle BAC$ and $BC$ bisects $\angle QPR$ [/list] Prove that $AB+AC=PQ+PR$.

2013 ELMO Shortlist, 13

Tags: geometry , circumcircle , ratio , asymptote , perpendicular bisector , angle bisector , power of a point

In $\triangle ABC$, $AB<AC$. $D$ and $P$ are the feet of the internal and external angle bisectors of $\angle BAC$, respectively. $M$ is the midpoint of segment $BC$, and $\omega$ is the circumcircle of $\triangle APD$. Suppose $Q$ is on the minor arc $AD$ of $\omega$ such that $MQ$ is tangent to $\omega$. $QB$ meets $\omega$ again at $R$, and the line through $R$ perpendicular to $BC$ meets $PQ$ at $S$. Prove $SD$ is tangent to the circumcircle of $\triangle QDM$. [i]Proposed by Ray Li[/i]

2015 Bosnia Herzegovina Team Selection Test, 6

Tags: geometry , incircle , incenter , perpendicular bisector , parallel

Let $D$, $E$ and $F$ be points in which incircle of triangle $ABC$ touches sides $BC$, $CA$ and $AB$, respectively, and let $I$ be a center of that circle.Furthermore, let $P$ be a foot of perpendicular from point $I$ to line $AD$, and let $M$ be midpoint of $DE$. If $\{N\}=PM\cap{AC}$, prove that $DN \parallel EF$

1995 Romania Team Selection Test, 1

Tags: incenter , circumcircle , ellipse , perpendicular bisector , geometry , conic

Let AD be the altitude of a triangle ABC and E , F be the incenters of the triangle ABD and ACD , respectively. line EF meets AB and AC at K and L. prove tht AK=AL if and only if AB=AC or A=90

2002 Italy TST, 1

Tags: circumcircle , perpendicular bisector , geometry

A scalene triangle $ABC$ is inscribed in a circle $\Gamma$. The bisector of angle $A$ meets $BC$ at $E$. Let $M$ be the midpoint of the arc $BAC$. The line $ME$ intersects $\Gamma$ again at $D$. Show that the circumcentre of triangle $AED$ coincides with the intersection point of the tangent to $\Gamma$ at $D$ and the line $BC$.

2014 China Girls Math Olympiad, 1

Tags: geometry , circumcircle , perpendicular bisector

In the figure of [url]http://www.artofproblemsolving.com/Forum/download/file.php?id=50643&mode=view[/url] $\odot O_1$ and $\odot O_2$ intersect at two points $A$, $B$. The extension of $O_1A$ meets $\odot O_2$ at $C$, and the extension of $O_2A$ meets $\odot O_1$ at $D$, and through $B$ draw $BE \parallel O_2A$ intersecting $\odot O_1$ again at $E$. If $DE \parallel O_1A$, prove that $DC \perp CO_2$.

2000 India Regional Mathematical Olympiad, 5

Tags: circumcircle , perpendicular bisector , angle bisector , geometry

The internal bisector of angle $A$ in a triangle $ABC$ with $AC > AB$ meets the circumcircle $\Gamma$ of the triangle in $D$. Join$D$ to the center $O$ of the circle $\Gamma$ and suppose that $DO$ meets $AC$ in $E$, possibly when extended. Given that $BE$ is perpendicular to $AD$, show that $AO$ is parallel to $BD$.

2019 Mexico National Olympiad, 2

Tags: orthocenter , perpendicular lines , perpendicular bisector , geometry

Let $H$ be the orthocenter of acute-angled triangle $ABC$ and $M$ be the midpoint of $AH$. Line $BH$ cuts $AC$ at $D$. Consider point $E$ such that $BC$ is the perpendicular bisector of $DE$. Segments $CM$ and $AE$ intersect at $F$. Show that $BF$ is perpendicular to $CM$. [i]Proposed by Germán Puga[/i]