Olympiad problems

2006 Iran MO (3rd Round), 1

Tags: geometry , geometric transformation , homothety , radical axis , ratio , angle bisector , incenter

Prove that in triangle $ABC$, radical center of its excircles lies on line $GI$, which $G$ is Centroid of triangle $ABC$, and $I$ is the incenter.

2010 Junior Balkan MO, 3

Tags: trigonometry , perpendicular bisector , angle bisector , geometry , circumcircle

Let $AL$ and $BK$ be angle bisectors in the non-isosceles triangle $ABC$ ($L$ lies on the side $BC$, $K$ lies on the side $AC$). The perpendicular bisector of $BK$ intersects the line $AL$ at point $M$. Point $N$ lies on the line $BK$ such that $LN$ is parallel to $MK$. Prove that $LN = NA$.

2006 National Olympiad First Round, 9

Tags: angle bisector , geometry

$ABC$ is a triangle with $|AB|=6$, $|BC|=7$, and $|AC|=8$. Let the angle bisector of $\angle A$ intersect $BC$ at $D$. If $E$ is a point on $[AC]$ such that $|CE|=2$, what is $|DE|$? $ \textbf{(A)}\ 3 \qquad\textbf{(B)}\ \frac {17}5 \qquad\textbf{(C)}\ \frac 72 \qquad\textbf{(D)}\ 2\sqrt 3 \qquad\textbf{(E)}\ 3\sqrt 2 $

2006 India National Olympiad, 1

Tags: trigonometry , similar triangles , incenter , angle bisector , geometry , circumcircle

In a non equilateral triangle $ABC$ the sides $a,b,c$ form an arithmetic progression. Let $I$ be the incentre and $O$ the circumcentre of the triangle $ABC$. Prove that (1) $IO$ is perpendicular to $BI$; (2) If $BI$ meets $AC$ in $K$, and $D$, $E$ are the midpoints of $BC$, $BA$ respectively then $I$ is the circumcentre of triangle $DKE$.

1999 Turkey Team Selection Test, 2

Tags: reflection , circumcircle , geometric transformation , cyclic quadrilateral , ratio , geometry , angle bisector

Let $L$ and $N$ be the mid-points of the diagonals $[AC]$ and $[BD]$ of the cyclic quadrilateral $ABCD$, respectively. If $BD$ is the bisector of the angle $ANC$, then prove that $AC$ is the bisector of the angle $BLD$.

1999 IMO Shortlist, 8

Tags: circumcircle , triangle , geometry , angle bisector , incenter

Given a triangle $ABC$. The points $A$, $B$, $C$ divide the circumcircle $\Omega$ of the triangle $ABC$ into three arcs $BC$, $CA$, $AB$. Let $X$ be a variable point on the arc $AB$, and let $O_{1}$ and $O_{2}$ be the incenters of the triangles $CAX$ and $CBX$. Prove that the circumcircle of the triangle $XO_{1}O_{2}$ intersects the circle $\Omega$ in a fixed point.

2004 Junior Balkan Team Selection Tests - Romania, 1

Tags: angle bisector , geometry , trigonometry

Let $ABC$ be a triangle, having no right angles, and let $D$ be a point on the side $BC$. Let $E$ and $F$ be the feet of the perpendiculars drawn from the point $D$ to the lines $AB$ and $AC$ respectively. Let $P$ be the point of intersection of the lines $BF$ and $CE$. Prove that the line $AP$ is the altitude of the triangle $ABC$ from the vertex $A$ if and only if the line $AD$ is the angle bisector of the angle $CAB$.

Ukraine Correspondence MO - geometry, 2005.4

Tags: geometry , angle , angle bisector

The bisectors of the angles $A$ and $B$ of the triangle $ABC$ intersect the sides $BC$ and $AC$ at points $D$ and $E$. It is known that $AE + BD = AB$. Find the angle $\angle C$.

2008 Sharygin Geometry Olympiad, 4

Tags: geometry , symmetry , angle bisector , parallelogram

(F.Nilov, A.Zaslavsky) Let $ CC_0$ be a median of triangle $ ABC$; the perpendicular bisectors to $ AC$ and $ BC$ intersect $ CC_0$ in points $ A'$, $ B'$; $ C_1$ is the meet of lines $ AA'$ and $ BB'$. Prove that $ \angle C_1CA \equal{} \angle C_0CB$.

2021 Brazil EGMO TST, 3

Tags: geometry , perpendicular bisector , circumcircle , angle bisector

Let $ABC$ be an acute-angled triangle with $AC>AB$, and $\Omega$ is your circumcircle. Let $P$ be the midpoint of the arc $BC$ of $\Omega$ (not containing $A$) and $Q$ be the midpoint of the arc $BC$ of $\Omega$(containing the point $A$). Let $M$ be the foot of perpendicular of $Q$ on the line $AC$. Prove that the circumcircle of $\triangle AMB$ cut the segment $AP$ in your midpoint.

2018 Pan-African Shortlist, G4

Tags: angle bisector , geometry , collinear

Let $ABC$ be a triangle and $\Gamma$ be the circle with diameter $[AB]$. The bisectors of $\angle BAC$ and $\angle ABC$ cut the circle $\Gamma$ again at $D$ and $E$, respectively. The incicrcle of the triangle $ABC$ cuts the lines $BC$ and $AC$ in $F$ and $G$, respectively. Show that the points $D, E, F$ and $G$ lie on the same line.

2014 Oral Moscow Geometry Olympiad, 6

Tags: excenter , concurrency , incenter , geometry , angle bisector

A convex quadrangle $ABCD$ is given. Let $I$ and $J$ be the circles of circles inscribed in the triangles $ABC$ and $ADC$, respectively, and $I_a$ and $J_a$ are the centers of the excircles circles of triangles $ABC$ and $ADC$, respectively (inscribed in the angles $BAC$ and $DAC$, respectively). Prove that the intersection point $K$ of the lines $IJ_a$ and $JI_a$ lies on the bisector of the angle $BCD$.

2012 NZMOC Camp Selection Problems, 5

Tags: geometry , parallel , equal angles , angle bisector

Let $ABCD$ be a quadrilateral in which every angle is smaller than $180^o$. If the bisectors of angles $\angle DAB$ and $\angle DCB$ are parallel, prove that $\angle ADC = \angle ABC$

2018 Bosnia And Herzegovina - Regional Olympiad, 5

Tags: orthocenter , collinear , angle bisector , geometry

Let $H$ be an orhocenter of an acute triangle $ABC$ and $M$ midpoint of side $BC$. If $D$ and $E$ are foots of perpendicular of $H$ on internal and external angle bisector of angle $\angle BAC$, prove that $M$, $D$ and $E$ are collinear

2020 Brazil National Olympiad, 1

Tags: geometry , angle bisector

Let $ABC$ be an acute triangle and $AD$ a height. The angle bissector of $\angle DAC$ intersects $DC$ at $E$. Let $F$ be a point on $AE$ such that $BF$ is perpendicular to $AE$. If $\angle BAE=45º$, find $\angle BFC$.

2005 Junior Balkan Team Selection Tests - Moldova, 5

Tags: geometry , trig identities , law of sines , perpendicular bisector , circumcircle , trigonometry , angle bisector

Let $ABC$ be an acute-angled triangle, and let $F$ be the foot of its altitude from the vertex $C$. Let $M$ be the midpoint of the segment $CA$. Assume that $CF=BM$. Then the angle $MBC$ is equal to angle $FCA$ if and only if the triangle $ABC$ is equilateral.

2007 Junior Balkan Team Selection Tests - Romania, 3

Tags: reflection , angle bisector , geometric transformation , power of a point , radical axis , trigonometry , geometry

Let $ABC$ be a right triangle with $A = 90^{\circ}$ and $D \in (AC)$. Denote by $E$ the reflection of $A$ in the line $BD$ and $F$ the intersection point of $CE$ with the perpendicular in $D$ to $BC$. Prove that $AF, DE$ and $BC$ are concurrent.

2019 IFYM, Sozopol, 4

Tags: angle bisector , radical axis , geometry

The inscribed circle of an acute $\Delta ABC$ is tangent to $AB$ and $AC$ in $K$ and $L$ respectively. The altitude $AH$ intersects the angle bisectors of $\angle ABC$ and $\angle ACB$ in $P$ and $Q$ respectively. Prove that the middle point $M$ of $AH$ lies on the radical axis of the circumscribed circles of $\Delta KPB$ and $\Delta LQC$.

1998 AMC 12/AHSME, 28

Tags: relatively prime , trigonometry , pythagorean theorem , angle bisector , number theory , geometry , ratio

In triangle $ ABC$, angle $ C$ is a right angle and $ CB > CA$. Point $ D$ is located on $ \overline{BC}$ so that angle $ CAD$ is twice angle $ DAB$. If $ AC/AD \equal{} 2/3$, then $ CD/BD \equal{} m/n$, where $ m$ and $ n$ are relatively prime positive integers. Find $ m \plus{} n$. $ \textbf{(A)}\ 10\qquad \textbf{(B)}\ 14\qquad \textbf{(C)}\ 18\qquad \textbf{(D)}\ 22\qquad \textbf{(E)}\ 26$

1989 India National Olympiad, 6

Tags: perpendicular bisector , incenter , angle bisector , geometry

Triangle $ ABC$ has incentre $ I$ and the incircle touches $ BC, CA$ at $ D, E$ respectively. Let $ BI$ meet $ DE$ at $ G$. Show that $ AG$ is perpendicular to $ BG$.

Ukrainian From Tasks to Tasks - geometry, 2011.8

Tags: isosceles , angle bisector , parallel , geometry

On the median $AD$ of the isosceles triangle $ABC$, point $E$ is marked. Point $F$ is the projection of point $E$ on the line $BC$, point $M$ lies on the segment $EF$, points $N$ and $P$ are projections of point $M$ on the lines $AC$ and $AB$, respectively. Prove that the bisectors of the angles $PMN$ and $PEN$ are parallel.

2002 Estonia Team Selection Test, 2

Tags: isosceles , trigonometry , angle bisector , geometry

Consider an isosceles triangle $KL_1L_2$ with $|KL_1|=|KL_2|$ and let $KA, L_1B_1,L_2B_2$ be its angle bisectors. Prove that $\cos \angle B_1AB_2 < \frac35$

2009 Ukraine National Mathematical Olympiad, 4

Tags: geometry , angle bisector

In the triangle $ABC$ given that $\angle ABC = 120^\circ .$ The bisector of $\angle B$ meet $AC$ at $M$ and external bisector of $\angle BCA$ meet $AB$ at $P.$ Segments $MP$ and $BC$ intersects at $K$. Prove that $\angle AKM = \angle KPC .$

2014 Contests, 3

Tags: geometry , angle bisector

In obtuse triangle $ABC$, with the obtuse angle at $A$, let $D$, $E$, $F$ be the feet of the altitudes through $A$, $B$, $C$ respectively. $DE$ is parallel to $CF$, and $DF$ is parallel to the angle bisector of $\angle BAC$. Find the angles of the triangle.

2011 Canada National Olympiad, 2

Tags: angle bisector , geometry , cyclic quadrilateral

Let $ABCD$ be a cyclic quadrilateral with opposite sides not parallel. Let $X$ and $Y$ be the intersections of $AB,CD$ and $AD,BC$ respectively. Let the angle bisector of $\angle AXD$ intersect $AD,BC$ at $E,F$ respectively, and let the angle bisectors of $\angle AYB$ intersect $AB,CD$ at $G,H$ respectively. Prove that $EFGH$ is a parallelogram.