Olympiad problems

2022 Iranian Geometry Olympiad, 4

Let $AD$ be the internal angle bisector of triangle $ABC$. The incircles of triangles $ABC$ and $ACD$ touch each other externally. Prove that $\angle ABC > 120^{\circ}$. (Recall that the incircle of a triangle is a circle inside the triangle that is tangent to its three sides.) [i]Proposed by Volodymyr Brayman (Ukraine)[/i]

2009 Serbia Team Selection Test, 1

Tags: trigonometry , angle bisector , derivative , geometry , inequalities , calculus

Let $ \alpha$ and $ \beta$ be the angles of a non-isosceles triangle $ ABC$ at points $ A$ and $ B$, respectively. Let the bisectors of these angles intersect opposing sides of the triangle in $ D$ and $ E$, respectively. Prove that the acute angle between the lines $ DE$ and $ AB$ isn't greater than $ \frac{|\alpha\minus{}\beta|}3$.

2001 Hungary-Israel Binational, 5

Tags: angle bisector , geometry , incenter , circumcircle

In a triangle $ABC$ , $B_{1}$ and $C_{1}$ are the midpoints of $AC$ and $AB$ respectively, and $I$ is the incenter. The lines $B_{1}I$ and $C_{1}I$ meet $AB$ and $AC$ respectively at $C_{2}$ and $B_{2}$ . If the areas of $\Delta ABC$ and $\Delta AB_{2}C_{2}$ are equal, ﬁnd $\angle{BAC}$ .

1999 North Macedonia National Olympiad, 3

Tags: geometry , tangent , angle bisector

Let the two tangents from a point $A$ outside a circle $k$ touch $k$ at $M$ and $N$. A line $p$ through $A$ intersects $k$ at $B$ and $C$, and $D$ is the midpoint of $MN$. Prove that $MN$ bisects the angle $BDC$

2000 Turkey Team Selection Test, 2

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

Points $M,\ N,\ K,\ L$ are taken on the sides $AB,\ BC,\ CD,\ DA$ of a rhombus $ABCD,$ respectively, in such a way that $MN\parallel LK$ and the distance between $MN$ and $KL$ is equal to the height of $ABCD.$ Show that the circumcircles of the triangles $ALM$ and $NCK$ intersect each other, while those of $LDK$ and $MBN$ do not.

2020 Abels Math Contest (Norwegian MO) Final, 4b

Tags: angle bisector , circumcircle , geometry

The triangle $ABC$ has a right angle at $A$. The centre of the circumcircle is called $O$, and the base point of the normal from $O$ to $AC$ is called $D$. The point $E$ lies on $AO$ with $AE = AD$. The angle bisector of $\angle CAO$ meets $CE$ in $Q$. The lines $BE$ and $OQ$ intersect in $F$. Show that the lines $CF$ and $OE$ are parallel.

1988 Balkan MO, 1

Tags: angle bisector , geometry , trigonometry , ratio

Let $ABC$ be a triangle and let $M,N,P$ be points on the line $BC$ such that $AM,AN,AP$ are the altitude, the angle bisector and the median of the triangle, respectively. It is known that $\frac{[AMP]}{[ABC]}=\frac{1}{4}$ and $\frac{[ANP]}{[ABC]}=1-\frac{\sqrt{3}}{2}$. Find the angles of triangle $ABC$.

1939 Moscow Mathematical Olympiad, 047

Tags: geometric inequality , angle bisector , geometry

Prove that for any triangle the bisector lies between the median and the height drawn from the same vertex.

1983 AMC 12/AHSME, 19

Tags: trigonometry , geometry , trig identities , law of cosines , angle bisector

Point $D$ is on side $CB$ of triangle $ABC$. If \[ \angle{CAD} = \angle{DAB} = 60^\circ,\quad AC = 3\quad\mbox{ and }\quad AB = 6, \] then the length of $AD$ is $\text{(A)} \ 2 \qquad \text{(B)} \ 2.5 \qquad \text{(C)} \ 3 \qquad \text{(D)} \ 3.5 \qquad \text{(E)} \ 4$

1987 India National Olympiad, 9

Tags: angle bisector , geometry , trigonometry

Prove that any triangle having two equal internal angle bisectors (each measured from a vertex to the opposite side) is isosceles.

2005 Morocco TST, 4

Tags: geometric transformation , reflection , geometry , angle bisector , perpendicular bisector

A convex quadrilateral $ABCD$ has an incircle. In each corner a circle is inscribed that also externally touches the two circles inscribed in the adjacent corners. Show that at least two circles have the same size.

2019 Romanian Master of Mathematics Shortlist, G2

Tags: parallel , angle bisector , geometry , perpendicular bisector

Let $ABC$ be an acute-angled triangle. The line through $C$ perpendicular to $AC$ meets the external angle bisector of $\angle ABC$ at $D$. Let $H$ be the foot of the perpendicular from $D$ onto $BC$. The point $K$ is chosen on $AB$ so that $KH \parallel AC$. Let $M$ be the midpoint of $AK$. Prove that $MC = MB + BH$. Giorgi Arabidze, Georgia,

2001 Junior Balkan MO, 2

Tags: power of a point , perpendicular bisector , symmetry , angle bisector , circumcircle , geometry , trigonometry

Let $ABC$ be a triangle with $\angle C = 90^\circ$ and $CA \neq CB$. Let $CH$ be an altitude and $CL$ be an interior angle bisector. Show that for $X \neq C$ on the line $CL$, we have $\angle XAC \neq \angle XBC$. Also show that for $Y \neq C$ on the line $CH$ we have $\angle YAC \neq \angle YBC$. [i]Bulgaria[/i]

2023 Sharygin Geometry Olympiad, 10

Tags: angle bisector , geometry , tangent circle

Altitudes $BE$ and $CF$ of an acute-angled triangle $ABC$ meet at point $H$. The perpendicular from $H$ to $EF$ meets the line $\ell$ passing through $A$ and parallel to $BC$ at point $P$. The bisectors of two angles between $\ell$ and $HP$ meet $BC$ at points $S$ and $T$. Prove that the circumcircles of triangles $ABC$ and $PST$ are tangent.

2007 All-Russian Olympiad, 3

Tags: asymptote , angle bisector , geometry , circumcircle , cyclic quadrilateral

$BB_{1}$ is a bisector of an acute triangle $ABC$. A perpendicular from $B_{1}$ to $BC$ meets a smaller arc $BC$ of a circumcircle of $ABC$ in a point $K$. A perpendicular from $B$ to $AK$ meets $AC$ in a point $L$. $BB_{1}$ meets arc $AC$ in $T$. Prove that $K$, $L$, $T$ are collinear. [i]V. Astakhov[/i]

2014 Saint Petersburg Mathematical Olympiad, 7

Tags: angle bisector , geometry , circumcircle , incenter

$I$ - incenter , $M$- midpoint of arc $BAC$ of circumcircle, $AL$ - angle bisector of triangle $ABC$. $MI$ intersect circumcircle in $K$. Circumcircle of $AKL$ intersect $BC$ at $L$ and $P$. Prove that $\angle AIP=90$

2002 Canada National Olympiad, 4

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

Let $\Gamma$ be a circle with radius $r$. Let $A$ and $B$ be distinct points on $\Gamma$ such that $AB < \sqrt{3}r$. Let the circle with centre $B$ and radius $AB$ meet $\Gamma$ again at $C$. Let $P$ be the point inside $\Gamma$ such that triangle $ABP$ is equilateral. Finally, let the line $CP$ meet $\Gamma$ again at $Q$. Prove that $PQ = r$.

2020 Bosnia and Herzegovina Junior BMO TST, 3

Tags: geometry , angle bisector

The angle bisector of $\angle ABC$ of triangle $ABC$ ($AB>BC$) cuts the circumcircle of that triangle in $K$. The foot of the perpendicular from $K$ to $AB$ is $N$, and $P$ is the midpoint of $BN$. The line through $P$ parallel to $BC$ cuts line $BK$ in $T$. Prove that the line $NT$ passes through the midpoint of $AC$.

2010 Postal Coaching, 2

Tags: geometry , angle bisector

In a circle with centre at $O$ and diameter $AB$, two chords $BD$ and $AC$ intersect at $E$. $F$ is a point on $AB$ such that $EF \perp AB$. $FC$ intersects $BD$ in $G$. If $DE = 5$ and $EG =3$, determine $BG$.

2008 Indonesia TST, 3

Tags: geometry , angle bisector , tangent circle , cyclic quadrilateral

Let $\Gamma_1$ and $\Gamma_2$ be two circles that tangents each other at point $N$, with $\Gamma_2$ located inside $\Gamma_1$. Let $A, B, C$ be distinct points on $\Gamma_1$ such that $AB$ and $AC$ tangents $\Gamma_2$ at $D$ and $E$, respectively. Line $ND$ cuts $\Gamma_1$ again at $K$, and line $CK$ intersects line $DE$ at $I$. (i) Prove that $CK$ is the angle bisector of $\angle ACB$. (ii) Prove that $IECN$ and $IBDN$ are cyclic quadrilaterals.

2008 Denmark MO - Mohr Contest, 4

Tags: angle bisector , geometry

In triangle $ABC$ we have $AB = 2, AC = 6$ and $\angle A = 120^o$ . The bisector of angle $A$ intersects the side BC at the point $D$. Determine the length of $AD$. The answer must be given as a fraction with integer numerator and denominator.

2011 National Olympiad First Round, 9

Tags: parallelogram , angle bisector , geometry

Let $ABCD$ be a convex quadrilateral with $m(\widehat{ADC}) = 90^{\circ}$. The line through $D$ which is parallel to $BC$ meets $AB$ at $E$. If $m(\widehat{DAC}) = m(\widehat{DAE})$, $|AB|=3$ and $|AC|=4$, then $|AE| = ?$ $\textbf{(A)}\ \frac56 \qquad\textbf{(B)}\ \frac13 \qquad\textbf{(C)}\ \frac12 \qquad\textbf{(D)}\ 1 \qquad\textbf{(E)}\ \frac34$

1949-56 Chisinau City MO, 25

Tags: geometry , angle bisector

Show that the straight lines passing through the feet of the altitudes of an acute-angled triangle form a triangle in which the altitudes of the original triangle are angle bisectors.

2002 Moldova National Olympiad, 3

Tags: angle bisector , geometry , circumcircle

In a triangle $ ABC$, the angle bisector at $ B$ intersects $ AC$ at $ D$ and the circumcircle again at $ E$. The circumcircle of the triangle $ DAE$ meets the segment $ AB$ again at $ F$. Prove that the triangles $ DBC$ and $ DBF$ are congruent.

2014 IMO Shortlist, G3

Tags: geometry , angle bisector

Let $\Omega$ and $O$ be the circumcircle and the circumcentre of an acute-angled triangle $ABC$ with $AB > BC$. The angle bisector of $\angle ABC$ intersects $\Omega$ at $M \ne B$. Let $\Gamma$ be the circle with diameter $BM$. The angle bisectors of $\angle AOB$ and $\angle BOC$ intersect $\Gamma$ at points $P$ and $Q,$ respectively. The point $R$ is chosen on the line $P Q$ so that $BR = MR$. Prove that $BR\parallel AC$. (Here we always assume that an angle bisector is a ray.) [i]Proposed by Sergey Berlov, Russia[/i]