Olympiad problems

2019 Kosovo National Mathematical Olympiad, 3

Let $ABC$ be a triangle with $\angle CAB=60^{\circ}$ and with incenter $I$. Let points $D,E$ be on sides $AC,AB$, respectively, such that $BD$ and $CE$ are angle bisectors of angles $\angle ABC$ and $\angle BCA$, respectively. Show that $ID=IE$.

2020 Dutch IMO TST, 2

Tags: reflection , circumcircle , geometry , angle bisector , fixed point , fixed

Given is a triangle $ABC$ with its circumscribed circle and $| AC | <| AB |$. On the short arc $AC$, there is a variable point $D\ne A$. Let $E$ be the reflection of $A$ wrt the inner bisector of $\angle BDC$. Prove that the line $DE$ passes through a fixed point, regardless of point $D$.

1991 ITAMO, 1

Tags: circumcircle , geometry , angle bisector

For every triangle $ABC$ inscribed in a circle $\Gamma$ , let $A',B',C'$ be the intersections of the bisectors of the angles at $A,B,C$ with $\Gamma$ . Consider the triangle $A'B'C'$ . (a) Do triangles $A'B'C'$ go over all possible triangles inscribed in $\Gamma$ as $\vartriangle ABC$ varies? If not, what are the constraints? (b) Prove that the angle bisectors of $\vartriangle ABC$ are the altitudes of $\vartriangle A',B',C'$ .

2022 Bulgarian Spring Math Competition, Problem 8.2

Tags: angle bisector , equal lengths , geometry

Let $\triangle ABC$ have $AB = 1$ cm, $BC = 2$ cm and $AC = \sqrt{3}$ cm. Points $D$, $E$ and $F$ lie on segments $AB$, $AC$ and $BC$ respectively are such that $AE = BD$ and $BF = AD$. The angle bisector of $\angle BAC$ intersects the circumcircle of $\triangle ADE$ for the second time at $M$ and the angle bisector of $\angle ABC$ intersects the circumcircle of $\triangle BDF$ at $N$. Determine the length of $MN$.

2009 AIME Problems, 5

Tags: parallelogram , angle bisector , geometry , similar triangles , ratio

Triangle $ ABC$ has $ AC \equal{} 450$ and $ BC \equal{} 300$. Points $ K$ and $ L$ are located on $ \overline{AC}$ and $ \overline{AB}$ respectively so that $ AK \equal{} CK$, and $ \overline{CL}$ is the angle bisector of angle $ C$. Let $ P$ be the point of intersection of $ \overline{BK}$ and $ \overline{CL}$, and let $ M$ be the point on line $ BK$ for which $ K$ is the midpoint of $ \overline{PM}$. If $ AM \equal{} 180$, find $ LP$.

2002 National Olympiad First Round, 29

Tags: angle bisector , geometry

In $\triangle ABC$, angle bisector ıf $\widehat{CAB}$ meets $BC$ at $L$, angle bisector of $\widehat{ABC}$ meets $AC$ at $N$. Lines $AL$ and $BN$ meet at $O$. If $|NL| = \sqrt 3$, what is$|ON| + |OL|$? $ \textbf{a)}\ 3\sqrt 3 \qquad\textbf{b)}\ 2\sqrt 3 \qquad\textbf{c)}\ 2 \qquad\textbf{d)}\ 3 \qquad\textbf{e)}\ 5 $

2010 Contests, 2

Tags: geometry , angle bisector , rectangle , ratio

Let $ABCD$ be a rectangle of centre $O$, such that $\angle DAC=60^{\circ}$. The angle bisector of $\angle DAC$ meets $DC$ at $S$. Lines $OS$ and $AD$ meet at $L$, and lines $BL$ and $AC$ meet at $M$. Prove that lines $SM$ and $CL$ are parallel.

2010 Sharygin Geometry Olympiad, 3

Tags: angle bisector , equal segments , bijection , geometry

Let $ABCD$ be a convex quadrilateral and $K$ be the common point of rays $AB$ and $DC$. There exists a point $P$ on the bisectrix of angle $AKD$ such that lines $BP$ and $CP$ bisect segments $AC$ and $BD$ respectively. Prove that $AB = CD$.

2012 National Olympiad First Round, 21

Tags: angle bisector , geometry

The angle bisector of vertex $A$ of $\triangle ABC$ cuts $[BC]$ at $D$. The circle passing through $A$ and touching to $BC$ at $D$ meets $[AB]$ and $[AC]$ at $P$ and $Q$, respectively. $AD$ and $PQ$ meet at $T$. If $|AB|=5, |BC|=6, |CA|=7$, then $\frac{|AT|}{|TD|}=?$ $ \textbf{(A)}\ \frac75 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ \frac72 \qquad \textbf{(E)}\ 4$

2020 Saint Petersburg Mathematical Olympiad, 3.

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

$BB_1$ is the angle bisector of $\triangle ABC$, and $I$ is its incenter. The perpendicular bisector of segment $AC$ intersects the circumcircle of $\triangle AIC$ at $D$ and $E$. Point $F$ is on the segment $B_1C$ such that $AB_1=CF$.Prove that the four points $B, D, E$ and $F$ are concyclic.

2009 Romania Team Selection Test, 1

Tags: angle bisector , symmetry , geometry , incenter

The quadrilateral $ ABCD$ inscribed in a circle wich has diameter $ BD$. Let $ A',B'$ are symmetric to $ A,B$ with respect to the line $ BD$ and $ AC$ respectively. If $ A'C \cap BD \equal{} P$ and $ AC\cap B'D \equal{} Q$ then prove that $ PQ \perp AC$

2006 Tournament of Towns, 4

Tags: equal angles , geometry , angle bisector

In triangle $ABC$ let $X$ be some fixed point on bisector $AA'$ while point $B'$ be intersection of $BX$ and $AC$ and point $C'$ be intersection of $CX$ and $AB$. Let point $P$ be intersection of segments $A'B'$ and $CC'$ while point $Q$ be intersection of segments $A'C'$ and $BB'$. Prove τhat $\angle PAC = \angle QAB$.

Swiss NMO - geometry, 2013.10

Tags: concurrency , tangential quadrilateral , angle bisector , geometry

Let $ABCD$ be a tangential quadrilateral with $BC> BA$. The point $P$ is on the segment $BC$, such that $BP = BA$ . Show that the bisector of $\angle BCD$, the perpendicular on line $BC$ through $P$ and the perpendicular on $BD$ through $A$, intersect at one point.

1951 AMC 12/AHSME, 46

Tags: perpendicular bisector , geometry , angle bisector

$ AB$ is a fixed diameter of a circle whose center is $ O$. From $ C$, any point on the circle, a chord $ CD$ is drawn perpendicular to $ AB$. Then, as $ C$ moves over a semicircle, the bisector of angle $ OCD$ cuts the circle in a point that always: $ \textbf{(A)}\ \text{bisects the arc } AB \qquad\textbf{(B)}\ \text{trisects the arc } AB \qquad\textbf{(C)}\ \text{varies}$ $ \textbf{(D)}\ \text{is as far from }AB \text{ as from } D \qquad\textbf{(E)}\ \text{is equidistant from }B \text{ and } C$

2007 Hong kong National Olympiad, 1

Tags: perimeter , geometric transformation , incenter , homothety , geometry , angle bisector

Let $ABC$ be a triangle and $D$ be a point on $BC$ such that $AB+BD=AC+CD$. The line $AD$ intersects the incircle of triangle $ABC$ at $X$ and $Y$ where $X$ is closer to $A$ than $Y$ i. Suppose $BC$ is tangent to the incircle at $E$, prove that: 1) $EY$ is perpendicular to $AD$; 2) $XD=2IM$ where $I$ is the incentre and $M$ is the midpoint of $BC$.

2004 National Olympiad First Round, 25

Tags: geometry , angle bisector

Let $D$ be the foot of the internal angle bisector of the angle $A$ of a triangle $ABC$. Let $E$ be a point on side $[AC]$ such that $|CE|= |CD|$ and $|AE|=6\sqrt 5$; let $F$ be a point on the ray $[AB$ such that $|DB|=|BF|$ and $|AB|<|AF| = 8\sqrt 5$. What is $|AD|$? $ \textbf{(A)}\ 10\sqrt 5 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 4\sqrt{15} \qquad\textbf{(D)}\ 7\sqrt 5 \qquad\textbf{(E)}\ \text{None of above} $

1959 IMO Shortlist, 5

Tags: geometry , angle bisector , circumcircle , perpendicular bisector

An arbitrary point $M$ is selected in the interior of the segment $AB$. The square $AMCD$ and $MBEF$ are constructed on the same side of $AB$, with segments $AM$ and $MB$ as their respective bases. The circles circumscribed about these squares, with centers $P$ and $Q$, intersect at $M$ and also at another point $N$. Let $N'$ denote the point of intersection of the straight lines $AF$ and $BC$. a) Prove that $N$ and $N'$ coincide; b) Prove that the straight lines $MN$ pass through a fixed point $S$ independent of the choice of $M$; c) Find the locus of the midpoints of the segments $PQ$ as $M$ varies between $A$ and $B$.

2023 Polish MO Finals, 2

Tags: geometry , incenter , angle bisector

Given an acute triangle $ABC$ with their incenter $I$. Point $X$ lies on $BC$ on the same side as $B$ wrt $AI$. Point $Y$ lies on the shorter arc $AB$ of the circumcircle $ABC$. It is given that $$\angle AIX = \angle XYA = 120^\circ.$$ Prove that $YI$ is the angle bisector of $XYA$.

2018 PUMaC Geometry B, 8

Tags: parallelogram , angle bisector , geometry

Let $ABCD$ be a parallelogram such that $AB = 35$ and $BC = 28$. Suppose that $BD \perp BC$. Let $\ell_1$ be the reflection of $AC$ across the angle bisector of $\angle BAD$, and let $\ell_2$ be the line through $B$ perpendicular to $CD$. $\ell_1$ and $\ell_2$ intersect at a point $P$. If $PD$ can be expressed in simplest form as $\frac{m}{n}$, find $m + n$.

2007 Bulgarian Autumn Math Competition, Problem 11.3

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

In $\triangle ABC$ we have that $CC_{1}$ is an angle bisector. The points $P\in C_{1}B$, $Q\in BC$, $R\in AC$, $S\in AC_{1}$ satisfy $C_{1}P=PQ=QC$ and $CR=RS=SC_{1}$. Prove that $CC_{1}$ bisects $\angle SCP$.

2003 Iran MO (3rd Round), 4

Tags: geometry , angle bisector , vector

XOY is angle in the plane.A,B are variable point on OX,OY such that 1/OA+1/OB=1/K (k is constant).draw two circles with diameter OA and OB.prove that common external tangent to these circles is tangent to the constant circle( ditermine the radius and the locus of its center).

Estonia Open Senior - geometry, 2017.2.5

Tags: diameter , geometry , angle bisector

The bisector of the exterior angle at vertex $C$ of the triangle $ABC$ intersects the bisector of the interior angle at vertex $B$ in point $K$. Consider the diameter of the circumcircle of the triangle $BCK$ whose one endpoint is $K$. Prove that $A$ lies on this diameter.

2019 Yasinsky Geometry Olympiad, p3

Tags: tangent circle , circles , angle bisector , geometry

Two circles $\omega_1$ and $\omega_2$ are tangent externally at the point $P$. Through the point $A$ of the circle $\omega_1$ is drawn a tangent to this circle, which intersects the circle $\omega_2$ at points $B$ and $C$ (see figure). Line $CP$ intersects again the circle $\omega_1$ to $D$. Prove that the $PA$ is a bisector of the angle $DPB$. [img]https://1.bp.blogspot.com/-nmKZGdBXfao/XOd51gRFuyI/AAAAAAAAKO0/EYo2SCW0eGcJsF64-Avo6w73ugkIIQ30ACK4BGAYYCw/s1600/Yasinsky%2B2019%2Bp2.png[/img]

2005 MOP Homework, 5

Tags: geometry , angle bisector

Let $ABC$ be a triangle. Points $D$ and $E$ lie on sides $BC$ and $CA$, respectively, such that $BD=AE$. Segments $AD$ and $BE$ meet at $P$. The bisector of angle $BCA$ meet segments $AD$ and $BE$ at $Q$ and $R$, respectively. Prove that $\frac{PQ}{AD}=\frac{PR}{BE}$.

2006 National Olympiad First Round, 25

Tags: geometry , incenter , angle bisector

Let $E$ be the midpoint of the side $[BC]$ of $\triangle ABC$ with $|AB|=7$, $|BC|=6$, and $|AC|=5$. The line, which passes through $E$ and is perpendicular to the angle bisector of $\angle A$, intersects $AB$ at $D$. What is $|AD|$? $ \textbf{(A)}\ 5 \qquad\textbf{(B)}\ 6 \qquad\textbf{(C)}\ \frac 92 \qquad\textbf{(D)}\ 3\sqrt 2 \qquad\textbf{(E)}\ \text{None of above} $