Olympiad problems

2004 Junior Balkan Team Selection Tests - Moldova, 7

Let the triangle $ABC$ have area $1$. The interior bisectors of the angles $\angle BAC,\angle ABC, \angle BCA$ intersect the sides $(BC), (AC), (AB) $ and the circumscribed circle of the respective triangle $ABC$ at the points $L$ and $G, N$ and $F, Q$ and $E$. The lines $EF, FG,GE$ intersect the bisectors $(AL), (CQ) ,(BN)$ respectively at points $P, M, R$. Determine the area of the hexagon $LMNPR$.

2013 Oral Moscow Geometry Olympiad, 2

Tags: common tangent , inscribed circle , angle bisector , equal angles , geometry

Inside the angle $AOD$, the rays $OB$ and $OC$ are drawn such that $\angle AOB = \angle COD.$ Two circles are inscribed inside the angles $\angle AOB$ and $\angle COD$ . Prove that the intersection point of the common internal tangents of these circles lies on the bisector of the angle $AOD$.

2012 National Olympiad First Round, 17

Tags: geometry , trigonometry , geometric transformation , reflection , symmetry , angle bisector

Let $D$ be a point inside $\triangle ABC$ such that $m(\widehat{BAD})=20^{\circ}$, $m(\widehat{DAC})=80^{\circ}$, $m(\widehat{ACD})=20^{\circ}$, and $m(\widehat{DCB})=20^{\circ}$. $m(\widehat{ABD})= ?$ $ \textbf{(A)}\ 5^{\circ} \qquad \textbf{(B)}\ 10^{\circ} \qquad \textbf{(C)}\ 15^{\circ} \qquad \textbf{(D)}\ 20^{\circ} \qquad \textbf{(E)}\ 25^{\circ}$

2015 Caucasus Mathematical Olympiad, 3

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

Let $AL$ be the angle bisector of the acute-angled triangle $ABC$. and $\omega$ be the circle circumscribed about it. Denote by $P$ the intersection point of the extension of the altitude $BH$ of the triangle $ABC$ with the circle $\omega$ . Prove that if $\angle BLA= \angle BAC$, then $BP = CP$.

2015 AIME Problems, 11

Tags: geometry , angle bisector

Triangle $ABC$ has positive integer side lengths with $AB=AC$. Let $I$ be the intersection of the bisectors of $\angle B$ and $\angle C$. Suppose $BI=8$. Find the smallest possible perimeter of $\triangle ABC$.

Swiss NMO - geometry, 2017.1

Tags: geometry , angle bisector , circles , equal segments

Let $A$ and $B$ be points on the circle $k$ with center $O$, so that $AB> AO$. Let $C$ be the intersection of the bisectors of $\angle OAB$ and $k$, different from $A$. Let $D$ be the intersection of the straight line $AB$ with the circumcircle of the triangle $OBC$, different from $B$. Show that $AD = AO$ .

1979 Chisinau City MO, 172

Tags: geometry , right triangle , angle bisector

Show that in a right-angled triangle the bisector of the right angle divides into equal parts the angle between the altitude and the median, drawn from the same vertex.

Novosibirsk Oral Geo Oly VII, 2020.6

Tags: angle bisector , angle , geometry

Angle bisectors $AA', BB'$and $CC'$ are drawn in triangle $ABC$ with angle $\angle B= 120^o$. Find $\angle A'B'C'$.

2010 Postal Coaching, 3

Tags: angle bisector , geometry

In a quadrilateral $ABCD$, we have $\angle DAB = 110^{\circ} , \angle ABC = 50^{\circ}$ and $\angle BCD = 70^{\circ}$ . Let $ M, N$ be the mid-points of $AB$ and $CD$ respectively. Suppose $P$ is a point on the segment $M N$ such that $\frac{AM}{CN} = \frac{MP}{PN}$ and $AP = CP$ . Find $\angle AP C$.

2024 Sharygin Geometry Olympiad, 1

Tags: geometry , incenter , angle bisector

Bisectors $AI$ and $CI$ meet the circumcircle of triangle $ABC$ at points $A_1, C_1$ respectively. The circumcircle of triangle $AIC_1$ meets $AB$ at point $C_0$; point $A_0$ is defined similarly. Prove that $A_0, A_1, C_0, C_1$ are collinear.

2013 Sharygin Geometry Olympiad, 5

Tags: geometry , angle bisector , median , altitude , concurrency

The altitude $AA'$, the median $BB'$, and the angle bisector $CC'$ of a triangle $ABC$ are concurrent at point $K$. Given that $A'K = B'K$, prove that $C'K = A'K$.

2008 Sharygin Geometry Olympiad, 4

Tags: symmetry , parallelogram , angle bisector , geometry

(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$.

1984 Tournament Of Towns, (067) T1

Tags: geometry , angle bisector , isosceles

In triangle $ABC$ the bisector of the angle at $B$ meets $AC$ at $D$ and the bisector of the angle at $C$ meets $AB$ at $E$. These bisectors intersect at $O$ and the lengths of $OD$ and $OE$ are equal. Prove that either $\angle BAC = 60^o$ or triangle $ABC$ is isosceles.

2021 Brazil EGMO TST, 3

Tags: geometry , circumcircle , angle bisector , perpendicular 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.

2003 IMO, 4

Tags: geometry , angle bisector , cyclic quadrilateral , quadrilateral , concurrency

Let $ABCD$ be a cyclic quadrilateral. Let $P$, $Q$, $R$ be the feet of the perpendiculars from $D$ to the lines $BC$, $CA$, $AB$, respectively. Show that $PQ=QR$ if and only if the bisectors of $\angle ABC$ and $\angle ADC$ are concurrent with $AC$.

2024 Yasinsky Geometry Olympiad, 5

Tags: geometry , angle bisector , median , parallelogram

Let $ AL $ be the bisector of triangle $ ABC $, $ O $ the center of its circumcircle, and $ D $ and $ E $ the midpoints of $ BL $ and $ CL $, respectively. Points $ P $ and $ Q $ are chosen on segments $ AD $ and $ AE $ such that $ APLQ $ is a parallelogram. Prove that $ PQ \perp AO $. [i]Proposed by Mykhailo Plotnikov[/i]

2023 Iranian Geometry Olympiad, 2

Tags: geometry , incenter , angle bisector

Let ${I}$ be the incenter of $\triangle {ABC}$ and ${BX}$, ${CY}$ are its two angle bisectors. ${M}$ is the midpoint of arc $\overset{\frown}{BAC}$. It is known that $MXIY$ are concyclic. Prove that the area of quadrilateral $MBIC$ is equal to that of pentagon $BXIYC$. [i]Proposed by Dominik Burek - Poland[/i]

2017 Latvia Baltic Way TST, 11

Tags: geometry , perimeter , angle bisector , geometric inequality

On the extension of the angle bisector $AL$ of the triangle $ABC$, a point $P$ is placed such that $P L = AL$. Prove that the perimeter of triangle $PBC$ does not exceed the perimeter of triangle $ABC$.

2019 Saudi Arabia Pre-TST + Training Tests, 4.3

Tags: geometry , circumcenter , angle bisector

Let $ABC$ be a triangle, let $D$ be the touch point of the side $BC$ and the incircle of the triangle $ABC$, and let $J_b$ and $J_c$ be the incentres of the triangles $ABD$ and $ACD$, respectively. Prove that the circumcentre of the triangle $AJ_bJ_c$ lies on the bisector of the angle $BAC$.

2012 Balkan MO Shortlist, G7

Tags: geometry , rectangle , angle bisector , cyclic quadrilateral , cyclic

$ABCD$ is a cyclic quadrilateral. The lines $AD$ and $BC$ meet at X, and the lines $AB$ and $CD$ meet at $Y$ . The line joining the midpoints $M$ and $N$ of the diagonals $AC$ and $BD$, respectively, meets the internal bisector of angle $AXB$ at $P$ and the external bisector of angle $BYC$ at $Q$. Prove that $PXQY$ is a rectangle

2015 Czech-Polish-Slovak Junior Match, 4

Tags: angle bisector , incenter , right triangle , midpoint , geometry

Let $ABC$ ne a right triangle with $\angle ACB=90^o$. Let $E, F$ be respecitvely the midpoints of the $BC, AC$ and $CD$ be it's altitude. Next, let $P$ be the intersection of the internal angle bisector from $A$ and the line $EF$. Prove that $P$ is the center of the circle inscribed in the triangle $CDE$ .

2024 JHMT HS, 8

Tags: geometry , angle bisector , circumcircle

Points $A$, $B$, $C$, and $D$ lie on a circle $\Gamma$, in that order, with $AB=5$ and $AD=3$. The angle bisector of $\angle ABC$ intersects $\Gamma$ at point $E$ on the opposite side of $\overleftrightarrow{CD}$ as $A$ and $B$. Assume that $\overline{BE}$ is a diameter of $\Gamma$ and $AC=AE$. Compute $DE$.

2017 CentroAmerican, 1

Tags: geometry , geometric transformation , reflection , angle bisector

$ABC$ is a right-angled triangle, with $\angle ABC = 90^{\circ}$. $B'$ is the reflection of $B$ over $AC$. $M$ is the midpoint of $AC$. We choose $D$ on $\overrightarrow{BM}$, such that $BD = AC$. Prove that $B'C$ is the angle bisector of $\angle MB'D$. NOTE: An important condition not mentioned in the original problem is $AB<BC$. Otherwise, $\angle MB'D$ is not defined or $B'C$ is the external bisector.

1959 IMO, 5

Tags: geometry , circumcircle , perpendicular bisector , angle 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 $B$.

2021 Polish Junior MO Finals, 4

Tags: geometry , angle bisector

On side $AB$ of a scalene triangle $ABC$ there are points $M$, $N$ such that $AN=AC$ and $BM=BC$. The line parallel to $BC$ through $M$ and the line parallel to $AC$ through $N$ intersect at $S$. Prove that $\measuredangle{CSM} = \measuredangle{CSN}$.