Olympiad problems

2023 Korea Summer Program Practice Test, P6

Tags: geometry , circumcircle , radical axis , power of a point , angle bisector

$AB < AC$ on $\triangle ABC$. The midpoint of arc $BC$ which doesn't include $A$ is $T$ and which includes $A$ is $S$. On segment $AB,AC$, $D,E$ exist so that $DE$ and $BC$ are parallel. The outer angle bisector of $\angle ABE$ and $\angle ACD$ meets $AS$ at $P$ and $Q$. Prove that the circumcircle of $\triangle PBE$ and $\triangle QCD$ meets on $AT$.

2024 Sharygin Geometry Olympiad, 1

Tags: angle bisector , geometry , incenter

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.

2009 China Team Selection Test, 2

Tags: law of sines , angle bisector , perpendicular bisector , geometry

In convex quadrilateral $ ABCD$, $ CB,DA$ are external angle bisectors of $ \angle DCA,\angle CDB$, respectively. Points $ E,F$ lie on the rays $ AC,BD$ respectively such that $ CEFD$ is cyclic quadrilateral. Point $ P$ lie in the plane of quadrilateral $ ABCD$ such that $ DA,CB$ are external angle bisectors of $ \angle PDE,\angle PCF$ respectively. $ AD$ intersects $ BC$ at $ Q.$ Prove that $ P$ lies on $ AB$ if and only if $ Q$ lies on segment $ EF$.

2018 Swedish Mathematical Competition, 1

Tags: cyclic quadrilateral , geometry , parallel , angle bisector

Let the $ABCD$ be a quadrilateral without parallel sides, inscribed in a circle. Let $P$ and $Q$ be the intersection points between the lines containing the quadrilateral opposite sides. Show that the bisectors to the angles at $P$ and $Q$ are parallel to the bisectors of the angles at the intersection point of the diagonals of the quadrilateral.

2019 Sharygin Geometry Olympiad, 6

Tags: angle chasing , circumcircle , geometry , angle bisector , median

Let $AK$ and $AT$ be the bisector and the median of an acute-angled triangle $ABC$ with $AC > AB$. The line $AT$ meets the circumcircle of $ABC$ at point $D$. Point $F$ is the reflection of $K$ about $T$. If the angles of $ABC$ are known, find the value of angle $FDA$.

2016 Bosnia And Herzegovina - Regional Olympiad, 2

Tags: geometry , angle bisector

Let $ABC$ be an isosceles triangle such that $\angle BAC = 100^{\circ}$. Let $D$ be an intersection point of angle bisector of $\angle ABC$ and side $AC$, prove that $AD+DB=BC$

2014 Sharygin Geometry Olympiad, 5

Tags: geometry , angle bisector

A triangle with angles of $30, 70$ and $80$ degrees is given. Cut it by a straight line into two triangles in such a way that an angle bisector in one of these triangles and a median in the other one drawn from two endpoints of the cutting segment are parallel to each other. (It suffices to find one such cutting.) (A. Shapovalov )

2004 India IMO Training Camp, 1

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

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 $AC$.

2004 India IMO Training Camp, 1

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

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 $AC$.

2019 New Zealand MO, 2

Tags: right angle , angle bisector , geometry

Let $X$ be the intersection of the diagonals $AC$ and $BD$ of convex quadrilateral $ABCD$. Let $P$ be the intersection of lines $AB$ and $CD$, and let $Q$ be the intersection of lines $PX$ and $AD$. Suppose that $\angle ABX = \angle XCD = 90^o$. Prove that $QP$ is the angle bisector of $\angle BQC$.

2013 Oral Moscow Geometry Olympiad, 4

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

Let $ABC$ be a triangle. On the extensions of sides $AB$ and $CB$ towards $B$, points $C_1$ and $A_1$ are taken, respectively, so that $AC = A_1C = AC_1$. Prove that circumscribed circles of triangles $ABA_1$ and $CBC_1$ intersect on the bisector of angle $B$.

2017 Brazil National Olympiad, 5.

Tags: circumcenter , angle bisector , geometry , orthocenter , incenter

[b]5.[/b] In triangle $ABC$, let $r_A$ be the line that passes through the midpoint of $BC$ and is perpendicular to the internal bisector of $\angle{BAC}$. Define $r_B$ and $r_C$ similarly. Let $H$ and $I$ be the orthocenter and incenter of $ABC$, respectively. Suppose that the three lines $r_A$, $r_B$, $r_C$ define a triangle. Prove that the circumcenter of this triangle is the midpoint of $HI$.

1959 AMC 12/AHSME, 28

Tags: angle bisector , geometry

In triangle $ABC$, $AL$ bisects angle $A$ and $CM$ bisects angle $C$. Points $L$ and $M$ are on $BC$ and $AB$, respectively. The sides of triangle $ABC$ are $a,b,$ and $c$. Then $\frac{\overline{AM}}{\overline{MB}}=k\frac{\overline{CL}}{\overline{LB}}$ where $k$ is: $ \textbf{(A)}\ 1 \qquad\textbf{(B)}\ \frac{bc}{a^2}\qquad\textbf{(C)}\ \frac{a^2}{bc}\qquad\textbf{(D)}\ \frac{c}{b}\qquad\textbf{(E)}\ \frac{c}{a} $

2021 Oral Moscow Geometry Olympiad, 3

Tags: angle bisector , geometry , circumcircle

Circle $(O)$ and its chord $BC$ are given. Point $A$ moves on the major arc $BC$. $AL$ is the angle bisector in a triangle $ABC$. Show that the disctance from the circumcenter of triangle $AOL$ to the line $BC$ does not depend on the position of point $A$.

1996 Vietnam National Olympiad, 2

Tags: symmetry , 3d geometry , tetrahedron , sphere , angle bisector , circumcircle , geometry

Given a trihedral angle Sxyz. A plane (P) not through S cuts Sx,Sy,Sz respectively at A,B,C. On the plane (P), outside triangle ABC, construct triangles DAB,EBC,FCA which are confruent to the triangles SAB,SBC,SCA respectively. Let (T) be the sphere lying inside Sxyz, but not inside the tetrahedron SABC, toucheing the planes containing the faces of SABC. Prove that (T) touches the plane (P) at the circumcenter of triangle DEF.

2014 Federal Competition For Advanced Students, 4

Tags: right triangle , angle bisector , geometry

We are given a right-angled triangle $MNP$ with right angle in $P$. Let $k_M$ be the circle with center $M$ and radius $MP$, and let $k_N$ be the circle with center $N$ and radius $NP$. Let $A$ and $B$ be the common points of $k_M$ and the line $MN$, and let $C$ and $D$ be the common points of $k_N$ and the line $MN$ with with $C$ between $A$ and $B$. Prove that the line $PC$ bisects the angle $\angle APB$.

1989 All Soviet Union Mathematical Olympiad, 504

Tags: equidistant , circumcenter , angle bisector , geometry

$ABC$ is a triangle. Points $D, E, F$ are chosen on $BC, CA, AB$ such that $B$ is equidistant from $D$ and $F$, and $C$ is equidistant from $D$ and $E$. Show that the circumcenter of $AEF$ lies on the bisector of $EDF$.

2006 France Team Selection Test, 1

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

Let $ABCD$ be a square and let $\Gamma$ be the circumcircle of $ABCD$. $M$ is a point of $\Gamma$ belonging to the arc $CD$ which doesn't contain $A$. $P$ and $R$ are respectively the intersection points of $(AM)$ with $[BD]$ and $[CD]$, $Q$ and $S$ are respectively the intersection points of $(BM)$ with $[AC]$ and $[DC]$. Prove that $(PS)$ and $(QR)$ are perpendicular.

1990 Swedish Mathematical Competition, 4

Tags: angle bisector , geometry , equal segments

$ABCD$ is a quadrilateral. The bisectors of $\angle A$ and $\angle B$ meet at $E$. The line through $E$ parallel to $CD$ meets $AD$ at $L$ and $BC$ at $M$. Show that $LM = AL + BM$.

2001 Federal Competition For Advanced Students, Part 2, 3

Tags: angle bisector , geometry

Let be given a semicircle with the diameter $AB$, and points $C,D$ on it such that $AC = CD$. The tangent at $C$ intersects the line $BD$ at $E$. The line $AE$ intersects the arc of the semicircle at $F$. Prove that $CF < FD$.

2013 National Olympiad First Round, 13

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

Let $D$ and $E$ be points on side $[BC]$ of a triangle $ABC$ with circumcenter $O$ such that $D$ is between $B$ and $E$, $|AD|=|DB|=6$, and $|AE|=|EC|=8$. If $I$ is the incenter of triangle $ADE$ and $|AI|=5$, then what is $|IO|$? $ \textbf{(A)}\ \dfrac {29}{5} \qquad\textbf{(B)}\ 5 \qquad\textbf{(C)}\ \dfrac {23}{5} \qquad\textbf{(D)}\ \dfrac {21}{5} \qquad\textbf{(E)}\ \text{None of above} $

2006 All-Russian Olympiad Regional Round, 11.4

Tags: geometry , angle bisector , parallel

The bisectors of angles $A$ and $C$ of triangle $ABC$ intersect its sides at points $A_1$ and $C_1$, and the circumcircle of this triangle is at points $A_0$ and $C_0$, respectively. Lines $A_1C_1$ and $A_0C_0$ intersect at point P. Prove that the segment connecting $P$ to the center of the incircle of triangle $ABC$ is parallel to $AC$.

2015 Romania National Olympiad, 3

Tags: midpoint , angle bisector , geometry , equal angles

In the convex quadrilateral $ABCD$ we have that $\angle BCD = \angle ADC \ge 90 ^o$. The bisectors of $\angle BAD$ and $\angle ABC$ intersect in $M$. Prove that if $M \in CD$, then $M$ is the middle of $CD$.

2022 Czech-Austrian-Polish-Slovak Match, 5

Tags: angle bisector , geometry

Let $ABC$ be a triangle with $AB < AC$ and circumcenter $O$. The angle bisector of $\angle BAC$ meets the side $BC$ at $D$. The line through $D$ perpendicular to $BC$ meets the segment $AO$ at $X$. Furthermore, let $Y$ be the midpoint of segment $AD$. Prove that points $B, C, X, Y$ are concyclic.

2012 Sharygin Geometry Olympiad, 3

Tags: angle bisector , construction , geometry

In triangle $ABC$, the bisector $CL$ was drawn. The incircles of triangles $CAL$ and $CBL$ touch $AB$ at points $M$ and $N$ respectively. Points $M$ and $N$ are marked on the picture, and then the whole picture except the points $A, L, M$, and $N$ is erased. Restore the triangle using a compass and a ruler. (V.Protasov)