Olympiad problems

1993 Bundeswettbewerb Mathematik, 3

In the triangle $ABC$, let $A'$ be the intersection of the perpendicular bisector of $AB$ and the angle bisector of $\angle BAC$ and define $B', C'$ analogously. Prove that a) The triangle $ABC$ is equilateral if and only if $A' =B'.$ b) If $A', B'$ and $C'$ are distinct, we have $\angle B' A' C' = 90^{\circ} - \frac{1}{2} \angle BAC.$

2016 Singapore Senior Math Olympiad, 1

Tags: geometry , angle bisector

Let $\triangle ABC$ be a triangle with $AB < AC$. Let the angle bisector of $\angle BAC$ meet $BC$ at $D$ , and let $M$ be the midpoint of $BC$ . Let $P$ be the foot of the perpendicular from $B$ to $AD$ . $Q$ the intersection of $BP$ and $AM$ . Show that : $(DQ) // (AB) $ .

2008 Mathcenter Contest, 1

Tags: ratio , cyclic quadrilateral , geometry , angle bisector

In a triangle $ABC$, the angle bisector at $A,B,C$ meet the opposite sides at $A_1,B_1,C_1$, respectively. Prove that if the quadrilateral $BA_1B_1C_1$ is cyclic, then $$\frac{AC}{AB+BC}=\frac{AB}{AC+CB}+\frac{BC}{BA+AC}.$$

2018 Iranian Geometry Olympiad, 4

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

Quadrilateral $ABCD$ is circumscribed around a circle. Diagonals $AC,BD$ are not perpendicular to each other. The angle bisectors of angles between these diagonals, intersect the segments $AB,BC,CD$ and $DA$ at points $K,L,M$ and $N$. Given that $KLMN$ is cyclic, prove that so is $ABCD$. Proposed by Nikolai Beluhov (Bulgaria)

2009 Belarus Team Selection Test, 1

Tags: geometry , circumcenter , median , angle bisector

In a triangle $ABC, AM$ is a median, $BK$ is a bisectrix, $L=AM\cap BK$. It is known that $BC=a, AB=c, a>c$. Given that the circumcenter of triangle $ABC$ lies on the line $CL$, find $AC$ I. Voronovich

2020 Peru Iberoamerican Team Selection Test, P3

Tags: geometry , angle bisector

Let $ABC$ be an acute triangle with $| AB | > | AC |$. Let $D$ be the foot of the altitude from $A$ to $BC$, let $K$ be the intersection of $AD$ with the internal bisector of angle $B$, Let $M$ be the foot of the perpendicular from $B$ to $CK$ (it could be in the extension of segment $CK$) and$ N$ the intersection of $BM$ and $AK$ (it could be in the extension of the segments). Let $T$ be the intersection of$ AC$ with the line that passes through $N$ and parallel to $DM$. Prove that $BM$ is the internal bisector of the angle $\angle TBC$

2004 Junior Balkan Team Selection Tests - Moldova, 3

Tags: parallelogram , geometry , cyclic , angle bisector

Let $ABCD$ be a parallelogram and point $M$ be the midpoint of $[AB]$ so that the quadrilateral $MBCD$ is cyclic. If $N$ is the point of intersection of the lines $DM$ and $BC$, and $P \in BC$, then prove that the ray $(DP$ is the angle bisector of $\angle ADM$ if and only if $PC = 4BC$.

2012 Dutch IMO TST, 5

Tags: geometry , circumcircle , angle bisector

Let $\Gamma$ be the circumcircle of the acute triangle $ABC$. The angle bisector of angle $ABC$ intersects $AC$ in the point $B_1$ and the short arc $AC$ of $\Gamma$ in the point $P$. The line through $B_1$ perpendicular to $BC$ intersects the short arc $BC$ of $\Gamma$ in $K$. The line through $B$ perpendicular to $AK$ intersects $AC$ in $L$. Prove that $K, L$ and $P$ lie on a line.

2007 Estonia Math Open Junior Contests, 7

Tags: geometry , right angle , square , angle bisector

The center of square $ABCD$ is $K$. The point $P$ is chosen such that $P \ne K$ and the angle $\angle APB$ is right . Prove that the line $PK$ bisects the angle between the lines $AP$ and $BP$.

2015 Bosnia And Herzegovina - Regional Olympiad, 3

Tags: angle bisector , geometry

Let $F$ be an intersection point of altitude $CD$ and internal angle bisector $AE$ of right angled triangle $ABC$, $\angle ACB = 90^{\circ}$. Let $G$ be an intersection point of lines $ED$ and $BF$. Prove that area of quadrilateral $CEFG$ is equal to area of triangle $BDG$

2019 Girls in Mathematics Tournament, 5

Tags: midpoint , equal segments , angle bisector , isosceles , projective geometry , geometry

Let $ABC$ be an isosceles triangle with $AB = AC$. Let $X$ and $K$ points over $AC$ and $AB$, respectively, such that $KX = CX$. Bisector of $\angle AKX$ intersects line $BC$ at $Z$. Show that $XZ$ passes through the midpoint of $BK$.

2010 Sharygin Geometry Olympiad, 1

Tags: tangent , circumcircle , geometry , angle bisector

For a nonisosceles triangle $ABC$, consider the altitude from vertex $A$ and two bisectrices from remaining vertices. Prove that the circumcircle of the triangle formed by these three lines touches the bisectrix from vertex $A$.

2015 Latvia Baltic Way TST, 6

Tags: geometry , bisects segment , angle bisector

$AM$ is the median of triangle $ABC$. A perpendicular $CC_1$ is drawn from point $C$ on the bisector of angle $\angle CMA$, a perpendicular $BB_1$ is drawn from point $B$ on the bisector of angle $\angle BMA$. Prove that line $AM$ intersects segment $B_1C_1$ at its midpoint.

2015 Switzerland - Final Round, 1

Tags: ratio , angle bisector , geometry , circumcircle

Let $ABC$ be an acute-angled triangle with $AB \ne BC$ and radius $k$. Let $P$ and $Q$ be the points of intersection of $k$ with the internal bisector and the external bisector of $\angle CBA$ respectively. Let $D$ be the intersection of $AC$ and $PQ$. Find the ratio $AD: DC$.

2014 Indonesia MO Shortlist, G4

Tags: concurrency , angle bisector , geometry

Given an acute triangle $ABC$ with $AB <AC$. Points $P$ and $Q$ lie on the angle bisector of $\angle BAC$ so that $BP$ and $CQ$ are perpendicular on that angle bisector. Suppose that point $E, F$ lie respectively at sides $AB$ and $AC$ respectively, in such a way that $AEPF$ is a kite. Prove that the lines $BC, PF$, and $QE$ intersect at one point.

2015 Sharygin Geometry Olympiad, 5

Tags: geometry , bijection , angle bisector

Two equal hard triangles are given. One of their angles is equal to $ \alpha$ (these angles are marked). Dispose these triangles on the plane in such a way that the angle formed by some three vertices would be equal to $ \alpha / 2$. [i](No instruments are allowed, even a pencil.)[/i] (E. Bakayev, A. Zaslavsky)

1999 Tournament Of Towns, 4

Tags: geometry , perimeter , bisects , excircle , midpoint , angle bisector

Points $K, L$ on sides $AC, CB$ respectively of a triangle $ABC$ are the points of contact of the excircles with the corresponding sides . Prove that the straight line through the midpoints of $KL$ and $AB$ (a) divides the perimeter of triangle $ABC$ in half, (b) is parallel to the bisector of angle $ACB$. ( L Emelianov)

2018 Azerbaijan Senior NMO, 3

Tags: geometry , tangent , angle bisector , circles

A circle $\omega$ and a point $T$ outside the circle is given. Let a tangent from $T$ to $\omega$ touch $\omega$ at $A$, and take points $B,C$ lying on $\omega$ such that $T,B,C$ are colinear. The bisector of $\angle ATC$ intersects $AB$ and $AC$ at $P$ and $Q$,respectively. Prove that $PA=\sqrt{PB\cdot QC}$

1995 Tournament Of Towns, (477) 1

Tags: parallelogram , geometry , angle bisector , locus

If P is a point inside a convex quadrilateral $ABCD$, let the angle bisectors of $\angle APB$, $\angle BPC$, $\angle CPD$ and $\angle DPA$ meet $AB$, $BC$, $CD$ and $DA$ at $K$, $L$, $M$ and $N$ respectively. (a) Find a point $P$ such that $KLMN$ is a parallelogram. (b) Find the locus of all such points $P$. (S Tokarev)

1998 Tournament Of Towns, 5

Tags: geometry , angle bisector , reflection , circles

A circle with center $O$ is inscribed in an angle. Let $A$ be the reflection of $O$ across one side of the angle. Tangents to the circle from $A$ intersect the other side of the angle at points $B$ and $C$. Prove that the circumcenter of triangle $ABC$ lies on the bisector of the original angle. (I.Sharygin)

2005 Germany Team Selection Test, 2

Tags: angle bisector , circumcircle , geometry

Let $ABC$ be a triangle satisfying $BC < CA$. Let $P$ be an arbitrary point on the side $AB$ (different from $A$ and $B$), and let the line $CP$ meet the circumcircle of triangle $ABC$ at a point $S$ (apart from the point $C$). Let the circumcircle of triangle $ASP$ meet the line $CA$ at a point $R$ (apart from $A$), and let the circumcircle of triangle $BPS$ meet the line $CB$ at a point $Q$ (apart from $B$). Prove that the excircle of triangle $APR$ at the side $AP$ is identical with the excircle of triangle $PQB$ at the side $PQ$ if and only if the point $S$ is the midpoint of the arc $AB$ on the circumcircle of triangle $ABC$.

2012 IMAR Test, 3

Tags: geometry , angle bisector , reflection , equal angles

Given a triangle $ABC$, let $D$ be a point different from $A$ on the external bisectrix $\ell$ of the angle $BAC$, and let $E$ be an interior point of the segment $AD$. Reflect $\ell$ in the internal bisectrices of the angles $BDC$ and $BEC$ to obtain two lines that meet at some point $F$. Show that the angles $ABD$ and $EBF$ are congruent.

2009 Serbia Team Selection Test, 3

Tags: geometry , angle bisector , geometric transformation , exterior angle , circumcircle , ratio , homothety

Let $ k$ be the inscribed circle of non-isosceles triangle $ \triangle ABC$, which center is $ S$. Circle $ k$ touches sides $ BC,CA,AB$ in points $ P,Q,R$ respectively. Line $ QR$ intersects $ BC$ in point $ M$. Let a circle which contains points $ B$ and $ C$ touch $ k$ in point $ N$. Circumscribed circle of $ \triangle MNP$ intersects line $ AP$ in point $ L$, different from $ P$. Prove that points $ S,L$ and $ M$ are collinear.

Ukraine Correspondence MO - geometry, 2021.11

Tags: angle bisector , geometry

Let $D$ be a point on the side $AB$ of the triangle $ABC$ such that $BD = CD$, and let the points $E$ on the side $BC$ and $F$ on the extension $AC$ beyond the point $C$ be such that $EF\parallel CD$. The lines $AE$ and $CD$ intersect at the point $G$. Prove that $BC$ is the bisector of the angle $FBG$.

2011 Benelux, 2

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

Let $ABC$ be a triangle with incentre $I$. The angle bisectors $AI$, $BI$ and $CI$ meet $[BC]$, $[CA]$ and $[AB]$ at $D$, $E$ and $F$, respectively. The perpendicular bisector of $[AD]$ intersects the lines $BI$ and $CI$ at $M$ and $N$, respectively. Show that $A$, $I$, $M$ and $N$ lie on a circle.