Olympiad problems

2008 Junior Balkan MO, 2

The vertices $ A$ and $ B$ of an equilateral triangle $ ABC$ lie on a circle $k$ of radius $1$, and the vertex $ C$ is in the interior of the circle $ k$. A point $ D$, different from $ B$, lies on $ k$ so that $ AD\equal{}AB$. The line $ DC$ intersects $ k$ for the second time at point $ E$. Find the length of the line segment $ CE$.

2011 USA TSTST, 2

Tags: perpendicular bisector

Two circles $\omega_1$ and $\omega_2$ intersect at points $A$ and $B$. Line $\ell$ is tangent to $\omega_1$ at $P$ and to $\omega_2$ at $Q$ so that $A$ is closer to $\ell$ than $B$. Let $X$ and $Y$ be points on major arcs $\overarc{PA}$ (on $\omega_1$) and $AQ$ (on $\omega_2$), respectively, such that $AX/PX = AY/QY = c$. Extend segments $PA$ and $QA$ through $A$ to $R$ and $S$, respectively, such that $AR = AS = c\cdot PQ$. Given that the circumcenter of triangle $ARS$ lies on line $XY$, prove that $\angle XPA = \angle AQY$.

2014 Lithuania Team Selection Test, 1

Tags: inradius , trapezoid , perpendicular bisector , geometry

Circle touches parallelogram‘s $ABCD$ borders $AB, BC$ and $CD$ respectively at points $K, L$ and $M$. Perpendicular is drawn from vertex $C$ to $AB$ . Prove, that the line $KL$ divides this perpendicular into two equal parts (with the same length).

JBMO Geometry Collection, 2001

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

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]

1985 Traian Lălescu, 2.1

Tags: geometry , bisector , perpendicular bisector , pure geometry

Let $ ABC $ be a triangle. The perpendicular in $ B $ of the bisector of the angle $ \angle ABC $ intersects the bisector of the angle $ \angle BAC $ in $ M. $ Show that $ MC $ is perpendicular to the bisector of $ \angle BCA. $

2017 Puerto Rico Team Selection Test, 2

Tags: perpendicular bisector , geometry , altitude

For an acute triangle $ ABC $ let $ H $ be the point of intersection of the altitudes $ AA_1 $, $ BB_1 $, $ CC_1 $. Let $ M $ and $ N $ be the midpoints of the $ BC $ and $ AH $ segments, respectively. Show that $ MN $ is the perpendicular bisector of segment $ B_1C_1 $.

2005 Junior Balkan Team Selection Tests - Moldova, 1

Tags: symmetry , perpendicular bisector , power of a point , radical axis , geometry

Let the triangle $ABC$ with $BC$ the smallest side. Let $P$ on ($AB$) such that angle $PCB$ equals angle $BAC$. and $Q$ on side ($AC$) such that angle $QBC$ equals angle $BAC$. Show that the line passing through the circumenters of triangles $ABC$ and $APQ$ is perpendicular on $BC$.

2020 OMMock - Mexico National Olympiad Mock Exam, 4

Tags: geometry , diameter , perpendicular bisector , vector

Let $ABC$ be a triangle. Suppose that the perpendicular bisector of $BC$ meets the circle of diameter $AB$ at a point $D$ at the opposite side of $BC$ with respect to $A$, and meets the circle through $A, C, D$ again at $E$. Prove that $\angle ACE=\angle BCD$. [i]Proposed by José Manuel Guerra and Victor Domínguez[/i]

2022 Germany Team Selection Test, 1

Tags: perpendicular bisector , triangle geometry , geometry

Given a triangle $ABC$ and three circles $x$, $y$ and $z$ such that $A \in y \cap z$, $B \in z \cap x$ and $C \in x \cap y$. The circle $x$ intersects the line $AC$ at the points $X_b$ and $C$, and intersects the line $AB$ at the points $X_c$ and $B$. The circle $y$ intersects the line $BA$ at the points $Y_c$ and $A$, and intersects the line $BC$ at the points $Y_a$ and $C$. The circle $z$ intersects the line $CB$ at the points $Z_a$ and $B$, and intersects the line $CA$ at the points $Z_b$ and $A$. (Yes, these definitions have the symmetries you would expect.) Prove that the perpendicular bisectors of the segments $Y_a Z_a$, $Z_b X_b$ and $X_c Y_c$ concur.

2010 Stars Of Mathematics, 2

Tags: geometric transformation , reflection , perpendicular bisector , geometry

Let $ABCD$ be a square and let the points $M$ on $BC$, $N$ on $CD$, $P$ on $DA$, be such that $\angle (AB,AM)=x,\angle (BC,MN)=2x,\angle (CD,NP)=3x$. 1) Show that for any $0\le x\le 22.5$, such a configuration uniquely exists, and that $P$ ranges over the whole segment $DA$; 2) Determine the number of angles $0\le x\le 22.5$ for which$\angle (DA,PB)=4x$. (Dan Schwarz)

1978 IMO Longlists, 51

Tags: perpendicular bisector , geometry

Find the relations among the angles of the triangle $ABC$ whose altitude $AH$ and median $AM$ satisfy $\angle BAH =\angle CAM$.

2004 Kazakhstan National Olympiad, 8

Tags: geometry , circumcircle , perpendicular bisector , convex quadrilateral

Let $ ABCD$ be a convex quadrilateral. The perpendicular bisectors of its sides $ AB$ and $ CD$ meet at $ Y$. Denote by $ X$ a point inside the quadrilateral $ ABCD$ such that $ \measuredangle ADX \equal{} \measuredangle BCX < 90^{\circ}$ and $ \measuredangle DAX \equal{} \measuredangle CBX < 90^{\circ}$. Show that $ \measuredangle AYB \equal{} 2\cdot\measuredangle ADX$.

2018 Bosnia And Herzegovina - Regional Olympiad, 4

Tags: geometry , circumcircle , perpendicular bisector

Let $P$ be a point on circumcircle of triangle $ABC$ on arc $\stackrel{\frown}{BC}$ which does not contain point $A$. Let lines $AB$ and $CP$ intersect at point $E$, and lines $AC$ and $BP$ intersect at $F$. If perpendicular bisector of side $AB$ intersects $AC$ in point $K$, and perpendicular bisector of side $AC$ intersects side $AB$ in point $J$, prove that: ${\left(\frac{CE}{BF}\right)}^2=\frac{AJ\cdot JE}{AK \cdot KF}$

2003 AIME Problems, 7

Tags: perimeter , inequalities , trigonometry , perpendicular bisector , pythagorean theorem , geometry

Point $B$ is on $\overline{AC}$ with $AB = 9$ and $BC = 21$. Point $D$ is not on $\overline{AC}$ so that $AD = CD$, and $AD$ and $BD$ are integers. Let $s$ be the sum of all possible perimeters of $\triangle ACD$. Find $s$.

2005 All-Russian Olympiad, 1

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

Given a parallelogram $ABCD$ with $AB<BC$, show that the circumcircles of the triangles $APQ$ share a second common point (apart from $A$) as $P,Q$ move on the sides $BC,CD$ respectively s.t. $CP=CQ$.

1998 IberoAmerican Olympiad For University Students, 4

Tags: rhombus , perpendicular bisector , geometry

Four circles of radius $1$ with centers $A,B,C,D$ are in the plane in such a way that each circle is tangent to two others. A fifth circle passes through the center of two of the circles and is tangent to the other two. Find the possible values of the area of the quadrilateral $ABCD$.

2015 China Team Selection Test, 1

Tags: perpendicular bisector , angle bisector , geometry

$\triangle{ABC}$ is isosceles with $AB = AC >BC$. Let $D$ be a point in its interior such that $DA = DB+DC$. Suppose that the perpendicular bisector of $AB$ meets the external angle bisector of $\angle{ADB}$ at $P$, and let $Q$ be the intersection of the perpendicular bisector of $AC$ and the external angle bisector of $\angle{ADC}$. Prove that $B,C,P,Q$ are concyclic.

2016 Czech-Polish-Slovak Junior Match, 4

Tags: geometry , equal segments , perpendicular bisector

We are given an acute-angled triangle $ABC$ with $AB < AC < BC$. Points $K$ and $L$ are chosen on segments $AC$ and $BC$, respectively, so that $AB = CK = CL$. Perpendicular bisectors of segments $AK$ and $BL$ intersect the line $AB$ at points $P$ and $Q$, respectively. Segments $KP$ and $LQ$ intersect at point $M$. Prove that $AK + KM = BL + LM$. Poland

2006 IMO Shortlist, 5

Tags: geometry , circumcircle , perpendicular bisector , angle chasing

In triangle $ABC$, let $J$ be the center of the excircle tangent to side $BC$ at $A_{1}$ and to the extensions of the sides $AC$ and $AB$ at $B_{1}$ and $C_{1}$ respectively. Suppose that the lines $A_{1}B_{1}$ and $AB$ are perpendicular and intersect at $D$. Let $E$ be the foot of the perpendicular from $C_{1}$ to line $DJ$. Determine the angles $\angle{BEA_{1}}$ and $\angle{AEB_{1}}$. [i]Proposed by Dimitris Kontogiannis, Greece[/i]

2000 Turkey MO (2nd round), 1

Tags: trigonometry , perpendicular bisector , geometry

A circle with center $O$ and a point $A$ in this circle are given. Let $P_{B}$ is the intersection point of $[AB]$ and the internal bisector of $\angle AOB$ where $B$ is a point on the circle such that $B$ doesn't lie on the line $OA$, Find the locus of $P_{B}$ as $B$ varies.

Brazil L2 Finals (OBM) - geometry, 2009.2

Tags: circumcircle , perpendicular bisector , geometry

Let $ A$ be one of the two points of intersection of two circles with centers $ X, Y$ respectively.The tangents at $ A$ to the two circles meet the circles again at $ B, C$. Let a point $ P$ be located so that $ PXAY$ is a parallelogram. Show that $ P$ is also the circumcenter of triangle $ ABC$.

2023 Stanford Mathematics Tournament, 8

Tags: geometry , perpendicular bisector , circumcircle

In acute triangle $\triangle ABC$, point $R$ lies on the perpendicular bisector of $AC$ such that $\overline{CA}$ bisects $\angle BAR$. Let $Q$ be the intersection of lines $AC$ and $BR$. The circumcircle of $\triangle ARC$ intersects segment $\overline{AB}$ at $P\neq A$, with $AP=1$, $PB=5$, and $AQ=2$. Compute $AR$.

2006 Tournament of Towns, 1

Tags: geometry , perpendicular bisector

Let $\angle A$ in a triangle $ABC$ be $60^\circ$. Let point $N$ be the intersection of $AC$ and perpendicular bisector to the side $AB$ while point $M$ be the intersection of $AB$ and perpendicular bisector to the side $AC$. Prove that $CB = MN$. [i](3 points)[/i]

2008 Hungary-Israel Binational, 1

Tags: inequalities , triangle inequality , perpendicular bisector , geometry

Find the largest value of n, such that there exists a polygon with n sides, 2 adjacent sides of length 1, and all his diagonals have an integer length.

2004 Turkey Team Selection Test, 2

Tags: circumcircle , incenter , geometric transformation , reflection , projective geometry , perpendicular bisector , geometry

Let $\triangle ABC$ be an acute triangle, $O$ be its circumcenter, and $D$ be a point different that $A$ and $C$ on the smaller $AC$ arc of its circumcircle. Let $P$ be a point on $[AB]$ satisfying $\widehat{ADP} = \widehat {OBC}$ and $Q$ be a point on $[BC]$ satisfying $\widehat{CDQ}=\widehat {OBA}$. Show that $\widehat {DPQ} = \widehat {DOC}$.