Olympiad problems

2024 Israel TST, P3

Let $ABCD$ be a parallelogram. Let $\omega_1$ be the circle passing through $D$ tangent to $AB$ at $A$. Let $\omega_2$ be the circle passing through $A$ tangent to $CD$ at $D$. The tangents from $B$ to $\omega_1$ touch it at $A$ and $P$. The tangents from $C$ to $\omega_2$ touch it at $D$ and $Q$. Lines $AP$ and $DQ$ intersect at $X$. The perpendicular bisector of $BC$ intersects $AD$ at $R$. Show that the circumcircles of triangles $\triangle PQX$, $\triangle BCR$ are concentric.

2016 Sharygin Geometry Olympiad, P20

Tags: geometry , incircle , perpendicular bisector

The incircle $\omega$ of a triangle $ABC$ touches $BC, AC$ and $AB$ at points $A_0, B_0$ and $C_0$ respectively. The bisectors of angles $B$ and $C$ meet the perpendicular bisector to segment $AA_0$ at points $Q$ and $P$ respectively. Prove that $PC_0$ and $QB_0$ meet on $\omega$ .

Indonesia MO Shortlist - geometry, g11

Tags: geometry , perpendicular bisector , ratio

Given triangle $ABC$ and point $P$ on the circumcircle of triangle $ABC$. Suppose the line $CP$ intersects line $AB$ at point $E$ and line $BP$ intersect line $AC$ at point $F$. Suppose also the perpendicular bisector of $AB$ intersects $AC$ at point $K$ and the perpendicular bisector of $AC$ intersects $AB$ at point $J$. Prove that $$\left( \frac{CE}{BF}\right)^2= \frac{AJ \cdot JE }{ AK \cdot KF}$$

2021 All-Russian Olympiad, 6

Tags: geometry , 3d geometry , tetrahedron , perpendicular bisector

In tetrahedron $ABCS$ no two edges have equal length. Point $A'$ in plane $BCS$ is symmetric to $S$ with respect to the perpendicular bisector of $BC$. Points $B'$ and $C'$ are defined analagously. Prove that planes $ABC, AB'C', A'BC'$ abd $A'B'C$ share a common point.

2008 Junior Balkan MO, 2

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

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 China Girls Math Olympiad, 2

Tags: geometry , circumcircle , trapezoid , rectangle , projective geometry , perpendicular bisector , geometric transformation

The diagonals $AC,BD$ of the quadrilateral $ABCD$ intersect at $E$. Let $M,N$ be the midpoints of $AB,CD$ respectively. Let the perpendicular bisectors of the segments $AB,CD$ meet at $F$. Suppose that $EF$ meets $BC,AD$ at $P,Q$ respectively. If $MF\cdot CD=NF\cdot AB$ and $DQ\cdot BP=AQ\cdot CP$, prove that $PQ\perp BC$.

2005 Cono Sur Olympiad, 1

Tags: incenter , geometric transformation , reflection , symmetry , perpendicular bisector , geometry

Let $ABC$ be a isosceles triangle, with $AB=AC$. A line $r$ that pass through the incenter $I$ of $ABC$ touches the sides $AB$ and $AC$ at the points $D$ and $E$, respectively. Let $F$ and $G$ be points on $BC$ such that $BF=CE$ and $CG=BD$. Show that the angle $\angle FIG$ is constant when we vary the line $r$.

2013 ELMO Problems, 4

Tags: geometry , incenter , angle bisector , perpendicular bisector

Triangle $ABC$ is inscribed in circle $\omega$. A circle with chord $BC$ intersects segments $AB$ and $AC$ again at $S$ and $R$, respectively. Segments $BR$ and $CS$ meet at $L$, and rays $LR$ and $LS$ intersect $\omega$ at $D$ and $E$, respectively. The internal angle bisector of $\angle BDE$ meets line $ER$ at $K$. Prove that if $BE = BR$, then $\angle ELK = \tfrac{1}{2} \angle BCD$. [i]Proposed by Evan Chen[/i]

JBMO Geometry Collection, 1999

Tags: geometry , circumcircle , trigonometry , perpendicular bisector

Let $ABC$ be a triangle with $AB=AC$. Also, let $D\in[BC]$ be a point such that $BC>BD>DC>0$, and let $\mathcal{C}_1,\mathcal{C}_2$ be the circumcircles of the triangles $ABD$ and $ADC$ respectively. Let $BB'$ and $CC'$ be diameters in the two circles, and let $M$ be the midpoint of $B'C'$. Prove that the area of the triangle $MBC$ is constant (i.e. it does not depend on the choice of the point $D$). [i]Greece[/i]

2014 Puerto Rico Team Selection Test, 1

Tags: geometry , parallelogram , perpendicular bisector

Let $ABCD$ be a parallelogram with $AB>BC$ and $\angle DAB$ less than $\angle ABC$. The perpendicular bisectors of sides $AB$ and $BC$ intersect at the point $M$ lying on the extension of $AD$. If $\angle MCD=15^{\circ}$, find the measure of $\angle ABC$

2003 Baltic Way, 12

Tags: circumcircle , ratio , perpendicular bisector , geometry

Points $M$ and $N$ are taken on the sides $BC$ and $CD$ respectively of a square $ABCD$ so that $\angle MAN=45^{\circ}$. Prove that the circumcentre of $\triangle AMN$ lies on $AC$.

2010 Contests, 2

Tags: circumcircle , rotation , analytic geometry , perpendicular bisector , lalala , geometry

$AB$ is a diameter of a circle with center $O$. Let $C$ and $D$ be two different points on the circle on the same side of $AB$, and the lines tangent to the circle at points $C$ and $D$ meet at $E$. Segments $AD$ and $BC$ meet at $F$. Lines $EF$ and $AB$ meet at $M$. Prove that $E,C,M$ and $D$ are concyclic.

1972 IMO Longlists, 43

Tags: parallelogram , rectangle , trapezoid , pythagorean theorem , perpendicular bisector , geometry

A fixed point $A$ inside a circle is given. Consider all chords $XY$ of the circle such that $\angle XAY$ is a right angle, and for all such chords construct the point $M$ symmetric to $A$ with respect to $XY$ . Find the locus of points $M$.

2014 Sharygin Geometry Olympiad, 6

Tags: perpendicular bisector , geometry

Given a circle with center $O$ and a point $P$ not lying on it, let $X$ be an arbitrary point on this circle and $Y$ be a common point of the bisector of angle $POX$ and the perpendicular bisector to segment $PX$. Find the locus of points $Y$.

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.

2017 Puerto Rico Team Selection Test, 2

Tags: altitude , perpendicular bisector , geometry

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 $.

2011 China Team Selection Test, 2

Tags: inequalities , geometry , perpendicular bisector , combinatorics

Let $S$ be a set of $n$ points in the plane such that no four points are collinear. Let $\{d_1,d_2,\cdots ,d_k\}$ be the set of distances between pairs of distinct points in $S$, and let $m_i$ be the multiplicity of $d_i$, i.e. the number of unordered pairs $\{P,Q\}\subseteq S$ with $|PQ|=d_i$. Prove that $\sum_{i=1}^k m_i^2\leq n^3-n^2$.

2009 Junior Balkan Team Selection Test, 3

Tags: parallelogram , perpendicular bisector , geometry

Let $ ABCD$ be a convex quadrilateral, such that $ \angle CBD\equal{}2\cdot\angle ADB, \angle ABD\equal{}2\cdot\angle CDB$ and $ AB\equal{}CB$. Prove that quadrilateral $ ABCD$ is a kite.

2010 Contests, 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)

2007 China Team Selection Test, 1

Tags: circumcircle , parallelogram , perpendicular bisector , geometry

Let convex quadrilateral $ ABCD$ be inscribed in a circle centers at $ O.$ The opposite sides $ BA,CD$ meet at $ H$, the diagonals $ AC,BD$ meet at $ G.$ Let $ O_{1},O_{2}$ be the circumcenters of triangles $ AGD,BGC.$ $ O_{1}O_{2}$ intersects $ OG$ at $ N.$ The line $ HG$ cuts the circumcircles of triangles $ AGD,BGC$ at $ P,Q$, respectively. Denote by $ M$ the midpoint of $ PQ.$ Prove that $ NO \equal{} NM.$

2014 Contests, 1

Tags: analytic geometry , trigonometry , trapezoid , perpendicular bisector , cyclic quadrilateral , trig identities , geometry

In a triangle $ABC$, let $D$ be the point on the segment $BC$ such that $AB+BD=AC+CD$. Suppose that the points $B$, $C$ and the centroids of triangles $ABD$ and $ACD$ lie on a circle. Prove that $AB=AC$.

2003 AIME Problems, 7

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

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 $s$.

2020 Australian Maths Olympiad, 3

Tags: geometry , right triangle , perpendicular bisector , tangent , midpoint , circumcircle , parallel

Let $ABC$ be a triangle with $\angle ACB=90^{\circ}$. Suppose that the tangent line at $C$ to the circle passing through $A,B,C$ intersects the line $AB$ at $D$. Let $E$ be the midpoint of $CD$ and let $F$ be a point on $EB$ such that $AF$ is parallel to $CD$. Prove that the lines $AB$ and $CF$ are perpendicular.

1967 Swedish Mathematical Competition, 2

Tags: geometry , angle bisector , construction , geometric construction , ruler , perpendicular bisector

You are given a ruler with two parallel straight edges a distance $d$ apart. It may be used (1) to draw the line through two points, (2) given two points a distance $\ge d$ apart, to draw two parallel lines, one through each point, (3) to draw a line parallel to a given line, a distance d away. One can also (4) choose an arbitrary point in the plane, and (5) choose an arbitrary point on a line. Show how to construct : (A) the bisector of a given angle, and (B) the perpendicular to the midpoint of a given line segment.

2002 ITAMO, 3

Tags: circumcircle , trigonometry , perpendicular bisector , geometry

Let $A$ and $B$ are two points on a plane, and let $M$ be the midpoint of $AB$. Let $r$ be a line and let $R$ and $S$ be the projections of $A$ and $B$ onto $r$. Assuming that $A$, $M$, and $R$ are not collinear, prove that the circumcircle of triangle $AMR$ has the same radius as the circumcircle of $BSM$.