Olympiad problems

2010 India IMO Training Camp, 7

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

Let $ABCD$ be a cyclic quadrilaterla and let $E$ be the point of intersection of its diagonals $AC$ and $BD$. Suppose $AD$ and $BC$ meet in $F$. Let the midpoints of $AB$ and $CD$ be $G$ and $H$ respectively. If $\Gamma $ is the circumcircle of triangle $EGH$, prove that $FE$ is tangent to $\Gamma $.

2006 China Team Selection Test, 2

Tags: geometry , circumcircle , geometric transformation , reflection , parallelogram , ratio , analytic geometry

Let $\omega$ be the circumcircle of $\triangle{ABC}$. $P$ is an interior point of $\triangle{ABC}$. $A_{1}, B_{1}, C_{1}$ are the intersections of $AP, BP, CP$ respectively and $A_{2}, B_{2}, C_{2}$ are the symmetrical points of $A_{1}, B_{1}, C_{1}$ with respect to the midpoints of side $BC, CA, AB$. Show that the circumcircle of $\triangle{A_{2}B_{2}C_{2}}$ passes through the orthocentre of $\triangle{ABC}$.

2020 European Mathematical Cup, 1

Tags: geometry , parallelogram , concurrency

Let $ABCD$ be a parallelogram such that $|AB| > |BC|$. Let $O$ be a point on the line $CD$ such that $|OB| = |OD|$. Let $\omega$ be a circle with center $O$ and radius $|OC|$. If $T$ is the second intersection of $\omega$ and $CD$, prove that $AT, BO$ and $\omega$ are concurrent. [i]Proposed by Ivan Novak[/i]

1989 IMO Longlists, 6

Tags: parallelogram , geometry

The circles $ c_1$ and $ c_2$ are tangent at the point $ A.$ A straight line $ l$ through $ A$ intersects $ c_1$ and $ c_2$ at points $ C_1$ and $ C_2$ respectively. A circle $ c,$ which contains $ C_1$ and $ C_2,$ meets $ c_1$ and $ c_2$ at points $ B_1$ and $ B_2$ respectively. Let $ \omega$ be the circle circumscribed around triangle $ AB_1B_2.$ The circle $ k$ tangent to $ \omega$ at the point $ A$ meets $ c_1$ and $ c_2$ at the points $ D_1$ and $ D_2$ respectively. Prove that [b](a)[/b] the points $ C_1,C_2,D_1,D_2$ are concyclic or collinear, [b](b)[/b] the points $ B_1,B_2,D_1,D_2$ are concyclic if and only if $ AC_1$ and $ AC_2$ are diameters of $ c_1$ and $ c_2.$

2009 AIME Problems, 4

Tags: geometry , parallelogram , ratio , rectangle , analytic geometry

In parallelogram $ ABCD$, point $ M$ is on $ \overline{AB}$ so that $ \frac{AM}{AB} \equal{} \frac{17}{1000}$ and point $ N$ is on $ \overline{AD}$ so that $ \frac{AN}{AD} \equal{} \frac{17}{2009}$. Let $ P$ be the point of intersection of $ \overline{AC}$ and $ \overline{MN}$. Find $ \frac{AC}{AP}$.

2019 BAMO, B

Tags: geometry , parallelogram , area

In the figure below, parallelograms $ABCD$ and $BFEC$ have areas $1234$ cm$^2$ and $2804$ cm$^2$, respectively. Points $M$ and $N$ are chosen on sides $AD$ and $FE$, respectively, so that segment $MN$ passes through $B$. Find the area of $\vartriangle MNC$. [img]https://cdn.artofproblemsolving.com/attachments/b/6/8b57b632191bdb3a27ab7c59e2376dab23950b.png[/img]

Kyiv City MO 1984-93 - geometry, 1985.9.5

Tags: geometry , parallelogram , equilateral

Outside the parallelogram $ABCD$ on its sides $AB$ and $BC$ are constructed equilateral triangles $ABK$, and $BCM$. Prove that the triangle $KMD$ is equilateral.

2010 Contests, 2

Tags: geometry , circumcircle , geometric transformation , reflection , parallelogram , calculus , trigonometry

Let $ABC$ be an acute triangle, $H$ its orthocentre, $D$ a point on the side $[BC]$, and $P$ a point such that $ADPH$ is a parallelogram. Show that $\angle BPC > \angle BAC$.

2001 Baltic Way, 9

Tags: geometry , rhombus , parallelogram , circumcircle , hyperbola , cyclic quadrilateral , conic

Given a rhombus $ABCD$, find the locus of the points $P$ lying inside the rhombus and satisfying $\angle APD+\angle BPC=180^{\circ}$.

2002 Romania National Olympiad, 1

Tags: parallelogram , geometric transformation , rotation , geometry

Let $X,Y,Z,T$ be four points in the plane. The segments $[XY]$ and $[ZT]$ are said to be [i]connected[/i], if there is some point $O$ in the plane such that the triangles $OXY$ and $OZT$ are right-angled at $O$ and isosceles. Let $ABCDEF$ be a convex hexagon such that the pairs of segments $[AB],[CE],$ and $[BD],[EF]$ are [i]connected[/i]. Show that the points $A,C,D$ and $F$ are the vertices of a parallelogram and $[BC]$ and $[AE]$ are [i]connected[/i].

2009 Oral Moscow Geometry Olympiad, 1

Tags: geometry , parallelogram , rhombus , construct

The figure shows a parallelogram and the point $P$ of intersection of its diagonals is marked. Draw a straight line through $P$ so that it breaks the parallelogram into two parts, from which you can fold a rhombus. [img]https://1.bp.blogspot.com/-Df2tIBthcmI/X2ZwIx3R4vI/AAAAAAAAMhQ/8Zkxfq30H8MSCdc66tm33n6jt-QKfGMowCLcBGAsYHQ/s0/2009%2Boral%2Bmoscow%2Bj1.png[/img]

2007 Romania Team Selection Test, 1

Tags: geometry , parallelogram , geometric transformation , rotation , reflection , analytic geometry , perpendicular bisector

Let $ ABCD$ be a parallelogram with no angle equal to $ 60^{\textrm{o}}$. Find all pairs of points $ E, F$, in the plane of $ ABCD$, such that triangles $ AEB$ and $ BFC$ are isosceles, of basis $ AB$, respectively $ BC$, and triangle $ DEF$ is equilateral. [i]Valentin Vornicu[/i]

2013 China Team Selection Test, 3

Tags: geometry , parallelogram , combinatorial geometry , combinatorics

Let $A$ be a set consisting of 6 points in the plane. denoted $n(A)$ as the number of the unit circles which meet at least three points of $A$. Find the maximum of $n(A)$

2013 Czech And Slovak Olympiad IIIA, 5

Tags: parallelogram , parallel , angle , geometry

Given the parallelogram $ABCD$ such that the feet $K, L$ of the perpendiculars from point $D$ on the sides $AB, BC$ respectively are internal points. Prove that $KL \parallel AC$ when $|\angle BCA| + |\angle ABD| = |\angle BDA| + |\angle ACD|$.

2007 Macedonia National Olympiad, 2

Tags: parallelogram , trapezoid , circumcircle , geometry

In a trapezoid $ABCD$ with a base $AD$, point $L$ is the orthogonal projection of $C$ on $AB$, and $K$ is the point on $BC$ such that $AK$ is perpendicular to $AD$. Let $O$ be the circumcenter of triangle $ACD$. Suppose that the lines $AK , CL$ and $DO$ have a common point. Prove that $ABCD$ is a parallelogram.

2006 Bulgaria Team Selection Test, 1

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

[b]Problem 1.[/b] Points $D$ and $E$ are chosen on the sides $AB$ and $AC$, respectively, of a triangle $\triangle ABC$ such that $DE\parallel BC$. The circumcircle $k$ of triangle $\triangle ADE$ intersects the lines $BE$ and $CD$ at the points $M$ and $N$ (different from $E$ and $D$). The lines $AM$ and $AN$ intersect the side $BC$ at points $P$ and $Q$ such that $BC=2\cdot PQ$ and the point $P$ lies between $B$ and $Q$. Prove that the circle $k$ passes through the point of intersection of the side $BC$ and the angle bisector of $\angle BAC$. [i]Nikolai Nikolov[/i]

2020 Polish Junior MO Second Round, 2.

Tags: geometry , parallelogram , perpendicular bisector

Let $ABCD$ be the parallelogram, such that angle at vertex $A$ is acute. Perpendicular bisector of the segment $AB$ intersects the segment $CD$ in the point $X$. Let $E$ be the intersection point of the diagonals of the parallelogram $ABCD$. Prove that $XE = \frac{1}{2}AD$.

2018 PUMaC Geometry B, 8

Tags: geometry , parallelogram , angle bisector

Let $ABCD$ be a parallelogram such that $AB = 35$ and $BC = 28$. Suppose that $BD \perp BC$. Let $\ell_1$ be the reflection of $AC$ across the angle bisector of $\angle BAD$, and let $\ell_2$ be the line through $B$ perpendicular to $CD$. $\ell_1$ and $\ell_2$ intersect at a point $P$. If $PD$ can be expressed in simplest form as $\frac{m}{n}$, find $m + n$.

2006 Mediterranean Mathematics Olympiad, 2

Tags: inequalities , parallelogram , geometry

Let $P$ be a point inside a triangle $ABC$, and $A_1B_2,B_1C_2,C_1A_2$ be segments passing through $P$ and parallel to $AB, BC, CA$ respectively, where points $A_1, A_2$ lie on $BC, B_1, B_2$ on $CA$, and $C_1,C_2$ on $AB$. Prove that \[ \text{Area}(A_1A_2B_1B_2C_1C_2) \ge \frac{1}{2}\text{Area}(ABC)\]

2008 Oral Moscow Geometry Olympiad, 3

Tags: midpoint , geometry , parallelogram , equal segments

Given a quadrilateral $ABCD$. $A ', B', C'$ and $D'$ are the midpoints of the sides $BC, CB, BA$ and $AB$, respectively. It is known that $AA'= CC'$, $BB'= DD'$. Is it true that $ABCD$ is a parallelogram? (M. Volchkevich)

2020-IMOC, G5

Tags: geometry , parallelogram , concyclic , concurrency

Let $O, H$ be the circumcentor and the orthocenter of a scalene triangle $ABC$. Let $P$ be the reflection of $A$ w.r.t. $OH$, and $Q$ is a point on $\odot (ABC)$ such that $AQ, OH, BC$ are concurrent. Let $A'$ be a points such that $ABA'C$ is a parallelogram. Show that $A', H, P, Q$ are concylic. (ltf0501).

2005 All-Russian Olympiad, 3

Tags: incenter , parallelogram , trigonometry , complex number , geometry

A quadrilateral $ABCD$ without parallel sides is circumscribed around a circle with centre $O$. Prove that $O$ is a point of intersection of middle lines of quadrilateral $ABCD$ (i.e. barycentre of points $A,\,B,\,C,\,D$) iff $OA\cdot OC=OB\cdot OD$.

2001 Federal Math Competition of S&M, Problem 4

Tags: geometry , 3d geometry , pyramid , parallelogram

Parallelogram $ABCD$ is the base of a pyramid $SABCD$. Planes determined by triangles $ASC$ and $BSD$ are mutually perpendicular. Find the area of the side $ASD$, if areas of sides $ASB,BSC$ and $CSD$ are equal to $x,y$ and $z$, respectively.

2013 Costa Rica - Final Round, G2

Tags: parallelogram , equal angles , geometry

Consider the triangle $ABC$. Let $P, Q$ inside the angle $A$ such that $\angle BAP=\angle CAQ$ and $PBQC$ is a parallelogram. Show that $\angle ABP=\angle ACP.$

2009 IMO Shortlist, 5

Tags: geometry , parallelogram , geometric inequality , area , polygon

Let $P$ be a polygon that is convex and symmetric to some point $O$. Prove that for some parallelogram $R$ satisfying $P\subset R$ we have \[\frac{|R|}{|P|}\leq \sqrt 2\] where $|R|$ and $|P|$ denote the area of the sets $R$ and $P$, respectively. [i]Proposed by Witold Szczechla, Poland[/i]