Olympiad problems

2018 Romania National Olympiad, 3

Tags: geometry , equilateral , rhombus , collinear , parallelogram

On the sides $[AB]$ and $[BC]$ of the parallelogram $ABCD$ are constructed the equilateral triangles $ABE$ and $BCF,$ so that the points $D$ and $E$ are on the same side of the line $AB$, and $F$ and $D$ on different sides of the line $BC$. If the points $E,D$ and $F$ are collinear, then prove that $ABCD$ is rhombus.

2010 Iran MO (3rd Round), 5

Tags: parallelogram , reflection , geometry , incenter , geometric transformation

In a triangle $ABC$, $I$ is the incenter. $D$ is the reflection of $A$ to $I$. the incircle is tangent to $BC$ at point $E$. $DE$ cuts $IG$ at $P$ ($G$ is centroid). $M$ is the midpoint of $BC$. prove that a) $AP||DM$.(15 points) b) $AP=2DM$. (10 points)

2023-IMOC, G1

Tags: parallelogram , geometry

Triangle $ABC$ has circumcenter $O$. $M$ is the midpoint of arc $BC$ not containing $A$. $S$ is a point on $(ABC)$ such that $AS$ and $BC$ intersect on the line passing through $O$ and perpendicular to $AM$. $D$ is a point such that $ABDC$ is a parallelogram. Prove that $D$ lies on the line $SM$.

1998 Belarus Team Selection Test, 3

Tags: geometry , area , equilateral , parallelogram , hexagon

Let $ABCDEF$ be a convex hexagon such that $BCEF$ is a parallelogram and $ABF$ an equilateral triangle. Given that $BC = 1, AD = 3, CD+DE = 2$, compute the area of $ABCDEF$

2011 All-Russian Olympiad, 2

Tags: reflection , symmetry , geometry , circumcircle , geometric transformation , parallelogram

On side $BC$ of parallelogram $ABCD$ ($A$ is acute) lies point $T$ so that triangle $ATD$ is an acute triangle. Let $O_1$, $O_2$, and $O_3$ be the circumcenters of triangles $ABT$, $DAT$, and $CDT$ respectively. Prove that the orthocenter of triangle $O_1O_2O_3$ lies on line $AD$.

2014 China Northern MO, 5

Tags: equal angles , parallelogram , geometry

As shown in the figure, in the parallelogram $ABCD$, $I$ is the incenter of $\vartriangle BCD$, and $H$ is the orthocenter of $\vartriangle IBD$. Prove that $\angle HAB=\angle HAD$. [img]https://cdn.artofproblemsolving.com/attachments/4/3/5fa16c208ef3940443854756ae7bdb9c4272ed.png[/img]

2012 France Team Selection Test, 2

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

Let $ABC$ be an acute-angled triangle with $AB\not= AC$. Let $\Gamma$ be the circumcircle, $H$ the orthocentre and $O$ the centre of $\Gamma$. $M$ is the midpoint of $BC$. The line $AM$ meets $\Gamma$ again at $N$ and the circle with diameter $AM$ crosses $\Gamma$ again at $P$. Prove that the lines $AP,BC,OH$ are concurrent if and only if $AH=HN$.

2014 Bundeswettbewerb Mathematik, 3

Tags: geometry , parallelogram

A regular hexagon with side length $1$ is given. Using a ruler construct points in such a way that among the given and constructed points there are two such points that the distance between them is $\sqrt7$. Notes: ''Using a ruler construct points $\ldots$'' means: Newly constructed points arise only as the intersection of straight lines connecting two points that are given or already constructed. In particular, no length can be measured by the ruler.

1967 Kurschak Competition, 3

Tags: parallelogram , geometry

For a vertex $X$ of a quadrilateral, let $h(X)$ be the sum of the distances from $X$ to the two sides not containing $X$. Show that if a convex quadrilateral $ABCD$ satisfies $h(A) = h(B) = h(C) = h(D)$, then it must be a parallelogram.

1963 AMC 12/AHSME, 12

Tags: analytic geometry , parallelogram , geometry

Three vertices of parallelogram $PQRS$ are $P(-3,-2)$, $Q(1,-5)$, $R(9,1)$ with $P$ and $R$ diagonally opposite. The sum of the coordinates of vertex $S$ is: $\textbf{(A)}\ 13 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 11 \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 9$

2014 Puerto Rico Team Selection Test, 1

Tags: parallelogram , geometry , 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 $\angle ABC$

2010 Germany Team Selection Test, 2

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

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]

2013 Czech And Slovak Olympiad IIIA, 5

Tags: parallelogram , angle , geometry , parallel

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 $|\angle BCA| + |\angle ABD| = |\angle BDA| + |\angle ACD|$.

2018 Switzerland - Final Round, 9

Tags: geometry , parallelogram , inequalities

Let $n$ be a positive integer and let $G$ be the set of points $(x, y)$ in the plane such that $x$ and $y$ are integers with $1 \leq x, y \leq n$. A subset of $G$ is called [i]parallelogram-free[/i] if it does not contains four non-collinear points, which are the vertices of a parallelogram. What is the largest number of elements a parallelogram-free subset of $G$ can have?

2010 Sharygin Geometry Olympiad, 4

Tags: concentric , concentric circles , projection , inscribed , geometry , parallelogram

Projections of two points to the sidelines of a quadrilateral lie on two concentric circles (projections of each point form a cyclic quadrilateral and the radii of circles are different). Prove that this quadrilateral is a parallelogram.

2005 Postal Coaching, 16

Tags: circumcircle , parallelogram , trapezoid , cyclic quadrilateral , geometry

The diagonals AC and BD of a cyclic ABCD intersect at E. Let O be circumcentre of ABCD. If midpoints of AB, CD, OE are collinear prove that AD=BC. Bomb [color=red][Moderator edit: The problem is wrong. See also http://www.mathlinks.ro/Forum/viewtopic.php?t=53090 .][/color]

1966 IMO Longlists, 22

Tags: geometry , parallelogram , vector

Let $P$ and $P^{\prime }$ be two parallelograms with equal area, and let their sidelengths be $a,$ $b$ and $a^{\prime },$ $b^{\prime }.$ Assume that $a^{\prime }\leq a\leq b\leq b^{\prime },$ and moreover, it is possible to place the segment $b^{\prime }$ such that it completely lies in the interior of the parallelogram $P.$ Show that the parallelogram $P$ can be partitioned into four polygons such that these four polygons can be composed again to form the parallelogram $% P^{\prime }.$

2007 IMAC Arhimede, 2

Tags: function , geometry , parallelogram

Let $ABCD$ be a parallelogram that is not rhombus. We draw the symmetrical half-line of $(DC$ with respect to line $BD$. Similarly we draw the symmetrical half- line of $(AB$ with respect to $AC$. These half- lines intersect each other in $P$. If $\frac{AP}{DP}= q$ find the value of $\frac{AC}{BD}$ in function of $q$.

1997 Moldova Team Selection Test, 2

Tags: parallelogram , pentagon , ratio , geometry

In a convex pentagon every diagonal is parallel to one side. Show that the ratios between the lengths of diagonals and the sides parallel to them are equal and find their value.

2013 APMO, 1

Tags: circumcircle , ratio , equal area , trigonometry , analytic geometry , parallelogram , geometry

Let $ABC$ be an acute triangle with altitudes $AD$, $BE$, and $CF$, and let $O$ be the center of its circumcircle. Show that the segments $OA$, $OF$, $OB$, $OD$, $OC$, $OE$ dissect the triangle $ABC$ into three pairs of triangles that have equal areas.

1992 Tournament Of Towns, (340) 2

Tags: geometry , circumcircle , parallelogram

On each side of a parallelogram an arbitrary point is chosen. Each pair of chosen points on neighbouring sides (i.e. sides with a common vertex) are connected by a line segment. Prove that the centres of the circumscribed circles of the four triangles so created are themselves vertices of a parallelogram. (ED Kulanin)

2008 Vietnam National Olympiad, 7

Tags: circumcircle , graphing lines , slope , analytic geometry , parallelogram , geometric transformation , geometry

Let $ AD$ is centroid of $ ABC$ triangle. Let $ (d)$ is the perpendicular line with $ AD$. Let $ M$ is a point on $ (d)$. Let $ E, F$ are midpoints of $ MB, MC$ respectively. The line through point $ E$ and perpendicular with $ (d)$ meet $ AB$ at $ P$. The line through point $ F$ and perpendicular with $ (d)$ meet $ AC$ at $ Q$. Let $ (d')$ is a line through point $ M$ and perpendicular with $ PQ$. Prove $ (d')$ always pass a fixed point.

2006 Germany Team Selection Test, 3

Tags: parallelogram , spiral similarity , exterior angle , geometry , homothety , incenter

Let $ABCD$ be a parallelogram. A variable line $g$ through the vertex $A$ intersects the rays $BC$ and $DC$ at the points $X$ and $Y$, respectively. Let $K$ and $L$ be the $A$-excenters of the triangles $ABX$ and $ADY$. Show that the angle $\measuredangle KCL$ is independent of the line $g$. [i]Proposed by Vyacheslev Yasinskiy, Ukraine[/i]

2000 CentroAmerican, 3

Tags: graphing lines , complex number , parallelogram , analytic geometry , trigonometry , geometry , slope

Let $ ABCDE$ be a convex pentagon. If $ P$, $ Q$, $ R$ and $ S$ are the respective centroids of the triangles $ ABE$, $ BCE$, $ CDE$ and $ DAE$, show that $ PQRS$ is a parallelogram and its area is $ 2/9$ of that of $ ABCD$.

Kvant 2024, M2793

Tags: parallelogram , tangent circle , circumcircle , geometry

In acute triangle $ABC$ ($AB<AC$) point $O$ is center of its circumcircle $\Omega$. Let the tangent to $\Omega$ drawn at point $A$ intersect the line $BC$ at point $D$. Let the line $DO$ intersects the segments $AB$ and $AC$ at points $E$ and $F$, respectively. Point $G$ is constructed such that $AEGF$ is a parallelogram. Let $K$ and $H$ be points of intersection of segment $BC$ with segments $EG$ and $FG$, respectively. Prove that the circle $(GKH)$ touches the circle $\Omega$. [i] Proposed by Dong Luu [/i]