Olympiad problems

2015 Chile National Olympiad, 1

Tags: geometry , construction , ruler , symmetric , parallelogram

On the plane, there is drawn a parallelogram $P$ and a point $X$ outside of $P$. Using only an ungraded rule, determine the point $W$ that is symmetric to $X$ with respect to the center $O$ of $P$.

2006 AIME Problems, 10

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

This is the one with the 8 circles? I made each circle into the square in which the circle is inscribed, then calculated it with that. It got the right answer but I don't think that my method is truly valid...

2007 Estonia Math Open Junior Contests, 2

Tags: midpoint , equal area , geometry , parallelogram

The sides $AB, BC, CD$ and $DA$ of the convex quadrilateral $ABCD$ have midpoints $E, F, G$ and $H$. Prove that the triangles $EFB, FGC, GHD$ and $HEA$ can be put together into a parallelogram equal to $EFGH$.

2021 Serbia JBMO TSTs, 4

Tags: geometry , orthocenter , parallelogram , circumcenter , circumcircle

On sides $AB$ and $AC$ of an acute triangle $\Delta ABC$, with orthocenter $H$ and circumcenter $O$, are given points $P$ and $Q$ respectively such that $APHQ$ is a parallelogram. Prove the following equality: \begin{align*} \frac{PB\cdot PQ}{QA\cdot QO}=2 \end{align*}

2010 Sharygin Geometry Olympiad, 4

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

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.

2013 Online Math Open Problems, 17

Tags: analytic geometry , geometry , parallelogram

Let $ABXC$ be a parallelogram. Points $K,P,Q$ lie on $\overline{BC}$ in this order such that $BK = \frac{1}{3} KC$ and $BP = PQ = QC = \frac{1}{3} BC$. Rays $XP$ and $XQ$ meet $\overline{AB}$ and $\overline{AC}$ at $D$ and $E$, respectively. Suppose that $\overline{AK} \perp \overline{BC}$, $EK-DK=9$ and $BC=60$. Find $AB+AC$. [i]Proposed by Evan Chen[/i]

2007 ITAMO, 3

Tags: circumcircle , rhombus , trapezoid , parallelogram , geometry

Let ABC be a triangle, G its centroid, M the midpoint of AB, D the point on the line $AG$ such that $AG = GD, A \neq D$, E the point on the line $BG$ such that $BG = GE, B \neq E$. Show that the quadrilateral BDCM is cyclic if and only if $AD = BE$.

1976 Chisinau City MO, 133

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

A triangle with a parallelogram inside was placed in a square. Prove that the area of a parallelogram is not more than a quarter of a square.

2014 Tuymaada Olympiad, 7

Tags: parallelogram , incenter , analytic geometry , geometry

A parallelogram $ABCD$ is given. The excircle of triangle $\triangle{ABC}$ touches the sides $AB$ at $L$ and the extension of $BC$ at $K$. The line $DK$ meets the diagonal $AC$ at point $X$; the line $BX$ meets the median $CC_1$ of trianlge $\triangle{ABC}$ at ${Y}$. Prove that the line $YL$, median $BB_1$ of triangle $\triangle{ABC}$ and its bisector $CC^\prime$ have a common point. [i](A. Golovanov)[/i]

2001 District Olympiad, 3

Tags: symmetry , parallelogram , geometry

Consider a triangle $\Delta ABC$ and three points $D,E,F$ such that: $B$ and $E$ are on different side of the line $AC$, $C$ and $D$ are on different sides of $AB$, $A$ and $F$ are on the same side of the line $BC$. Also $\Delta ADB \sim \Delta CEA \sim \Delta CFB$. Let $M$ be the middle point of $AF$. Prove that: a)$\Delta BDF \sim \Delta FEC$. b) $M$ is the middle point of $DE$. [i]Dan Branzei[/i]

CVM 2020, Problem 4

Tags: geometry , parallelogram

Consider an $ABCD$ parallelogram with $\overline{AD}$ $=$ $\overline{BD}$. Point E lies in segment $\overline{BD}$ in such a way that $\overline{AE}$ $=$ $\overline{DE}$. The extension of line $\overline{AE}$ cuts segment $\overline{BC}$ and $F$. if line $\overline{DF}$ is the bisector of the $\angle CED$. Find the value of the $\angle ABD$ $\textbf{4.1.}$ Point $E$ lies in segment $\overline{BD}$ means that exits a point $E$ in the segment $\overline{BD}$ in other words lies refers to the same thing found [i]Proposed by Alicia Smith, Francisco Morazan[/i]

2008 Cuba MO, 2

Tags: geometry , parallelogram

Consider the parallelogram $ABCD$. A circle is drawn that passes through $A$ and intersects side $AD$ at $N$, side $AB$ at $M$ and diagonal $AC$ in $P$ such that points $A, M, N, P$ are different. Prove that $$AP\cdot AC = AM \cdot AB + AN \cdot AD.$$

Estonia Open Senior - geometry, 2002.2.3

Tags: geometry , parallelogram , rhombus , equilateral

Let $ABCD$ be a rhombus with $\angle DAB = 60^o$. Let $K, L$ be points on its sides $AD$ and $DC$ and $M$ a point on the diagonal $AC$ such that $KDLM$ is a parallelogram. Prove that triangle $BKL$ is equilateral.

2002 Federal Competition For Advanced Students, Part 1, 4

Tags: parallelogram , geometry

Let $A,C, P$ be three distinct points in the plane. Construct all parallelograms $ABCD$ such that point $P$ lies on the bisector of angle $DAB$ and $\angle APD = 90^\circ$.

2013 Bogdan Stan, 3

Tags: geometry , parallelogram

$ O $ is the center of a parallelogram $ ABCD. $ Let $ G $ on the segment $ OB $ (excluding its endpoints), $ N $ on the line $ DC $ and $ M $ on the segment $ AD $ (excluding its endpoints) such that $ CN>ND, AM=6MD $ and so that there exists a natural number $ n\ge 3 $ such that $ OB=nGO. $ Show that $ G,M,N $ are collinear if and only if $$ \left( \frac{CN}{ND} -6 \right) (n+1)=2. $$

1992 India Regional Mathematical Olympiad, 4

Tags: geometry , circumcircle , trigonometry , parallelogram , rectangle , cyclic quadrilateral , power of a point

$ABCD$ is a cyclic quadrilateral with $AC \perp BD$; $AC$ meets $BD$ at $E$. Prove that \[ EA^2 + EB^2 + EC^2 + ED^2 = 4 R^2 \] where $R$ is the radius of the circumscribing circle.

2005 Federal Competition For Advanced Students, Part 1, 4

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

We're given two congruent, equilateral triangles $ABC$ and $PQR$ with parallel sides, but one has one vertex pointing up and the other one has the vertex pointing down. One is placed above the other so that the area of intersection is a hexagon $A_1A_2A_3A_4A_5A_6$ (labelled counterclockwise). Prove that $A_1A_4$, $A_2A_5$ and $A_3A_6$ are concurrent.

1990 Hungary-Israel Binational, 2

Tags: parallelogram , geometric transformation , rotation , geometry

Let $ ABC$ be a triangle where $ \angle ACB\equal{}90^{\circ}$. Let $ D$ be the midpoint of $ BC$ and let $ E$, and $ F$ be points on $ AC$ such that $ CF\equal{}FE\equal{}EA$. The altitude from $ C$ to the hypotenuse $ AB$ is $ CG$, and the circumcentre of triangle $ AEG$ is $ H$. Prove that the triangles $ ABC$ and $ HDF$ are similar.

Kyiv City MO 1984-93 - geometry, 1991.9.4

Tags: geometry , parallelogram , inscribed , area

A parallelogram is inscribed in a quadrilateral, two opposite vertices of which are the midpoints of the opposite sides of the quadrilateral. Determine the area of such a parallelogram if the area of the quadrilateral is equal to $S_o$.

1995 Tournament Of Towns, (477) 1

Tags: angle bisector , locus , parallelogram , geometry

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)

2022 Iranian Geometry Olympiad, 2

Tags: trapezoid , geometry , parallelogram

An isosceles trapezoid $ABCD$ $(AB \parallel CD)$ is given. Points $E$ and $F$ lie on the sides $BC$ and $AD$, and the points $M$ and $N$ lie on the segment $EF$ such that $DF = BE$ and $FM = NE$. Let $K$ and $L$ be the foot of perpendicular lines from $M$ and $N$ to $AB$ and $CD$, respectively. Prove that $EKFL$ is a parallelogram. [i]Proposed by Mahdi Etesamifard[/i]

2010 Moldova Team Selection Test, 3

Tags: ratio , parallelogram , perpendicular bisector , geometry

Let $ ABC$ be an acute triangle. $ H$ is the orthocenter and $ M$ is the middle of the side $ BC$. A line passing through $ H$ and perpendicular to $ HM$ intersect the segment $ AB$ and $ AC$ in $ P$ and $ Q$. Prove that $ MP \equal{} MQ$

2005 IMO Shortlist, 3

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

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]

2014 Iran MO (3rd Round), 4

Tags: geometry , circumcircle , incenter , parallelogram , rectangle

$D$ is an arbitrary point lying on side $BC$ of $\triangle{ABC}$. Circle $\omega_1$ is tangent to segments $AD$ , $BD$ and the circumcircle of $\triangle{ABC}$ and circle $\omega_2$ is tangent to segments $AD$ , $CD$ and the circumcircle of $\triangle{ABC}$. Let $X$ and $Y$ be the intersection points of $\omega_1$ and $\omega_2$ with $BC$ respectively and take $M$ as the midpoint of $XY$. Let $T$ be the midpoint of arc $BC$ which does not contain $A$. If $I$ is the incenter of $\triangle{ABC}$, prove that $TM$ goes through the midpoint of $ID$.

2001 Moldova National Olympiad, Problem 6

Tags: parallelogram , geometry

Two sides of a quadrilateral $ABCD$ are parallel. Let $M$ and $N$ be the midpoints of $BC$ and $CD$ respectively, and $P$ be the intersection point of $AN$ and $DM$. Prove that if $AP=4PN$, then $ABCD$ is a parallelogram.