Olympiad problems

1976 Chisinau City MO, 133

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.

2012 NIMO Problems, 8

Tags: perimeter , parallelogram , geometry , reflection , geometric transformation , rotation

A convex 2012-gon $A_1A_2A_3 \dots A_{2012}$ has the property that for every integer $1 \le i \le 1006$, $\overline{A_iA_{i+1006}}$ partitions the polygon into two congruent regions. Show that for every pair of integers $1 \le j < k \le 1006$, quadrilateral $A_jA_kA_{j+1006}A_{k+1006}$ is a parallelogram. [i]Proposed by Lewis Chen[/i]

May Olympiad L2 - geometry, 2014.2

Tags: parallelogram , geometry

In a convex quadrilateral $ABCD$, let $M$, $N$, $P$, and $Q$ be the midpoints of $AB$, $BC$, $CD$, and $DA$ respectively. If $MP$ and $NQ$ divide $ABCD$ in four quadrilaterals with the same area, prove that $ABCD$ is a parallelogram.

2024 AMC 10, 10

Tags: parallelogram , geometry

Quadrilateral $ABCD$ is a parallelogram, and $E$ is the midpoint of the side $\overline{AD}$. Let $F$ be the intersection of lines $EB$ and $AC$. What is the ratio of the area of quadrilateral $CDEF$ to the area of triangle $CFB$? $\textbf{(A) } 5 : 4 \qquad \textbf{(B) } 4 : 3 \qquad \textbf{(C) } 3 : 2 \qquad \textbf{(D) } 5 : 3 \qquad \textbf{(E) } 2 : 1$

2003 Federal Competition For Advanced Students, Part 1, 4

Tags: parallelogram , geometry , ratio

In a parallelogram $ABCD$, points $E$ and $F$ are the midpoints of $AB$ and $BC$, respectively, and $P$ is the intersection of $EC$ and $FD$. Prove that the segments $AP,BP,CP$ and $DP$ divide the parallelogram into four triangles whose areas are in the ratio $1 : 2 : 3 : 4$.

2004 India IMO Training Camp, 1

Tags: complex number , geometry , circumcircle , parallelogram

A set $A_1 , A_2 , A_3 , A_4$ of 4 points in the plane is said to be [i]Athenian[/i] set if there is a point $P$ of the plane satsifying (*) $P$ does not lie on any of the lines $A_i A_j$ for $1 \leq i < j \leq 4$; (**) the line joining $P$ to the mid-point of the line $A_i A_j$ is perpendicular to the line joining $P$ to the mid-point of $A_k A_l$, $i,j,k,l$ being distinct. (a) Find all [i]Athenian[/i] sets in the plane. (b) For a given [i]Athenian[/i] set, find the set of all points $P$ in the plane satisfying (*) and (**)

1974 IMO Longlists, 3

Tags: geometry , parallelogram

Let $ABCD$ be an arbitrary quadrilateral. Let squares $ABB_1A_2, BCC_1B_2, CDD_1C_2, DAA_1D_2$ be constructed in the exterior of the quadrilateral. Furthermore, let $AA_1PA_2$ and $CC_1QC_2$ be parallelograms. For any arbitrary point $P$ in the interior of $ABCD$, parallelograms $RASC$ and $RPTQ$ are constructed. Prove that these two parallelograms have two vertices in common.

2007 Oral Moscow Geometry Olympiad, 3

Tags: midpoint , construction , parallelogram , geometry

Construct a parallelogram $ABCD$, if three points are marked on the plane: the midpoints of its altitudes $BH$ and $BP$ and the midpoint of the side $AD$.

1957 Moscow Mathematical Olympiad, 352

Tags: area , diagonal , minimum , parallelogram , geometry

Of all parallelograms of a given area find the one with the shortest possible longer diagonal.

2021 Polish Junior MO Second Round, 4

Tags: parallelogram , perpendicular , geometry

Points $K$ and $L$ are on the sides $BC$ and $CD$, respectively of the parallelogram $ABCD$, such that $AB + BK = AD + DL$. Prove that the bisector of angle $BAD$ is perpendicular to the line $KL$.

2013 Brazil Team Selection Test, 2

Tags: moving points , ptolemy sinus lemma , parallelogram , reflection , geometry

Let $ABCD$ be a cyclic quadrilateral whose diagonals $AC$ and $BD$ meet at $E$. The extensions of the sides $AD$ and $BC$ beyond $A$ and $B$ meet at $F$. Let $G$ be the point such that $ECGD$ is a parallelogram, and let $H$ be the image of $E$ under reflection in $AD$. Prove that $D,H,F,G$ are concyclic.

2009 Germany Team Selection Test, 3

Tags: geometry , parallelogram

Let $ A,B,C,M$ points in the plane and no three of them are on a line. And let $ A',B',C'$ points such that $ MAC'B, MBA'C$ and $ MCB'A$ are parallelograms: (a) Show that \[ \overline{MA} \plus{} \overline{MB} \plus{} \overline{MC} < \overline{AA'} \plus{} \overline{BB'} \plus{} \overline{CC'}.\] (b) Assume segments $ AA', BB'$ and $ CC'$ have the same length. Show that $ 2 \left(\overline{MA} \plus{} \overline{MB} \plus{} \overline{MC} \right) \leq \overline{AA'} \plus{} \overline{BB'} \plus{} \overline{CC'}.$ When do we have equality?

2018 AMC 12/AHSME, 22

Tags: parallelogram , complex number , geometry

The solutions to the equations $z^2=4+4\sqrt{15}i$ and $z^2=2+2\sqrt 3i,$ where $i=\sqrt{-1},$ form the vertices of a parallelogram in the complex plane. The area of this parallelogram can be written in the form $p\sqrt q-r\sqrt s,$ where $p,$ $q,$ $r,$ and $s$ are positive integers and neither $q$ nor $s$ is divisible by the square of any prime number. What is $p+q+r+s?$ $\textbf{(A) } 20 \qquad \textbf{(B) } 21 \qquad \textbf{(C) } 22 \qquad \textbf{(D) } 23 \qquad \textbf{(E) } 24 $

2015 IFYM, Sozopol, 1

Tags: trig identities , law of cosines , inequalities , geometry , parallelogram , trigonometry

Let ABCD be a convex quadrilateral such that $AB + CD = \sqrt{2}AC$ and $BC + DA = \sqrt{2}BD$. Prove that ABCD is a parallelogram.

2007 Romania Team Selection Test, 1

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

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]

JOM 2015 Shortlist, G1

Tags: geometry , parallelogram

Given a triangle $ABC$, and let $ E $ and $ F $ be the feet of altitudes from vertices $ B $ and $ C $ to the opposite sides. Denote $ O $ and $ H $ be the circumcenter and orthocenter of triangle $ ABC $. Given that $ FA=FC $, prove that $ OEHF $ is a parallelogram.

1997 Canada National Olympiad, 4

Tags: angle chasing , geometry , parallelogram

The point $O$ is situated inside the parallelogram $ABCD$ such that $\angle AOB+\angle COD=180^{\circ}$. Prove that $\angle OBC=\angle ODC$.

1989 Tournament Of Towns, (232) 6

Tags: regular , equal area , hexagon , geometry , parallelogram , combinatorial geometry

A regular hexagon is cut up into $N$ parallelograms of equal area. Prove that $N$ is divisible by three. (V. Prasolov, I. Sharygin, Moscow)

1990 Hungary-Israel Binational, 2

Tags: parallelogram , geometric transformation , geometry , rotation

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.

1997 Czech And Slovak Olympiad IIIA, 6

Tags: cyclic , locus , geometry , diagonal , parallelogram

In a parallelogram $ABCD$, triangle $ABD$ is acute-angled and $\angle BAD = \pi /4$. Consider all possible choices of points $K,L,M,N$ on sides $AB,BC, CD,DA$ respectively, such that $KLMN$ is a cyclic quadrilateral whose circumradius equals those of triangles $ANK$ and $CLM$. Find the locus of the intersection of the diagonals of $KLMN$

2023 All-Russian Olympiad Regional Round, 10.8

Tags: parallelogram , geometry

The bisector of $\angle BAD$ of a parallelogram $ABCD$ meets $BC$ at $K$. The point $L$ lies on $AB$ such that $AL=CK$. The lines $AK$ and $CL$ meet at $M$. Let $(ALM)$ meet $AD$ after $D$ at $N$. Prove that $\angle CNL=90^{o}$

1999 Israel Grosman Mathematical Olympiad, 6

Tags: midpoint , 3d geometry , coplanar , parallelogram , geometry

Let $A,B,C,D,E,F$ be points in space such that the quadrilaterals $ABDE,BCEF, CDFA$ are parallelograms. Prove that the six midpoints of the sides $AB,BC,CD,DE,EF,FA$ are coplanar

2009 Middle European Mathematical Olympiad, 9

Tags: circumcircle , geometry , parallelogram

Let $ ABCD$ be a parallelogram with $ \angle BAD \equal{} 60$ and denote by $ E$ the intersection of its diagonals. The circumcircle of triangle $ ACD$ meets the line $ BA$ at $ K \ne A$, the line $ BD$ at $ P \ne D$ and the line $ BC$ at $ L\ne C$. The line $ EP$ intersects the circumcircle of triangle $ CEL$ at points $ E$ and $ M$. Prove that triangles $ KLM$ and $ CAP$ are congruent.

1996 All-Russian Olympiad, 6

Tags: angle bisector , geometry , circumcircle , parallelogram , incenter

In isosceles triangle $ABC$ ($AB = BC$) one draws the angle bisector $CD$. The perpendicular to $CD$ through the center of the circumcircle of $ABC$ intersects $BC$ at $E$. The parallel to $CD$ through $E$ meets $AB$ at $F$. Show that $BE$ = $FD$. [i]M. Sonkin[/i]

1988 Polish MO Finals, 3

Tags: symmetry , geometry , parallelogram

$W$ is a polygon which has a center of symmetry $S$ such that if $P$ belongs to $W$, then so does $P'$, where $S$ is the midpoint of $PP'$. Show that there is a parallelogram $V$ containing $W$ such that the midpoint of each side of $V$ lies on the border of $W$.