Found problems: 1049
2011 N.N. Mihăileanu Individual, 4
Let be a convex quadrilateral $ ABCD $ and the points $ M,N,P,Q $ such that $ MAB\sim NBC\sim PCD\sim QDA. $
[b]a)[/b] Prove that $ ABCD $ is a parallelogram if and only if $ MNPQ $ is a parallelogram.
[b]b)[/b] Show that if the diagonals of $ MNPQ $ are congruent and perpendicular, then the diagonals of $ ABCD $ are congruent and perpendicular, or $ MAB $ is a right isosceles triangle.
[i]Nelu Chichirim[/i]
2001 Croatia National Olympiad, Problem 1
Let $O$ and $P$ be fixed points on a plane, and let $ABCD$ be any parallelogram with center $O$. Let $M$ and $N$ be the midpoints of $AP$ and $BP$ respectively. Lines $MC$ and $ND$ meet at $Q$. Prove that the point $Q$ lies on the lines $OP$, and show that it is independent of the choice of the parallelogram $ABCD$.
2005 Korea - Final Round, 3
In a trapezoid $ABCD$ with $AD \parallel BC , O_{1}, O_{2}, O_{3}, O_{4}$ denote the circles with diameters $AB, BC, CD, DA$, respectively. Show that there exists a circle with center inside the trapezoid which is tangent to all the four circles $O_{1},..., O_{4}$ if and only if $ABCD$ is a parallelogram.
2014 Iran Geometry Olympiad (senior), 5:
Two points $P$ and $Q$ lying on side $BC$ of triangle $ABC$ and their distance from the midpoint of $BC$ are equal.The perpendiculars from $P$ and $Q$ to $BC$ intersect $AC$ and $AB$ at $E$ and $F$,respectively.$M$ is point of intersection $PF$ and $EQ$.If $H_1$ and $H_2$ be the orthocenters of triangles $BFP$ and $CEQ$, respectively, prove that $ AM\perp H_1H_2 $.
Author:Mehdi E'tesami Fard , Iran
1979 IMO Shortlist, 22
Two circles in a plane intersect. $A$ is one of the points of intersection. Starting simultaneously from $A$ two points move with constant speed, each travelling along its own circle in the same sense. The two points return to $A$ simultaneously after one revolution. Prove that there is a fixed point $P$ in the plane such that the two points are always equidistant from $P.$
2016 PAMO, 6
Consider an $n\times{n}$ grid formed by $n^2$ unit squares. We define the centre of a unit square as the intersection of its diagonals.
Find the smallest integer $m$ such that, choosing any $m$ unit squares in the grid, we always get four unit squares among them whose centres are vertices of a parallelogram.
2003 AMC 8, 10
$\textbf{Bake Sale}$
Four friends, Art, Roger, Paul and Trisha, bake cookies, and all cookies have the same thickness. The shapes of the cookies differ, as shown.
$\circ$ Art's cookies are trapezoids:
[asy]size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8));
draw(origin--(5,0)--(5,3)--(2,3)--cycle);
draw(rightanglemark((5,3), (5,0), origin));
label("5 in", (2.5,0), S);
label("3 in", (5,1.5), E);
label("3 in", (3.5,3), N);[/asy]
$\circ$ Roger's cookies are rectangles:
[asy]size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8));
draw(origin--(4,0)--(4,2)--(0,2)--cycle);
draw(rightanglemark((4,2), (4,0), origin));
draw(rightanglemark((0,2), origin, (4,0)));
label("4 in", (2,0), S);
label("2 in", (4,1), E);[/asy]
$\circ$ Paul's cookies are parallelograms:
[asy]size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8));
draw(origin--(3,0)--(2.5,2)--(-0.5,2)--cycle);
draw((2.5,2)--(2.5,0), dashed);
draw(rightanglemark((2.5,2),(2.5,0), origin));
label("3 in", (1.5,0), S);
label("2 in", (2.5,1), W);[/asy]
$\circ$ Trisha's cookies are triangles:
[asy]size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8));
draw(origin--(3,0)--(3,4)--cycle);
draw(rightanglemark((3,4),(3,0), origin));
label("3 in", (1.5,0), S);
label("4 in", (3,2), E);[/asy]
Each friend uses the same amount of dough, and Art makes exactly 12 cookies. How many cookies will be in one batch of Trisha's cookies?
$ \textbf{(A)}\ 10\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 24$
2006 Sharygin Geometry Olympiad, 8.3
A parallelogram $ABCD$ is given. Two circles with centers at the vertices $A$ and $C$ pass through $B$. The straight line $\ell$ that passes through $B$ and crosses the circles at second time at points $X, Y$ respectively. Prove that $DX = DY$.
2013 AMC 10, 20
A unit square is rotated $45^\circ$ about its center. What is the area of the region swept out by the interior of the square?
$ \textbf{(A)}\ 1-\frac{\sqrt2}2+\frac\pi4\qquad\textbf{(B)}\ \frac12+\frac\pi4\qquad\textbf{(C)}\ 2-\sqrt2+\frac\pi4\qquad\textbf{(D)}\ \frac{\sqrt2}2+\frac\pi4\qquad\textbf{(E)}\ 1+\frac{\sqrt2}4+\frac\pi8 $
1976 Kurschak Competition, 1
$ABCD$ is a parallelogram. $P$ is a point outside the parallelogram such that angles $\angle PAB$ and $\angle PCB$ have the same value but opposite orientation. Show that $\angle APB = \angle DPC$.
2002 Switzerland Team Selection Test, 2
A point$ O$ inside a parallelogram $ABCD$ satisfies $\angle AOB + \angle COD = \pi$. Prove that $\angle CBO = \angle CDO$.
2005 Postal Coaching, 16
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]
2001 Tournament Of Towns, 3
Let $AH_A$, $BH_B$ and $CH_C$ be the altitudes of triangle $\triangle ABC$. Prove that the triangle whose vertices are the intersection points of the altitudes of $\triangle AH_BH_C$, $\triangle BH_AH_C$ and $\triangle CH_AH_B$ is congruent to $\triangle H_AH_BH_C$.
2020 Poland - Second Round, 3.
Let $M$ be the midpoint of the side $BC$ of a acute triangle $ABC$. Incircle of the triangle $ABM$ is tangent to the side $AB$ at the point $D$. Incircle of the triangle $ACM$ is tangent to the side $AC$ at the point $E$. Let $F$ be the such point, that the quadrilateral $DMEF$ is a parallelogram. Prove that $F$ lies on the bisector of $\angle BAC$.
2014 Taiwan TST Round 3, 3
Let $M$ be any point on the circumcircle of triangle $ABC$. Suppose the tangents from $M$ to the incircle meet $BC$ at two points $X_1$ and $X_2$. Prove that the circumcircle of triangle $MX_1X_2$ intersects the circumcircle of $ABC$ again at the tangency point of the $A$-mixtilinear incircle.
2021 Sharygin Geometry Olympiad, 9
Points $E$ and $F$ lying on sides $BC$ and $AD$ respectively of a parallelogram $ABCD$ are such that $EF=ED=DC$. Let $M$ be the midpoint of $BE$ and $MD$ meet $EF$ at $G$. Prove that $\angle EAC=\angle GBD$.
2020 Iranian Geometry Olympiad, 2
A parallelogram $ABCD$ is given ($AB \neq BC$). Points $E$ and $G$ are chosen on the line $\overline{CD}$ such that $\overline{AC}$ is the angle bisector of both angles $\angle EAD$ and $\angle BAG$. The line $\overline{BC}$ intersects $\overline{AE}$ and $\overline{AG}$ at $F$ and $H$, respectively. Prove that the line $\overline{FG}$ passes through the midpoint of $HE$.
[i]Proposed by Mahdi Etesamifard[/i]
1986 Miklós Schweitzer, 9
Consider a latticelike packing of translates of a convex region $K$. Let $t$ be the area of the fundamental parallelogram of the lattice defining the packing, and let $t_{\min} (K)$ denote the minimal value of $t$ taken for all latticelike packings. Is there a natural number $N$ such that for any $n>N$ and for any $K$ different from a parallelogram, $nt_{\min} (K)$ is smaller that the area of any convex domain in which $n$ translates to $K$ can be placed without overlapping? (By a [i]latticelike packing[/i] of $K$ we mean a set of nonoverlapping translates of $K$ obtained from $K$ by translations with all vectors of a lattice.) [G. and L. Fejes-Toth]
1984 Bulgaria National Olympiad, Problem 6
Let there be given a pyramid $SABCD$ whose base $ABCD$ is a parallelogram. Let $N$ be the midpoint of $BC$. A plane $\lambda$ intersects the lines $SC,SA,AB$ at points $P,Q,R$ respectively such that $\overline{CP}/\overline{CS}=\overline{SQ}/\overline{SA}=\overline{AR}/\overline{AB}$. A point $M$ on the line $SD$ is such that the line $MN$ is parallel to $\lambda$. Show that the locus of points $M$, when $\lambda$ takes all possible positions, is a segment of the length $\frac{\sqrt5}2SD$.
2015 IFYM, Sozopol, 1
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.
2019 IMEO, 1
Let $ABC$ be a scalene triangle with circumcircle $\omega$. The tangent to $\omega$ at $A$ meets $BC$ at $D$. The $A$-median of triangle $ABC$ intersects $BC$ and $\omega$ at $M$ and $N$, respectively. Suppose that $K$ is a point such that $ADMK$ is a parallelogram. Prove that $KA = KN$.
[i]Proposed by Alexandru Lopotenco (Moldova)[/i]
2004 USA Team Selection Test, 1
Suppose $a_1, a_2, \ldots, a_n$ and $b_1, b_2, \ldots, b_n$ are real numbers such that \[ (a_1 ^ 2 + a_2 ^ 2 + \cdots + a_n ^ 2 -1)(b_1 ^ 2 + b_2 ^ 2 + \cdots + b_n ^ 2 - 1) > (a_1 b_1 + a_2 b_2 + \cdots + a_n b_n - 1)^2. \] Prove that $a_1 ^ 2 + a_2 ^ 2 + \cdots + a_n ^ 2 > 1$ and $b_1 ^ 2 + b_2 ^ 2 + \cdots + b_n ^ 2 > 1$.
2013 Balkan MO Shortlist, G2
Let $ABCD$ be a quadrilateral, let $O$ be the intersection point of diagonals $AC$ and $BD$, and let $P$ be the intersection point of sides $AB$ and $CD$. Consider the parallelograms $AODE$ and $BOCF$. Prove that $E, F$ and $P$ are collinear.
1999 Tuymaada Olympiad, 4
A right parallelepiped (i.e. a parallelepiped one of whose edges is perpendicular to a face) is given. Its vertices have integral coordinates, and no other points with integral coordinates lie on its faces or edges. Prove that the volume of this parallelepiped is a sum of three perfect squares.
[i]Proposed by A. Golovanov[/i]
2010 Iran MO (3rd Round), 5
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)