Found problems: 1049
Let $ABCD$ be a quadrilateral with parallel sides $AD$ and $BC$, $M$ and $N$ are the midpoints of its sides $AB$ and $CD$, respectively. The straight line $MN$ bisects the segment connecting the centers of the circumcircles of triangles $ABC$ and $ADC$. Prove that $ABCD$ is a parallelogram.
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]
Let $n\ge 1$ be a positive integer. A square of side length $n$ is divided by lines parallel to each side into $n^2$ squares of side length $1$. Find the number of parallelograms which have vertices among the vertices of the $n^2$ squares of side length $1$, with both sides smaller or equal to $2$, and which have tha area equal to $2$.
(Greece)
In parallelogram $ABCD$, angles $B$ and $D$ are acute while angles $A$ and $C$ are obtuse. The perpendicular from $C$ to $AB$ and the perpendicular from $A$ to $BC$ intersect at a point $P$ inside the parallelogram. If $PB=700$ and $PD=821$, what is $AC$?
Let $ C_1$ and $ C_2$ be two congruent circles centered at $ O_1$ and $ O_2$, which intersect at $ A$ and $ B$. Take a point $ P$ on the arc $ AB$ of $ C_2$ which is contained in $ C_1$. $ AP$ meets $ C_1$ at $ C$, $ CB$ meets $ C_2$ at $ D$ and the bisector of $ \angle CAD$ intersects $ C_1$ and $ C_2$ at $ E$ and $ L$, respectively. Let $ F$ be the symmetric point of $ D$ with respect to the midpoint of $ PE$. Prove that there exists a point $ X$ satisfying $ \angle XFL \equal{} \angle XDC \equal{} 30^\circ$ and $ CX \equal{} O_1O_2$.
[i]
Author: Arnoldo Aguilar (El Salvador)[/i]
A figure is an equiangular parallelogram if and only if it is a
$ \textbf{(A)}\ \text{rectangle}\qquad\textbf{(B)}\ \text{regular polygon}\qquad\textbf{(C)}\ \text{rhombus}\qquad\textbf{(D)}\ \text{square}\qquad\textbf{(E)}\ \text{trapezoid} $
On a plane are given three non-collinear points $A, B, C$. We are given a disk of diameter different from that of the circle passing through $A, B, C$ large enough to cover all three points. Construct the fourth vertex of the parallelogram $ABCD$ using only this disk (The disk is to be used as a circular ruler, for constructing a circle passing through two given points).
Given a rhombus $ABCD$, find the locus of the points $P$ lying inside the rhombus and satisfying $\angle APD+\angle BPC=180^{\circ}$.
Show that the four perpendiculars dropped from the midpoints of the sides of a cyclic quadrilateral to the respective opposite sides are concurrent.
[b]Note by Darij:[/b] A [i]cyclic quadrilateral [/i]is a quadrilateral inscribed in a circle.
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].
Let $ABCD$ be a quadrilateral such that $\angle ABC = \angle ADC = 90º$ and $\angle BCD$ > $90º$. Let $P$ be a point inside of the $ABCD$ such that $BCDP$ is parallelogram, the line $AP$ intersects $BC$ in $M$. If $BM = 2, MC = 5, CD = 3$. Find the length of $AM$.
Let $ABC$ be an isosceles triangle with base $AC$. Points $D$ and $E$ are chosen on the sides $AC$ and $BC$, respectively, such that $CD = DE$. Let $H, J,$ and $K$ be the midpoints of $DE, AE,$ and $BD$, respectively. The circumcircle of triangle $DHK$ intersects $AD$ at point $F$, whereas the circumcircle of triangle $HEJ$ intersects $BE$ at $G$. The line through $K$ parallel to $AC$ intersects $AB$ at $I$. Let $IH \cap GF =$ {$M$}. Prove that $J, M,$ and $K$ are collinear points.
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.
$ \Delta ABC$ is a triangle such that $ AB \neq AC$. The incircle of $ \Delta ABC$ touches $ BC, CA, AB$ at $ D, E, F$ respectively. $ H$ is a point on the segment $ EF$ such that $ DH \bot EF$. Suppose $ AH \bot BC$, prove that $ H$ is the orthocentre of $ \Delta ABC$.
Remark: the original question has missed the condition $ AB \neq AC$
Let $ABCD$ be a parallelogram and $H$ the Orthocentre of $\triangle{ABC}$. The line parallel to $AB$ through $H$ intersects $BC$ at $P$ and $AD$ at $Q$ while the line parallel to $BC$ through $H$ intersects $AB$ at $R$ and $CD$ at $S$. Show that $P$, $Q$, $R$ and $S$ are concyclic.
[i](Swiss Mathematical Olympiad 2011, Final round, problem 8)[/i]
A circle touches sides $AB, BC, CD$ of a parallelogram $ABCD$ at points $K, L, M$ respectively. Prove that the line $KL$ bisects the height of the parallelogram drawn from the vertex $C$ to $AB$.
A cube with side length $ 1$ is sliced by a plane that passes through two diagonally opposite vertices $ A$ and $ C$ and the midpoints $ B$ and $ D$ of two opposite edges not containing $ A$ and $ C$, ac shown. What is the area of quadrilateral $ ABCD$?
[asy]import three;
size(200);
defaultpen(fontsize(8)+linewidth(0.7));
currentprojection=obliqueX;
dotfactor=4;
draw((0.5,0,0)--(0,0,0)--(0,0,1)--(0,0,0)--(0,1,0),linetype("4 4"));
draw((0.5,0,1)--(0,0,1)--(0,1,1)--(0.5,1,1)--(0.5,0,1)--(0.5,0,0)--(0.5,1,0)--(0.5,1,1));
draw((0.5,1,0)--(0,1,0)--(0,1,1));
dot((0.5,0,0));
label("$A$",(0.5,0,0),WSW);
dot((0,1,1));
label("$C$",(0,1,1),NE);
dot((0.5,1,0.5));
label("$D$",(0.5,1,0.5),ESE);
dot((0,0,0.5));
label("$B$",(0,0,0.5),NW);[/asy]$ \textbf{(A)}\ \frac {\sqrt6}{2} \qquad \textbf{(B)}\ \frac {5}{4} \qquad \textbf{(C)}\ \sqrt2 \qquad \textbf{(D)}\ \frac {3}{2} \qquad \textbf{(E)}\ \sqrt3$
Let $ABCD$ be a trapezoid (quadrilateral with one pair of parallel sides) such that $AB < CD$. Suppose that $AC$ and $BD$ meet at $E$ and $AD$ and $BC$ meet at $F$. Construct the parallelograms $AEDK$ and $BECL$. Prove that $EF$ passes through the midpoint of the segment $KL$.
In the plane, there are $n$ points ($n\ge 4$) where no 3 of them are collinear. Let $A(n)$ be the number of parallelograms whose vertices are those points with area $1$. Prove the following inequality:
$A(n)\leq \frac{n^2-3n}{4}$ for all $n\ge 4$
Let $ABCD$ be a convex quadrilateral, where $M$ is the midpoint of $DC$, $N$ is the midpoint of $BC$, and $O$ is the intersection of diagonals $AC$ and $BD$. Prove that $O$ is the centroid of the triangle $AMN$ if and only if $ABCD$ is a parallelogram.
Given is a parallelogram $ABCD$ with $\angle A < 90^o$ and $|AB| < |BC|$. The angular bisector of angle $A$ intersects side $BC$ in $M$ and intersects the extension of $DC$ in $N$. Point $O$ is the centre of the circle through $M, C$, and $N$. Prove that $\angle OBC = \angle ODC$.
[asy]
unitsize (1.2 cm);
pair A, B, C, D, M, N, O;
A = (0,0);
B = (2,0);
D = (1,3);
C = B + D - A;
M = extension(A, incenter(A,B,D), B, C);
N = extension(A, incenter(A,B,D), D, C);
O = circumcenter(C,M,N);
draw(D--A--B--C);
draw(interp(D,N,-0.1)--interp(D,N,1.1));
draw(A--interp(A,N,1.1));
draw(circumcircle(M,C,N));
label("$\circ$", A + (0.45,0.15));
label("$\circ$", A + (0.25,0.35));
dot("$A$", A, SW);
dot("$B$", B, SE);
dot("$C$", C, dir(90));
dot("$D$", D, dir(90));
dot("$M$", M, SE);
dot("$N$", N, dir(90));
dot("$O$", O, SE);
[/asy]
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)
The point P lies on the side $CD$ of the parallelogram $ABCD$ with $\angle DBA = \angle CBP$. Point $O$ is the center of the circle passing through the points $D$ and $P$ and tangent to the straight line $AD$ at point $D$. Prove that $AO = OC$.
Call an $n$-gon to be [i]lattice[/i] if its vertices are lattice points. Prove that inside every lattice convex pentagon there exists a lattice point.
Let $N$ be a point on the longest side $AC$ of a triangle $ABC$. The perpendicular bisectors of $AN$ and $NC$ intersect $AB$ and $BC$ respectively in $K$ and $M$. Prove that the circumcenter $O$ of $\triangle ABC$ lies on the circumcircle of triangle $KBM$.