Found problems: 1049
The opposite sidelines of quadrilateral $ ABCD$ intersect at points $ P$ and $ Q$. Two lines passing through these points meet the side of $ ABCD$ in four points which are the vertices of a parallelogram. Prove that the center of this parallelogram lies on the line passing through the midpoints of diagonals of $ ABCD$.
Triangle $ABC$ has sides $AB = 25$, $BC = 30$, and $CA=20$. Let $P,Q$ be the points on segments $AB,AC$, respectively, such that $AP=5$ and $AQ=4$. Suppose lines $BQ$ and $CP$ intersect at $R$ and the circumcircles of $\triangle{BPR}$ and $\triangle{CQR}$ intersect at a second point $S\ne R$. If the length of segment $SA$ can be expressed in the form $\frac{m}{\sqrt{n}}$ for positive integers $m,n$, where $n$ is not divisible by the square of any prime, find $m+n$.
[i]Victor Wang[/i]
$[AB]$ and $[CD]$ are not parallel in the convex quadrilateral $ABCD$. Let $E$ and $F$ be the midpoints of $[AD]$ and $[BC]$, respectively. If $|CD|=12$, $|AB|=22$, and $|EF|=x$, what is the sum of integer values of $x$?
$
\textbf{(A)}\ 110
\qquad\textbf{(B)}\ 114
\qquad\textbf{(C)}\ 118
\qquad\textbf{(D)}\ 121
\qquad\textbf{(E)}\ \text{None of the above}
$
Given acute triangle $ABC$ with $AB>AC$, let $M$ be the midpoint of $BC$. $P$ is a point in triangle $AMC$ such that $\angle MAB=\angle PAC$. Let $O,O_1,O_2$ be the circumcenters of $\triangle ABC,\triangle ABP,\triangle ACP$ respectively. Prove that line $AO$ passes through the midpoint of $O_1 O_2$.
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]
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 $AD$.
In the parallelogram $ABCD$, a line through $C$ intersects the diagonal $BD$ at $E$ and $AB$ at $F$. If $F$ is the midpoint of $AB$ and the area of $\vartriangle BEC$ is $100$, find the area of the quadrilateral $AFED$.
A $k\times \ell$ 'parallelogram' is drawn on a paper with hexagonal cells (it consists of $k$ horizontal rows of $\ell$ cells each). In this parallelogram a set of non-intersecting sides of hexagons is chosen; it divides all the vertices into pairs.
Juniors) How many vertical sides can there be in this set?
Seniors) How many ways are there to do that?
[asy]
size(120);
defaultpen(linewidth(0.8));
path hex = dir(30)--dir(90)--dir(150)--dir(210)--dir(270)--dir(330)--cycle;
for(int i=0;i<=3;i=i+1)
{
for(int j=0;j<=2;j=j+1)
{
real shiftx=j*sqrt(3)/2+i*sqrt(3),shifty=j*3/2;
draw(shift(shiftx,shifty)*hex);
}
}
[/asy]
[i](T. Doslic)[/i]
$AB$ is the diameter of semicircle $O$. $C$,$D$ are two arbitrary points on semicircle $O$. Point $P$ lies on line $CD$ such that line $PB$ is tangent to semicircle $O$ at $B$. Line $PO$ intersects line $CA$, $AD$ at point $E$, $F$ respectively. Prove that $OE$=$OF$.
Let $ABC$ be an acute scalene triangle, $H$ its orthocenter and $G$ its geocenter. The circumference with diameter $AH$ cuts the circumcircle of $BHC$ in $A'$ ($A' \neq H$). Points $B'$ and $C'$ are defined similarly. Show that the points $A'$, $B'$, $C'$, and $G$ lie in one circumference.
On side \( AC \) of triangle \( ABC \), a point \( P \) is chosen such that \( AP = \frac{1}{3} AC \), and on segment \( BP \), a point \( S \) is chosen such that \( CS \perp BP \). A point \( T \) is such that \( BCST \) is a parallelogram. Prove that \( AB = AT \).
[i]Proposed by Bohdan Zheliabovskyi[/i]
Let $(O)$ be a fixed circle with the radius $R$. Let $A$ and $B$ be fixed points in $(O)$ such that $A,B,O$ are not collinear. Consider a variable point $C$ lying on $(O)$ ($C\neq A,B$). Construct two circles $(O_1),(O_2)$ passing through $A,B$ and tangent to $BC,AC$ at $C$, respectively. The circle $(O_1)$ intersects the circle $(O_2)$ in $D$ ($D\neq C$). Prove that:
a) \[ CD\leq R \]
b) The line $CD$ passes through a point independent of $C$ (i.e. there exists a fixed point on the line $CD$ when $C$ lies on $(O)$).
Let $ABC$ be a triangle with circumcircle $k$. The points $A_1, B_1,$ and $C_1$ on $k$ are the midpoints of arcs $\widehat{BC}$ (not containing $A$), $\widehat{AC}$ (not containing $B$), and $\widehat{AB}$ (not containing $C$), respectively. The pairwise distinct points $A_2, B_2,$ and $C_2$ are chosen such that the quadrilaterals $AB_1A_2C_1, BA_1B_2C_1,$ and $CA_1C_2B_1$ are parallelograms. Prove that $k$ and the circumcircle of triangle $A_2B_2C_2$ have a common center.
[b]Comment.[/b] Point $A_2$ can also be defined as the reflection of $A$ with respect to the midpoint of $B_1C_1$, and analogous definitions can be used for $B_2$ and $C_2$.
Given a convex quadrilateral whose opposite sides are not parallel, and giving an internal point $P$. Find a parallelogram whose vertices are on the side lines of the rectangle and whose center is $P$. Give a method by which we can construct it (provided there is one).
[img]https://1.bp.blogspot.com/-t4aCJza0LxI/X9j1qbSQE4I/AAAAAAAAMz4/V9pr7Cd22G4F320nyRLZMRnz18hMw9NHQCLcBGAsYHQ/s0/2012%2BDurer%2BC3.png[/img]
Prove that the parallelogram $ABCD$ with relation $\angle ABD + \angle DAC = 90^o$, is either a rectangle or a rhombus.
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 $q$.
Let $ABC$ be a triangle. Let $\Omega$ be the brocard point. Prove that $\left(\frac{A\Omega}{BC}\right)^2+\left(\frac{B\Omega}{AC}\right)^2+\left(\frac{C\Omega}{AB}\right)^2\ge 1$
Let $ABCD$ be a convex quadrilateral whose sides $AD$ and $BC$ are not parallel. Suppose that the circles with diameters $AB$ and $CD$ meet at points $E$ and $F$ inside the quadrilateral. Let $\omega_E$ be the circle through the feet of the perpendiculars from $E$ to the lines $AB,BC$ and $CD$. Let $\omega_F$ be the circle through the feet of the perpendiculars from $F$ to the lines $CD,DA$ and $AB$. Prove that the midpoint of the segment $EF$ lies on the line through the two intersections of $\omega_E$ and $\omega_F$.
[i]Proposed by Carlos Yuzo Shine, Brazil[/i]
Let $ABCD$ be a parallelogram. Let $M$ be the midpoint of the arc $AC$ containing $B$ of the circumcircle of $ABC$ . Let $E$ be a point on segment $AD$ and $F$ a point on segment $CD$ such that $ME=MD=MF$. Show that $BMEF$ is cyclic.
[i]Proposed by A. Tereshin[/i]
A circle touches sides $DA$, $AB$, $BC$, $CD$ of a quadrilateral $ABCD$ at points $K$, $L$, $M$, $N$, respectively. Let $S_1$, $S_2$, $S_3$, $S_4$ respectively be the incircles of triangles $AKL$, $BLM$, $CMN$, $DNK$. The external common tangents distinct from the sides of $ABCD$ are drawn to $S_1$ and $S_2$, $S_2$ and $S_3$, $S_3$ and $S_4$, $S_4$ and $S_1$. Prove that these four tangents determine a rhombus.
A parallelopiped has vertices $A_1, A_2, ... , A_8$ and center $O$. Show that:
\[ 4 \sum_{i=1}^8 OA_i ^2 \leq \left(\sum_{i=1}^8 OA_i \right) ^2 \]
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]
Let $ABCDE$ be a convex pentagon such that $BC \parallel AE,$ $AB = BC + AE,$ and $\angle ABC = \angle CDE.$ Let $M$ be the midpoint of $CE,$ and let $O$ be the circumcenter of triangle $BCD.$ Given that $\angle DMO = 90^{\circ},$ prove that $2 \angle BDA = \angle CDE.$
[i]Proposed by Nazar Serdyuk, Ukraine[/i]
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$.
Let $ABC$ be an isosceles triangle with $AC=BC$. Let $P,Q,R$ be points on the sides $AB, BC, CA$ of the triangle such that $CQPR$ is a parallelogram. Show that the reflection of $P$ over $QR$ lies on the circumcircle of $ABC$.