Found problems: 1049
Let $M$ and $N$ are midpoint of edges $AB$ and $CD$ of the tetrahedron $ABCD$, $AN=DM$ and $CM=BN$. Prove that $AC=BD$.
S. Berlov
Let $ ABCD$ be a parallelogram and let $ \overrightarrow{AA^\prime}$, $ \overrightarrow{BB^\prime}$, $ \overrightarrow{CC^\prime}$, and $ \overrightarrow{DD^\prime}$ be parallel rays in space on the same side of the plane determined by $ ABCD$. If $ AA^\prime \equal{} 10$, $ BB^\prime \equal{} 8$, $ CC^\prime \equal{} 18$, $ DD^\prime \equal{} 22$, and $ M$ and $ N$ are the midpoints of $ \overline{A^{\prime}C^{\prime}}$ and $ \overline{B^{\prime}D^{\prime}}$, respectively, then $ MN \equal{}$
$ \textbf{(A)}\ 0\qquad
\textbf{(B)}\ 1\qquad
\textbf{(C)}\ 2\qquad
\textbf{(D)}\ 3\qquad
\textbf{(E)}\ 4$
Let $ ABCD$ be a convex cuadrilateral inscribed in a circumference centered at $ O$ such that $ AC$ is a diameter. Pararellograms $ DAOE$ and $ BCOF$ are constructed. Show that if $ E$ and $ F$ lie on the circumference then $ ABCD$ is a rectangle.
Let a triangle $ABC$ , $O$ it's circumcenter , $H$ ortocenter and $M$ the midpoint of $AH$. The perpendicular at $M$ to line $OM$ meets $AB$ and $AC$ at points $P$, respective $Q$. Prove that $MP=MQ$.
Babis
Let $ABCD$ be a parallelogram. Half-circles $\omega_1,\omega_2,\omega_3$ and $\omega_4$ with diameters $AB,BC,CD$ and $DA$, respectively, are erected on the exterior of $ABCD$. Line $l_1$ is parallel to $BC$ and cuts $\omega_1$ at $X$, segment $AB$ at $P$, segment $CD$ at $R$ and $\omega_3$ at $Z$. Line $l_2$ is parallel to $AB$ and cuts $\omega_2$ at $Y$, segment $BC$ at $Q$, segment $DA$ at $S$ and $\omega_4$ at $W$. If $XP\cdot RZ=YQ\cdot SW$, prove that $PQRS$ is cyclic.
[i]Proposed by José Alejandro Reyes González[/i]
Let $P$ be a polygon that is convex and symmetric to some point $O$. Prove that for some parallelogram $R$ satisfying $P\subset R$ we have \[\frac{|R|}{|P|}\leq \sqrt 2\]
where $|R|$ and $|P|$ denote the area of the sets $R$ and $P$, respectively.
[i]Proposed by Witold Szczechla, Poland[/i]
Let $ABC$ be an acute triangle, and let $X$ be a variable interior point on the minor arc $BC$ of its circumcircle. Let $P$ and $Q$ be the feet of the perpendiculars from $X$ to lines $CA$ and $CB$, respectively. Let $R$ be the intersection of line $PQ$ and the perpendicular from $B$ to $AC$. Let $\ell$ be the line through $P$ parallel to $XR$. Prove that as $X$ varies along minor arc $BC$, the line $\ell$ always passes through a fixed point. (Specifically: prove that there is a point $F$, determined by triangle $ABC$, such that no matter where $X$ is on arc $BC$, line $\ell$ passes through $F$.)
[i]Robert Simson et al.[/i]
Let $ABCD$ be a parallelogram. $E$ and $F$ are on $BC, CD$ respectively such that the triangles $ABE$ and $BCF$ have the same area. Let $BD$ intersect $AE, AF$ at $M, N$ respectively. Prove there exists a triangle whose side lengths are $BM, MN, ND$.
Let $ABC$ be a triangle, $H$ the orthocenter and $M$ the midpoint of $AC$. Let $\ell$ be the parallel through $M$ to the bisector of $\angle AHC$. Prove that $\ell$ divides the triangle in two parts of equal perimeters.
[i]Pedro Marrone, Panamá[/i]
Circles $\omega_1$ and $\omega_2$ with centers at points $O_1$ and $O_2$ intersect at points $A$ and $B$. A point $C$ is constructed such that $AO_2CO_1$ is a parallelogram. An arbitrary line is drawn through point $A$, which intersects the circles $\omega_1$ and $\omega_2$ for the second time at points $X$ and $Y$, respectively. Prove that $CX = CY$.
(Oleksii Masalitin)
A triangle $ ABC$ has an obtuse angle at $ B$. The perpindicular at $ B$ to $ AB$ meets $ AC$ at $ D$, and $ |CD| \equal{} |AB|$.
Prove that $ |AD|^2 \equal{} |AB|.|BC|$ if and only if $ \angle CBD \equal{} 30^\circ$.
Let $ABC$ be an acute triangle with $AB \neq AC$ . Also let $M$ be the midpoint of the side $BC$ , $H$ the orthocenter of the triangle $ABC$ , $O_1$ the midpoint of the segment $AH$ and $O_2$ the center of the circumscribed circle of the triangle $BCH$ . Prove that $O_1AMO_2$ is a parallelogram .
Given is the convex quadrilateral $ ABCD$. Assume that there exists a point $ P$ inside the quadrilateral for which the triangles $ ABP$ and $ CDP$ are both isosceles right triangles with the right angle at the common vertex $ P$. Prove that there exists a point $ Q$ for which the triangles $ BCQ$ and $ ADQ$ are also isosceles right triangles with the right angle at the common vertex $ Q$.
Let $ABCDEF$ be a convex hexagon such that $BCEF$ is a parallelogram and $ABF$ an equilateral triangle. Given that $BC = 1, AD = 3, CD+DE = 2$, compute the area of $ABCDEF$
Let $ABC$ be a triangle. The incircle of $ABC$ touches $BC,AC,AB$ at $D,E,F$, respectively. Each pair of the incircles of triangles $AEF, BDF,CED$ has two pair of common external tangents, one of them being one of the sides of $ABC$. Show that the other three tangents divide triangle $DEF$ into three triangles and three parallelograms.
Let $V_n=\sqrt{F_n^2+F_{n+2}^2}$, where $F_n$ is the Fibonacci sequence
($F_1=F_2=1,F_{n+2}=F_{n+1}+F_{n}$)
Show that $V_n,V_{n+1},V_{n+2}$ are the sides of a triangle with area $1/2$
Fedja used matches to put down the equally long sides of a parallelogram whose vertices are not on a common line. He figures out that exactly 7 or 9 matches, respectively, fit into the diagonals. How many matches compose the parallelogram's perimeter?
Let $s=\frac{AB+BC+AC}{2}$ be half-perimeter of triangle $ABC$. Let $L$ and $N$be a point's on ray's $AB$ and $CB$, for which $AL=CN=s$. Let $K$ is point, symmetric of point $B$ by circumcenter of $ABC$. Prove, that perpendicular from $K$ to $NL$ passes through incenter of $ABC$.
Solution for problem [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=125&t=365714&p=2011659#p2011659]here[/url]
Let $G$ be the barycenter of triangle $ABC$. Let $D$ be a point such that $AGDB$ is a parallelogram. Show that $BG \parallel CD$.
In the parallelogram $ABCD$, the diagonal $AC$ touches the incircles of triangles $ABC$ and $ADC$ at $W$ and $Y$ respectively, and the diagonal $BD$ touches the incircles of triangles $BAD$ and $BCD$ at $X$ and $Z$ respectively. Prove that either $W,X, Y$ and $Z$ coincide, or $WXYZ$ is a rectangle.
Let $ABCD$ be a parallelogram which is not a rectangle and $E$ be the point in its plane such that $AE \perp AB$ and $CE \perp CB$. Prove that $\angle DEA = \angle CEB$.
Point $P$ lies inside of parallelogram $ABCD$. Perpendicular lines to $PA,PB,PC$ and $PD$ through $A,B,C$ and $D$ construct convex quadrilateral $XYZT$. Prove that $S_{XYZT}\geq 2S_{ABCD}$.
[i]Proposed by Siamak Ahmadpour[/i]
Let $M$ be the intersection of diagonals of the convex quadrilateral $ABCD$, where $m(\widehat{AMB})=60^\circ$. Let the points $O_1$, $O_2$, $O_3$, $O_4$ be the circumcenters of the triangles $ABM$, $BCM$, $CDM$, $DAM$, respectively. What is $Area(ABCD)/Area(O_1O_2O_3O_4)$?
$
\textbf{(A)}\ \dfrac 12
\qquad\textbf{(B)}\ \dfrac 32
\qquad\textbf{(C)}\ \dfrac {\sqrt 3}2
\qquad\textbf{(D)}\ \dfrac {1+2\sqrt 3}2
\qquad\textbf{(E)}\ \dfrac {1+\sqrt 3}2
$
The bisectors $BB_1$ and $CC_1$ of the acute triangle $ABC$ intersect in point $I$. On the extensions of the segments $BB_1$ and $CC_1$, the points $B'$ and $C'$ are marked, respectively So, the quadrilateral $AB'IC'$ is a parallelogram. Prove that if $\angle BAC = 60^o$, then the straight line $B'C'$ passes through the intersection point of the circumscribed circles of the triangles $BC_1B'$ and $CB_1C'$.
In the acute-angled triangle $ABC$ all angles are greater than $45^o$. Let $AM$ and $BN$ be the heights of this triangle and let $X$ and $Y$ be the points on $MA$ and $NB$, respecively, such that $|MX| =|MB|$ and $|NY| =|NA|$. Prove that $MN$ and $XY$ are parallel.