Found problems: 1049
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.
Let $ABC$ be an acute triangle, $(O)$ be the circumcircle, and $AB<AC.$ Let $I$ be the midpoint of arc $BC$ (not containing $A$). $K$ lies on $AC,$ $K\ne C$ such that $IK=IC.$ $BK$ intersects $(O)$ at the second point $D,$ $D\ne B$ and intersects $AI$ at $E.$ $DI$ intersects $AC$ at $F.$
a) Prove that $EF=\frac{BC}{2}.$
b) $M$ lies on $DI$ such that $CM$ is parallel to $AD.$ $KM$ intersects $BC$ at $N.$ The circumcircle of triangle $BKN$ intersects $(O)$ at the second point $P.$ Prove that $PK$ passes through the midpoint of segment $AD.$
Circles $ \omega_{1}$ and $ \omega_{2}$ intersect at points $ A$ and $ B.$ Points $ C$ and $ D$ are on circles $ \omega_{1}$ and $ \omega_{2},$ respectively,
such that lines $ AC$ and $ AD$ are tangent to circles $ \omega_{2}$ and $ \omega_{1},$ respectively.
Let $ I_{1}$ and $ I_{2}$ be the incenters of triangles $ ABC$ and $ ABD,$ respectively. Segments $ I_{1}I_{2}$ and $ AB$ intersect at $ E$.
Prove that: $ \frac {1}{AE} \equal{} \frac {1}{AC} \plus{} \frac {1}{AD}$
Let ABC be a triangle, G its centroid, M the midpoint of AB, D the point on the line $AG$ such that $AG = GD, A \neq D$, E the point on the line $BG$ such that $BG = GE, B \neq E$. Show that the quadrilateral BDCM is cyclic if and only if $AD = BE$.
Consider the parallelogram $ ABCD, $ whose diagonals intersect at $ O. $ The bisector of the angle $ \angle DAC $ and that of $ \angle DBC $ intersect each other at $ T. $ Moreover, $ \overrightarrow{TD} +\overrightarrow{TC} =\overrightarrow{TO} . $
Find the angles of the triangle $ ABT. $
Let a quadrilateral $ABCD$ have an inscribed circle and let $K, L, M, N$ be the tangency points of the sides $AB, BC, CD$ and $DA$, respectively. Consider the orthocenters of each of the triangles $\vartriangle AKN, \vartriangle BLK, \vartriangle CML$ and $\vartriangle DNM$. Prove that these four points are the vertices of a 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.
The sides $AB, BC, CD$ and $DA$ of the convex quadrilateral $ABCD$ have midpoints $E, F, G$ and $H$. Prove that the triangles $EFB, FGC, GHD$ and $HEA$ can be put together into a parallelogram equal to $EFGH$.
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?
$[BD]$ is a median of $\triangle ABC$. $m(\widehat{ABD})=90^\circ$, $|AB|=2$, and $|AC|=6$. $|BC|=?$
$ \textbf{(A)}\ 3
\qquad\textbf{(B)}\ 3\sqrt2
\qquad\textbf{(C)}\ 5
\qquad\textbf{(D)}\ 4\sqrt2
\qquad\textbf{(E)}\ 2\sqrt6
$
Let $ABCD$ be a parallelogram. The interior angle bisector of $\angle ADC$ intersects the line $BC$ in $E$, and the perpendicular bisector of the side $AD$ intersects the line $DE$ in $M$. Let $F= AM \cap BC$. Prove that:
a) $DE=AF$;
b) $AD\cdot AB = DE\cdot DM$.
[i]Daniela and Marius Lobaza, Timisoara[/i]
Let $H$ be the orthocenter of the triangle $ABC$ and $P$ an arbitrary point on circumcircle of triangle. $BH$ meets $AC$ at $E$. $PAQB$ and $PARC$ are two parallelograms and $AQ$ meets $HR$ at $X$. Show that $EX \parallel AP$.
Trapezoid $ ABCD$ has $ AD\parallel{}BC$, $ BD \equal{} 1$, $ \angle DBA \equal{} 23^{\circ}$, and $ \angle BDC \equal{} 46^{\circ}$. The ratio $ BC: AD$ is $ 9: 5$. What is $ CD$?
$ \textbf{(A)}\ \frac {7}{9}\qquad \textbf{(B)}\ \frac {4}{5}\qquad \textbf{(C)}\ \frac {13}{15} \qquad \textbf{(D)}\ \frac {8}{9}\qquad \textbf{(E)}\ \frac {14}{15}$
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]
Let $ ABC$ be a tringle and let $ P$ be an interior point such that $ \angle BPC \equal{} 90 ,\angle BAP \equal{} \angle BCP$.Let $ M,N$ be the mid points of $ AC,BC$ respectively.Suppose $ BP \equal{} 2PM$.Prove that $ A,P,N$ are collinear.
Point $ P$ is taken inside a right angle $ KLM$. A circle $ S_1$ with center $ O_1$ is tangent to the rays $ LK,LP$ of angle $ KLP$ at $ A,D$ respectively. A circle $ S_2$ with center $ O_2$ is tangent to the rays of angle $ MLP$, touching $ LP$ at $ B$. Suppose $ A,B,O_1$ are collinear. Let $ O_2D,KL$ meet at $ C$. Prove that $ BC$ bisects angle $ ABD$.
Let $ABCD$ be a cyclic quadrilateral whose diagonals $AC$ and $BD$ are perpendicular. Let $O$ be the circumcenter of $ABCD$, $K$ the intersection of the diagonals, $ L\neq O $ the intersection of the circles circumscribed to $OAC$ and $OBD$, and $G$ the intersection of the diagonals of the quadrilateral whose vertices are the midpoints of the sides of $ABCD$. Prove that $O, K, L$ and $G$ are collinear
Let $ABCD$ be a convex quadrilateral, let $E$ and $F$ be the midpoints of the sides $AD$ and $BC$, respectively. The segment $CE$ meets $DF$ in $O$. Show that if the lines $AO$ and $BO$ divide the side $CD$ in 3 equal parts, then $ABCD$ is a parallelogram.
Let $ ABCDEF$ be a convex hexagon such that $ AD \equal{} BC \plus{} EF$, $ BE \equal{} AF \plus{} CD$, $ CF \equal{} DE \plus{} AB$. Prove that:
\[ \frac {AB}{DE} \equal{} \frac {CD}{AF} \equal{} \frac {EF}{BC}.
\]
$ABCD$ is a parallelogram in which angle $DAB$ is acute. Points $A, P, B, D$ lie on one circle in exactly this order. Lines $AP$ and $CD$ intersect in $Q$. Point $O$ is the circumcenter of the triangle $CPQ$. Prove that if $D \neq O$ then the lines $AD$ and $DO$ are perpendicular.
Let $I$ be the center of the incircle of non-isosceles triangle $ABC,A_{1}=AI\cap BC$ and $B_{1}=BI\cap AC.$ Let $l_{a}$ be the line through $A_{1}$ which is parallel to $AC$ and $l_{b}$ be the line through $B_{1}$ parallel to $BC.$ Let $l_{a}\cap CI=A_{2}$ and $l_{b}\cap CI=B_{2}.$ Also $N=AA_{2}\cap BB_{2}$ and $M$ is the midpoint of $AB.$ If $CN\parallel IM$ find $\frac{CN}{IM}$.
We are given a convex pentagon $ABCDE$ in the coordinate plane such that $A$, $B$, $C$, $D$, $E$ are lattice points. Let $Q$ denote the convex pentagon bounded by the five diagonals of the pentagon $ABCDE$ (so that the vertices of $Q$ are the interior points of intersection of diagonals of the pentagon $ABCDE$). Prove that there exists a lattice point inside of $Q$ or on the boundary of $Q$.
The area in square units of the region enclosed by parallelogram $ABCD$ is
[asy]
unitsize(24);
pair A,B,C,D;
A=(-1,0); B=(0,2); C=(4,2); D=(3,0);
draw(A--B--C--D); draw((0,-1)--(0,3)); draw((-2,0)--(6,0));
draw((-.25,2.75)--(0,3)--(.25,2.75)); draw((5.75,.25)--(6,0)--(5.75,-.25));
dot(origin); dot(A); dot(B); dot(C); dot(D); label("$y$",(0,3),N); label("$x$",(6,0),E);
label("$(0,0)$",origin,SE); label("$D (3,0)$",D,SE); label("$C (4,2)$",C,NE);
label("$A$",A,SW); label("$B$",B,NW);
[/asy]
$\text{(A)}\ 6 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 15 \qquad \text{(E)}\ 18$
Prove that the lines joining the middle-points of non-adjacent sides of an convex quadrilateral and the line joining the middle-points of diagonals, are concurrent. Prove that the intersection point is the middle point of the three given segments.
If $A, B, C$, and $D$ are four distinct points in space, prove that there is a plane $P$ on which the orthogonal projections of $A, B, C$, and $D$ form a parallelogram (possibly degenerate).