Found problems: 1049
Let $ABC$ be an acute triangle with $AB<AC$, and let $D$ and $E$ be points on side $BC$ such that $BD=CE$ and $D$ lies between $B$ and $E$. Suppose there exists a point $P$ inside $ABC$ such that $PD\parallel AE$ and $\angle PAB=\angle EAC$. Prove that $\angle PBA=\angle PCA$.
[i]Calvin Deng.[/i]
Let $ABC$ be an acute triangle with $AB<AC$, and let $D$ and $E$ be points on side $BC$ such that $BD=CE$ and $D$ lies between $B$ and $E$. Suppose there exists a point $P$ inside $ABC$ such that $PD\parallel AE$ and $\angle PAB=\angle EAC$. Prove that $\angle PBA=\angle PCA$.
[i]Calvin Deng.[/i]
Consider a parallelogram $ABCD$ such that the midpoint $M$ of the side $CD$ lies on the angle bisector of $\angle BAD$. Show that $\angle AMB$ is a right angle.
$ABC$ is an acute angle triangle such that $AB>AC$ and $\hat{BAC}=60^{\circ}$. Let's denote by $O$ the center of the circumscribed circle of the triangle and $H$ the intersection of altitudes of this triangle. Line $OH$ intersects $AB$ in point $P$ and $AC$ in point $Q$. Find the value of the ration $\frac{PO}{HQ}$.
A regular hexagon with side length $1$ is given. Using a ruler construct points in such a way that among the given and constructed points there are two such points that the distance between them is $\sqrt7$.
Notes: ''Using a ruler construct points $\ldots$'' means: Newly constructed points arise only as the intersection of straight lines connecting two points that are given or already constructed. In particular, no length can be measured by the ruler.
In triangle $ABC$, the altitude through $B$ intersects $AC$ at $E$ and the altitude through $C$ intersects $AB$ at $F$. Point $T$ is such that $AETF$ is a parallelogram and points $ A$ ,$T$ lie on different half-planes wrt the line $EF$. Point $D$ is such that $ABDC$ is a parallelogram and points $ A$ ,$D$ lie in different half-planes wrt line $BC$. Prove that $T, D$ and the orthocenter of $ABC$ are collinear.
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
In a $4 \times 6$ grid, all edges and diagonals are drawn (see attachment). Determine the number of parallelograms in the grid that uses only the line segments drawn and none of its four angles are right.
Let $K, L, M, N$ be the midpoints of sides $AB, BC, CD, DA$ of a cyclic quadrilateral $ABCD$. Prove that the orthocentres of triangles $ANK, BKL, CLM, DMN$ are the vertices of a parallelogram.
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.
The point $O$ is situated inside the parallelogram $ABCD$ such that $\angle AOB+\angle COD=180^{\circ}$. Prove that $\angle OBC=\angle ODC$.
$A, B, C$ are collinear with $B$ betweeen $A$ and $C$. $K_{1}$ is the circle with diameter $AB$, and $K_{2}$ is the circle with diameter $BC$. Another circle touches $AC$ at $B$ and meets $K_{1}$ again at $P$ and $K_{2}$ again at $Q$. The line $PQ$ meets $K_{1}$ again at $R$ and $K_{2}$ again at $S$. Show that the lines $AR$ and $CS$ meet on the perpendicular to $AC$ at $B$.
Let $ A,B,C,$ and $ D$ denote four points in space and $ AB$ the distance between $ A$ and $ B$, and so on. Show that \[ AC^2\plus{}BD^2\plus{}AD^2\plus{}BC^2 \ge AB^2\plus{}CD^2.\]
Let $ABC$ be a triangle. The incircle of $ABC$ touches the sides $AB$ and $AC$ at the points $Z$ and $Y$, respectively. Let $G$ be the point where the lines $BY$ and $CZ$ meet, and let $R$ and $S$ be points such that the two quadrilaterals $BCYR$ and $BCSZ$ are parallelogram.
Prove that $GR=GS$.
[i]Proposed by Hossein Karke Abadi, Iran[/i]
[b]8.[/b] Among all point lattices on the plane intersecting every closed convex region of unit width, which on's fundamental parallelogram has the largest area? ([b]G.36[/b])
[L. Fejes-Tóth]
Variable points $M$ and $N$ are considered on the arms $\left[ OX \right.$ and $\left[ OY \right.$ , respectively, of an angle $XOY$ so that $\left| OM \right|+\left| ON \right|$ is constant. Determine the locus of the midpoint of $\left[ MN \right]$.
Let $ABC$ be an acute triangle with circumcenter $O$, on the sides $BC, CA$ and $AB$ they take the points $D, E$ and $F$, respectively, in such a way that $BDEF$ is a parallelogram. Supposing that $DF^2 = AE\cdot EC <\frac{AC^2}{4}$ show that the circles circumscribed to the triangles $FBD$ and $AOC$ are tangent.
Let $ABCD$ be a convex quadrilateral such that $BC=AD$. Let $M$ and $N$ be the midpoints of $AB$ and $CD$, respectively. The lines $AD$ and $BC$ meet the line $MN$ at $P$ and $Q$, respectively. Prove that $CQ=DP$.
Let $ABCD$ be a cyclic quadrilateral in which internal angle bisectors $\angle ABC$ and $\angle ADC$ intersect on diagonal $AC$. Let $M$ be the midpoint of $AC$. Line parallel to $BC$ which passes through $D$ cuts $BM$ at $E$ and circle $ABCD$ in $F$ ($F \neq D$ ). Prove that $BCEF$ is parallelogram
[i]Proposed by Steve Dinh[/i]
The incircle of $ABC$ touches $BC$, $CA$, $AB$ at $K$, $L$, $M$ respectively. The line through $A$ parallel to $LK$ meets $MK$ at $P$, and the line through $A$ parallel to $MK$ meets $LK$ at $Q$. Show that the line $PQ$ bisects $AB$ and bisects $AC$.
Point $D$ lies inside triangle $ABC$ such that $\angle DAC = \angle DCA = 30^{\circ}$ and $\angle DBA = 60^{\circ}$. Point $E$ is the midpoint of segment $BC$. Point $F$ lies on segment $AC$ with $AF = 2FC$. Prove that $DE \perp EF$.
In a convex quadrilateral $ABCD$ with $\angle ABC = \angle ADC = 135^\circ$, points $M$ and $N$ are taken on the rays $AB$ and $AD$ respectively such that $\angle MCD = \angle NCB = 90^\circ$. The circumcircles of triangles $AMN$ and $ABD$ intersect at $A$ and $K$. Prove that $AK \perp KC.$
In a triangle $ABC$, let $H$ be the point of intersection of altitudes, $I$ the center of incircle, $O$ the center of circumcircle, $K$ the point where the incircle touches $BC$. Given that $IO$ is parallel to $BC$, prove that $AO$ is parallel to $HK$.
Let $ABCD$ be a parallelogram. Let $X$ and $Y$ be arbitrary points on sides $BC$ and $CD$, respectively. Segments $BY$ and $DX$ intersect at $P$.
Prove that the line going through the midpoints of segments $BD$ and $XY$ is either parallel to or coincides with line $AP$.
Take any point $P_1$ on the side $BC$ of a triangle $ABC$ and draw the following chain of lines: $P_1P_2$ parallel to $AC$; $P_2P_3$ parallel to $BC$; $P_3P_4$ parallel to $AB$ ; $P_4P_5$ parallel to $CA$; and $P_5P_6$ parallel to $BC$, Here, $P_2,P_5$ lie on $AB$; $P_3,P_6$ lie on $CA$ and $P_4$ on $BC$> Show that $P_6P_1$ is parallel to $AB$.