Found problems: 1049
Let $ABCD$ be a parallelogram. Suppose that there exists a point $P$ in the interior of the parallelogram which is on the perpendicular bisector of $AB$ and such that $\angle PBA = \angle ADP$
Show that $\angle CPD = 2 \angle BAP$
Let $ABCD$ be a parallelogram with $\angle ABC = 105^\circ$. Inside the parallelogram, there is a point $E$ such that triangle $BEC$ is equilateral and $\angle CED = 135^\circ$. Let $K$ be the midpoint of side $AB$. Determine the measure of angle $\angle BKC$.
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]
Consider an octagon with equal angles and with rational sides. Prove that it has a center of symmetry.
Consider a circle $S$, and a point $P$ outside it. The tangent lines from $P$ meet $S$ at $A$ and $B$, respectively. Let $M$ be the midpoint of $AB$. The perpendicular bisector of $AM$ meets $S$ in a point $C$ lying inside the triangle $ABP$. $AC$ intersects $PM$ at $G$, and $PM$ meets $S$ in a point $D$ lying outside the triangle $ABP$. If $BD$ is parallel to $AC$, show that $G$ is the centroid of the triangle $ABP$.
[i]Arnoldo Aguilar (El Salvador)[/i]
Let $ABCD$ be a parallelogram of center $O$. Points $M$ and $N$ are the midpoints of $BO$ and $CD$, respectively. Prove that if the triangles $ABC$ and $AMN$ are similar, then $ABCD$ is a square.
A point $B$ lies on a chord $AC$ of circle $\omega.$ Segments $AB$ and $BC$ are diameters of circles $\omega_1$ and $\omega_2$ centered at $O_1$ and $O_2$ respectively. These circles intersect $\omega$ for the second time in points $D$ and $E$ respectively. The rays $O_1D$ and $O_2E$ meet in a point $F,$ and the rays $AD$ and $CE$ do in a point $G.$ Prove that the line $FG$ passes through the midpoint of the segment $AC.$
(a) A regular $4k$-gon is cut into parallelograms. Prove that among these there are at least $k$ rectangles.
(b) Find the total area of the rectangles in (a) if the lengths of the sides of the $4k$-gon equal $a$.
(VV Proizvolov, Moscow)
Let $ABC$ be an arbitrary triangle. On the perpendicular bisector of $AB$, there is a point $P$ inside of triangle $ABC$. On the sides $BC$ and $CA$, triangles $BQC$ and $CRA$ are placed externally. These triangles satisfy $\vartriangle BPA \sim \vartriangle BQC \sim \vartriangle CRA$. (So $Q$ and $A$ lie on opposite sides of $BC$, and $R$ and $B$ lie on opposite sides of $AC$.) Show that the points $P, Q, C$ and $R$ form a parallelogram.
A convex quadrilateral intersects a circle at points $A_1,A_2,B_1,B_2,C_1,C_2,D_1,$ and $D_2$. (Note that for some letter $N$, points $N_1$ and $N_2$ are on one side of the quadrilateral. Also, the points lie in that specific order on the circle.) Prove that if $A_1B_2=B_1C_2=C_1D_2= D_1A_2$, then quadrilateral formed by these four segments is cyclic.
Given a parallelogram $ABCD$. The line perpendicular to $AC$ passing through $C$ and the line perpendicular to $BD$ passing through $A$ intersect at point $P$. The circle centered at point $P$ and radius $PC$ intersects the line $BC$ at point $X$, ($X \ne C$) and the line $DC$ at point $Y$ , ($Y \ne C$). Prove that the line $AX$ passes through the point $Y$ .
A triangle $ T$ is contained inside a point-symmetrical polygon $ M.$ The triangle $ T'$ is the mirror image of the triangle $ T$ with the reflection at one point $ P$, which inside the triangle $ T$ lies. Prove that at least one of the vertices of the triangle $ T'$ lies in inside or on the boundary of the polygon $ M.$
Prove that a triangle inscribed in a parallelogram has at most half the area of the parallelogram.
Let $ABCD$ be a cyclic quadrilateral and let $H_{A}, H_{B}, H_{C}, H_{D}$ be the orthocenters of the triangles $BCD$, $CDA$, $DAB$ and $ABC$ respectively. Show that the quadrilaterals $ABCD$ and $H_{A}H_{B}H_{C}H_{D}$ are congruent.
A convex polygon is partitioned into parallelograms. A vertex of the polygon is called [i]good[/i] if it belongs to exactly one parallelogram. Prove that there are more than two good vertices.
As shown in the figure, quadrilateral $ ABCD$ is inscribed in a circle with $ AC$ as its diameter, $ BD \perp AC,$ and $ E$ the intersection of $ AC$ and $ BD.$ Extend line segment $ DA$ and $ BA$ through $ A$ to $ F$ and $ G$ respectively, such that $ DG \parallel{} BF.$ Extend $ GF$ to $ H$ such that $ CH \perp GH.$ Prove that points $ B, E, F$ and $ H$ lie on one circle.
[asy]
defaultpen(linewidth(0.8)+fontsize(10));size(150);
real a=4, b=6.5, c=9, d=a*c/b, g=14, f=sqrt(a^2+b^2)*sqrt(a^2+d^2)/g;
pair E=origin, A=(0,a), B=(-b,0), C=(0,-c), D=(d,0), G=A+g*dir(B--A), F=A+f*dir(D--A), M=midpoint(G--C);
path c1=circumcircle(A,B,C), c2=Circle(M, abs(M-G));
pair Hf=F+10*dir(G--F), H=intersectionpoint(F--Hf, c2);
dot(A^^B^^C^^D^^E^^F^^G^^H);
draw(c1^^c2^^G--D--C--A--G--F--D--B--A^^F--H--C--B--F);
draw(H--B^^F--E^^G--C, linetype("2 2"));
pair point= E;
label("$A$", A, dir(point--A));
label("$B$", B, dir(point--B));
label("$C$", C, dir(point--C));
label("$D$", D, dir(point--D));
label("$F$", F, dir(point--F));
label("$G$", G, dir(point--G));
label("$H$", H, dir(point--H));
label("$E$", E, NE);[/asy]
Let $ABCD$ be a parallelogram $.$ Let $O$ be a point in its interior such that $\angle AOB + \angle DOC = 180^{\circ} . $
Show that $,\angle ODC = \angle OBC . $
$ AA_{3}$ and $ BB_{3}$ are altitudes of acute-angled $ \triangle ABC$. Points $ A_{1}$ and $ B_{1}$ are second points of intersection lines $ AA_{3}$ and $ BB_{3}$ with circumcircle of $ \triangle ABC$ respectively. $ A_{2}$ and $ B_{2}$ are points on $ BC$ and $ AC$ respectively. $ A_{1}A_{2}\parallel AC$, $ B_{1}B_{2}\parallel BC$. Point $ M$ is midpoint of $ A_{2}B_{2}$. $ \angle BCA \equal{} x$. Find $ \angle A_{3}MB_{3}$.
Starting from a given cyclic quadrilateral $\mathcal{Q}_0$, a sequence of quadrilaterals is constructed so that $\mathcal{Q}_{k + 1}$ is the circumscribed quadrilateral of $\mathcal{Q}_k$ for $k = 0,1,\dots$. The sequence terminates when a quadrilateral is reached that is not cyclic. (The circumscribed quadrilateral of a cylic quadrilateral $ABCD$ has sides that are tangent to the circumcircle of $ABCD$ at $A$, $B$, $C$ and $D$.) Prove that the sequence always terminates, except when $\mathcal{Q}_0$ is a square.
In a trapezoid $ABCD$ with a base $AD$, point $L$ is the orthogonal projection of $C$ on $AB$, and $K$ is the point on $BC$ such that $AK$ is perpendicular to $AD$. Let $O$ be the circumcenter of triangle $ACD$. Suppose that the lines $AK , CL$ and $DO$ have a common point. Prove that $ABCD$ is a parallelogram.
In plane,let a circle $(O)$ and two fixed points $B,C$ lies in $(O)$
such that $BC$ not is the diameter.Consider a point $A$ varies in
$(O)$ such that $A\neq B,C$ and $AB\neq AC$.Call $D$ and $E$
respective is intersect of $BC$ and internal and external bisector
of $\widehat{BAC}$,$I$ is midpoint of $DE$.The line that pass through
orthocenter of $\triangle ABC$
and perpendicular with $AI$ intersects $AD,AE$ respective at $M,N$.
1/Prove that $MN$ pass through a fixed point
2/Determint the place of $A$ such that $S_{AMN}$ has maxium value
$m$ and $n$ are two nonnegative integers. In the Philosopher's Chess, The chessboard is an infinite grid of identical regular hexagons and a new piece named the Donkey moves on it as follows:
Starting from one of the hexagons, the Donkey moves $m$ cells in one of the $6$ directions, then it turns $60$ degrees clockwise and after that moves $n$ cells in this new direction until it reaches it's final cell.
At most how many cells are in the Philosopher's chessboard such that one cannot go from anyone of them to the other with a finite number of movements of the Donkey?
[i]Proposed by Shayan Dashmiz[/i]
Consider an octogon with equal angles and rational side lengths. Prove that it has a symmetry center.
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.
In parallelogram $ABCD$ with acute angle $A$ a point $N$ is chosen on the segment $AD$, and a point $M$ on the segment $CN$ so that $AB = BM = CM$. Point $K$ is the reflection of $N$ in line $MD$. The line $MK$ meets the segment $AD$ at point $L$. Let $P$ be the common point of the circumcircles of $AMD$ and $CNK$ such that $A$ and $P$ share the same side of the line $MK$. Prove that $\angle CPM = \angle DPL$.