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 $Y$ .

It is known that $ABCD$ is a parallelogram. The point $E$ is taken so that $BCED$ is a cyclic quadrilateral. Let $\ell$ be a line that passes through $A$, intersects the segment $DC$ at point $F$ and intersects the extension of the line $BC$ at $G$. Given $EF = EG = EC$. Prove that $\ell$ is the bisector of the angle $\angle BAD$.

Let an acute-angled triangle $ABC$ with $AB<AC<BC$, inscribed in circle $c(O,R)$. The angle bisector $AD$ meets $c(O,R)$ at $K$. The circle $c_1(O_1,R_1)$(which passes from $A,D$ and has its center $O_1$ on $OA$) meets $AB$ at $E$ and $AC$ at $Z$. If $M,N$ are the midpoints of $ZC$ and $BE$ respectively, prove that: [b]a)[/b]the lines $ZE,DM,KC$ are concurrent at one point $T$. [b]b)[/b]the lines $ZE,DN,KB$ are concurrent at one point $X$. [b]c)[/b]$OK$ is the perpendicular bisector of $TX$.

Given is trapezoid $ABCD$, $M$ and $N$ being the midpoints of the bases of $AD$ and $BC$, respectively. a) Prove that the trapezoid is isosceles if it is known that the intersection point of perpendicular bisectors of the lateral sides belongs to the segment $MN$. b) Does the statement of point a) remain true if it is only known that the intersection point of perpendicular bisectors of the lateral sides belongs to the line $MN$?

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]

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 be a point $ P $ in the interior of a parallelogram $ ABCD $ such that $ \angle PAD=\angle PCD. $ Prove that the bisectors of $ \angle BAD $ and $ \angle BPD $ are parallel.

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

The points of intersection of $ xy \equal{} 12$ and $ x^2 \plus{} y^2 \equal{} 25$ are joined in succession. The resulting figure is: $ \textbf{(A)}\ \text{a straight line} \qquad\textbf{(B)}\ \text{an equilateral triangle} \qquad\textbf{(C)}\ \text{a parallelogram}$ $ \textbf{(D)}\ \text{a rectangle} \qquad\textbf{(E)}\ \text{a square}$

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 $DE \perp EF$.

$ABCD$ is a parallelogram. A point $M$ is found on the side $AB$ or its extension such that $\angle MAD = \angle AMO$ where $O$ is the intersection point of the diagonals of the parallelogram. Prove that $MD = MG$. (M Smurov)

Two circles intersect at two points $A$ and $B$. A line $\ell$ which passes through the point $A$ meets the two circles again at the points $C$ and $D$, respectively. Let $M$ and $N$ be the midpoints of the arcs $BC$ and $BD$ (which do not contain the point $A$) on the respective circles. Let $K$ be the midpoint of the segment $CD$. Prove that $\measuredangle MKN = 90^{\circ}$.

Let $O$ be the intersection point of the diagonals of a parallelogram $ABCD$ . Prove that if the line $BC$ is tangent to the circle passing through the points $A, B$, and $O$, then the line $CD$ is tangent to the circle passing through the points $B, C$ and $O$. (A Zaslavskiy)

Given integer $n > 1$ and real number $t \geq 1$. $P$ is a parallelogram with four vertices $(0, 0), (0, t), (tF_{2n+1}, tF_{2n}), (tF_{2n+1}, tF_{2n} + t)$. Here, ${F_n}$ is the $n$-th term of Fibonacci sequence defined by $F_0 = 0, F_1 = 1$ and $F_{m+1} = F_m + F_{m-1}$. Let $L$ be the number of integral points (whose coordinates are integers) interior to $P$, and $M$ be the area of $P$, which is $t^2F_{2n+1}.$ [b][i]i)[/i][/b] Prove that for any integral point $(a, b)$, there exists a unique pair of integers $(j, k)$ such that$ j(F_{n+1}, F_n) + k(F_n, F_{n-1}) = (a, b)$, that is,$ jF_{n+1} + kF_n = a$ and $jF_n + kF_{n-1} = b.$ [i][b]ii)[/b][/i] Using [i][b]i)[/b][/i] or not, prove that $|\sqrt L-\sqrt M| \leq \sqrt 2.$

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.

On side, $BC, AB$ of a parallelogram $ABCD$ lie points $M,N$ respectively such that $|AM| =|CN|$. Let $P$ be the intersection of $AM$ and $CN$. Prove that the angle bisector of $\angle APC$ passes through $D$.

The parallelogram $ABCD$ isn't a diamond. The ratio of the diagonal lengths $|AC|/|BD|$ equals to $k$. The $[AM)$ ray is symmetric to the $[AD)$ ray with respect to the $(AC)$ line. The $[BM)$ ray is symmetric to the $[BC)$ ray with respect to the $(BD)$ line. ($M$ point is those rays intersection.) Find the ratio $|AM|/|BM|$ .

A white cylindrical silo has a diameter of 30 feet and a height of 80 feet. A red stripe with a horizontal width of 3 feet is painted on the silo, as shown, making two complete revolutions around it. What is the area of the stripe in square feet? [asy] size(250);defaultpen(linewidth(0.8)); draw(ellipse(origin, 3, 1)); fill((3,0)--(3,2)--(-3,2)--(-3,0)--cycle, white); draw((3,0)--(3,16)^^(-3,0)--(-3,16)); draw((0, 15)--(3, 12)^^(0, 16)--(3, 13)); filldraw(ellipse((0, 16), 3, 1), white, black); draw((-3,11)--(3, 5)^^(-3,10)--(3, 4)); draw((-3,2)--(0,-1)^^(-3,1)--(-1,-0.89)); draw((0,-1)--(0,15), dashed); draw((3,-2)--(3,-4)^^(-3,-2)--(-3,-4)); draw((-7,0)--(-5,0)^^(-7,16)--(-5,16)); draw((3,-3)--(-3,-3), Arrows(6)); draw((-6,0)--(-6,16), Arrows(6)); draw((-2,9)--(-1,9), Arrows(3)); label("$3$", (-1.375,9.05), dir(260), fontsize(7)); label("$A$", (0,15), N); label("$B$", (0,-1), NE); label("$30$", (0, -3), S); label("$80$", (-6, 8), W);[/asy] $ \textbf{(A)}\; 120\qquad \textbf{(B)}\; 180\qquad \textbf{(C)}\; 240\qquad \textbf{(D)}\; 360\qquad \textbf{(E)}\; 480$

What is the ratio of the area of the shaded square to the area of the large square? (The figure is drawn to scale) [asy] draw((0,0)--(0,4)--(4,4)--(4,0)--cycle); draw((0,0)--(4,4)); draw((0,4)--(3,1)--(3,3)); draw((1,1)--(2,0)--(4,2)); fill((1,1)--(2,0)--(3,1)--(2,2)--cycle,black);[/asy] $ \text{(A)}\ \frac{1}{6}\qquad\text{(B)}\ \frac{1}{7}\qquad\text{(C)}\ \frac{1}{8}\qquad\text{(D)}\ \frac{1}{12}\qquad\text{(E)}\ \frac{1}{16} $

Let $P$ and $P^{\prime }$ be two parallelograms with equal area, and let their sidelengths be $a,$ $b$ and $a^{\prime },$ $b^{\prime }.$ Assume that $a^{\prime }\leq a\leq b\leq b^{\prime },$ and moreover, it is possible to place the segment $b^{\prime }$ such that it completely lies in the interior of the parallelogram $P.$ Show that the parallelogram $P$ can be partitioned into four polygons such that these four polygons can be composed again to form the parallelogram $% P^{\prime }.$

Let $ABC$ and $XYZ$ be two triangles. Define \[A_1=BC\cap ZX, A_2=BC\cap XY,\]\[B_1=CA\cap XY, B_2=CA\cap YZ,\]\[C_1=AB\cap YZ, C_2=AB\cap ZX.\] Hereby, the abbreviation $g\cap h$ means the point of intersection of two lines $g$ and $h$. Prove that $\frac{C_1C_2}{AB}=\frac{A_1A_2}{BC}=\frac{B_1B_2}{CA}$ holds if and only if $\frac{A_1C_2}{XZ}=\frac{C_1B_2}{ZY}=\frac{B_1A_2}{YX}$.

Let $ABC$ be an isosceles right triangle with $\angle BCA=90^{\circ}, BC=AC=10$. Let $P$ be a point on $AB$ that is a distance $x$ from $A$, $Q$ be a point on $AC$ such that $PQ$ is parallel to $BC$. Let $R$ and $S$ be points on $BC$ such that $QR$ is parallel to $AB$ and $PS$ is parallel to $AC$. The union of the quadrilaterals $PBRQ$ and $PSCQ$ determine a shaded area $f(x)$. Evaluate $f(2)$

Given a triangle $ ABC$ with $ \angle A \equal{} 120^{\circ}$. The points $ K$ and $ L$ lie on the sides $ AB$ and $ AC$, respectively. Let $ BKP$ and $ CLQ$ be equilateral triangles constructed outside the triangle $ ABC$. Prove that $ PQ \ge\frac{\sqrt 3}{2}\left(AB \plus{} AC\right)$.

Let $P$ be any point inside the parallelogram $ABCD$ and let $R$ be the radius of the circle through $A$, $B$, and $C$. Show that the distance from $P$ to the nearest vertex is not greater than $R$.