Found problems: 844
$ABCD$ 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.
Find the smallest $n \in \mathbb{N}$ such that if any 5 vertices of a regular $n$-gon are colored red, there exists a line of symmetry $l$ of the $n$-gon such that every red point is reflected across $l$ to a non-red point.
Given trapezoid $ ABCD$ with parallel sides $ AB$ and $ CD$, assume that there exist points $ E$ on line $ BC$ outside segment $ BC$, and $ F$ inside segment $ AD$ such that $ \angle DAE \equal{} \angle CBF$. Denote by $ I$ the point of intersection of $ CD$ and $ EF$, and by $ J$ the point of intersection of $ AB$ and $ EF$. Let $ K$ be the midpoint of segment $ EF$, assume it does not lie on line $ AB$. Prove that $ I$ belongs to the circumcircle of $ ABK$ if and only if $ K$ belongs to the circumcircle of $ CDJ$.
[i]Proposed by Charles Leytem, Luxembourg[/i]
Let $ABC$ be a triangle where $\angle BAC = 30^\circ$. Construct $D$ in $\triangle ABC$ such that $\angle ABD =
\angle ACD = 30^\circ$. Let the circumcircle of $\triangle ABD$ intersect $AC$ at $X$. Let the circumcircle of $\triangle ACD$ intersect $AB$ at $Y$. Given that $DB - DC = 10$ and $BC = 20$, find $AX \cdot AY$.
Let $ABCD$ be a trapezoid, with $AB \parallel CD$ (the vertices are listed in cyclic order). The diagonals of this trapezoid are perpendicular to one another and intersect at $O$. The base angles $\angle DAB$ and $\angle CBA$ are both acute. A point $M$ on the line sgement $OA$ is such that $\angle BMD = 90^o$, and a point $N$ on the line segment $OB$ is such that $\angle ANC = 90^o$. Prove that triangles $OMN$ and $OBA$ are similar.
Let $ABCD$ be a trapezium in which $AB$ is parallel to $CD$. The circles on $AD$ and $BC$ as diameters intersect at two distinct points $P$ and $Q$. Prove that the lines $PQ,AC,BD$ are concurrent.
Let $E$ be a point outside the square $ABCD$ such that $m(\widehat{BEC})=90^{\circ}$, $F\in [CE]$, $[AF]\perp [CE]$, $|AB|=25$, and $|BE|=7$. What is $|AF|$?
$
\textbf{(A)}\ 29
\qquad\textbf{(B)}\ 30
\qquad\textbf{(C)}\ 31
\qquad\textbf{(D)}\ 32
\qquad\textbf{(E)}\ 33
$
Let $ ABCD$ be a trapezoid with parallel sides $ AB > CD$. Points $ K$ and $ L$ lie on the line segments $ AB$ and $ CD$, respectively, so that $AK/KB=DL/LC$. Suppose that there are points $ P$ and $ Q$ on the line segment $ KL$ satisfying \[\angle{APB} \equal{} \angle{BCD}\qquad\text{and}\qquad \angle{CQD} \equal{} \angle{ABC}.\] Prove that the points $ P$, $ Q$, $ B$ and $ C$ are concyclic.
[i]Proposed by Vyacheslev Yasinskiy, Ukraine[/i]
Let $ABCD$ be a trapezoid with $AB\parallel CD$ and $\angle D=90^\circ$. Suppose that there is a point $E$ on $CD$ such that $AE=BE$ and that triangles $AED$ and $CEB$ are similar, but not congruent. Given that $\tfrac{CD}{AB}=2014$, find $\tfrac{BC}{AD}$.
$AB=CD,AD \parallel BC$ and $AD>BC$. $\Omega$ is circumcircle of $ABCD$. Point $E$ is on $\Omega$ such that $BE \perp AD$. Prove that $AE+BC>DE$
Given is a trapezoid $ ABCD$ where $ AB$ and $ CD$ are parallel, and $ A,B,C,D$ are clockwise in this order. Let $ \Gamma_1$ be the circle with center $ A$ passing through $ B$, $ \Gamma_2$ be the circle with center $ C$ passing through $ D$. The intersection of line $ BD$ and $ \Gamma_1$ is $ P$ $ ( \ne B,D)$. Denote by $ \Gamma$ the circle with diameter $ PD$, and let $ \Gamma$ and $ \Gamma_1$ meet at $ X$$ ( \ne P)$. $ \Gamma$ and $ \Gamma_2$ meet at $ Y$. If the circumcircle of triangle $ XBY$ and $ \Gamma_2$ meet at $ Q$, prove that $ B,D,Q$ are collinear.
Let $P_1$ be a regular $n$-gon, where $n\in\mathbb{N}$. We construct $P_2$ as the regular $n$-gon whose vertices are the midpoints of the edges of $P_1$. Continuing analogously, we obtain regular $n$-gons $P_3,P_4,\ldots ,P_m$. For $m\ge n^2-n+1$, find the maximum number $k$ such that for any colouring of vertices of $P_1,\ldots ,P_m$ in $k$ colours there exists an isosceles trapezium $ABCD$ whose vertices $A,B,C,D$ have the same colour.
[i]Radu Ignat[/i]
In the $xy$-plane, consider the L-shaped region bounded by horizontal and vertical segments with vertices at $(0,0), (0,3), (3,3), (3,1), (5,1)$ and $(5,0)$. The slope of the line through the origin that divides the area of this region exactly in half is
[asy]
size(200);
Label l;
l.p=fontsize(6);
xaxis("$x$",0,6,Ticks(l,1.0,0.5),EndArrow);
yaxis("$y$",0,4,Ticks(l,1.0,0.5),EndArrow);
draw((0,3)--(3,3)--(3,1)--(5,1)--(5,0)--(0,0)--cycle,black+linewidth(2));[/asy]
$ \textbf{(A)}\ \frac{2}{7} \qquad\textbf{(B)}\ \frac{1}{3} \qquad\textbf{(C)}\ \frac{2}{3} \qquad\textbf{(D)}\ \frac{3}{4} \qquad\textbf{(E)}\ \frac{7}{9} $
Circles $C_1$ and $C_2$ are externally tangent, and they are both internally tangent to circle $C_3$. The radii of $C_1$ and $C_2$ are $4$ and $10$, respectively, and the centers of the three circles are all collinear. A chord of $C_3$ is also a common external tangent of $C_1$ and $C_2$. Given that the length of the chord is $\frac{m\sqrt{n}}{p}$ where $m,n,$ and $p$ are positive integers, $m$ and $p$ are relatively prime, and $n$ is not divisible by the square of any prime, find $m+n+p$.
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.
Isosceles trapezoid $ ABCD$, with $ AB \parallel CD$, is such that there exists a circle $ \Gamma$ tangent to its four sides. Let $ T \equal{} \Gamma \cap BC$, and $ P \equal{} \Gamma \cap AT$ ($ P \neq T$).
If $ \frac{AP}{AT} \equal{} \frac{2}{5}$, compute $ \frac{AB}{CD}$.
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.
Triangle $ABC$ is inscribed in circle $O$. Tangent $PD$ is drawn from $A$, $D$ is on ray $BC$, $P$ is on ray $DA$. Line $PU$ ($U \in BD$) intersects circle $O$ at $Q$, $T$, and intersect $AB$ and $AC$ at $R$ and $S$ respectively. Prove that if $QR=ST$, then $PQ=UT$.
Each cell of a beehive is constructed from a right regular 6-angled prism, open at the bottom and closed on the top by a regular 3-sided pyramidical mantle. The edges of this pyramid are connected to three of the rising edges of the prism and its apex $T$ is on the perpendicular line through the center $O$ of the base of the prism (see figure). Let $s$ denote the side of the base, $h$ the height of the cell and $\theta$ the angle between the line $TO$ and $TV$.
(a) Prove that the surface of the cell consists of 6 congruent trapezoids and 3 congruent rhombi.
(b) the total surface area of the cell is given by the formula $6sh - \dfrac{9s^2}{2\tan\theta} + \dfrac{s^2 3\sqrt{3}}{2\sin\theta}$
[img]http://www.mathlinks.ro/Forum/album_pic.php?pic_id=286[/img]
A rectangle has sides of length $a$ and $36$. A hinge is installed at each vertex of the rectangle and at the midpoint of each side of length $36$. The sides of length $a$ can be pressed toward each other keeping those two sides parallel so the rectangle becomes a convex hexagon as shown. When the figure is a hexagon with the sides of length $a$ parallel and separated by a distance of $24,$ the hexagon has the same area as the original rectangle. Find $a^2$.
[asy]
pair A,B,C,D,E,F,R,S,T,X,Y,Z;
dotfactor = 2;
unitsize(.1cm);
A = (0,0);
B = (0,18);
C = (0,36);
// don't look here
D = (12*2.236, 36);
E = (12*2.236, 18);
F = (12*2.236, 0);
draw(A--B--C--D--E--F--cycle);
dot(" ",A,NW);
dot(" ",B,NW);
dot(" ",C,NW);
dot(" ",D,NW);
dot(" ",E,NW);
dot(" ",F,NW);
//don't look here
R = (12*2.236 +22,0);
S = (12*2.236 + 22 - 13.4164,12);
T = (12*2.236 + 22,24);
X = (12*4.472+ 22,24);
Y = (12*4.472+ 22 + 13.4164,12);
Z = (12*4.472+ 22,0);
draw(R--S--T--X--Y--Z--cycle);
dot(" ",R,NW);
dot(" ",S,NW);
dot(" ",T,NW);
dot(" ",X,NW);
dot(" ",Y,NW);
dot(" ",Z,NW);
// sqrt180 = 13.4164
// sqrt5 = 2.236
[/asy]
In trapezoid $ABCD$ angles $A$ and $B$ are right, $AB = AD, CD = BC + AD, BC < AD$. Prove that $\angle ADC = 2\angle ABE$, where $E$ is the midpoint of segment $AD$.
(V. Yasinsky)
Let $ A,B,C$ be collinear points on the plane with $ B$ between $ A$ and $ C$. Equilateral triangles $ ABD,BCE,CAF$ are constructed with $ D,E$ on one side of the line $ AC$ and $ F$ on the other side. Prove that the centroids of the triangles are the vertices of an equilateral triangle, and that the centroid of this triangle lies on the line $ AC$.
Let $ABCD$ be an isosceles trapezoid with $AD\parallel BC$. Points $P$ and $Q$ are placed on segments $\overline{CD}$ and $\overline{DA}$ respectively such that $AP\perp CD$ and $BQ\perp DA$, and point $X$ is the intersection of these two altitudes. Suppose that $BX=3$ and $XQ=1$. Compute the largest possible area of $ABCD$.
Points $A$ and $B$ are on a circle $\omega$ with center $O$ such that $\tfrac{\pi}{3}< \angle AOB <\tfrac{2\pi}{3}$. Let $C$ be the circumcenter of the triangle $AOB$. Let $l$ be a line passing through $C$ such that the angle between $l$ and the segment $OC$ is $\tfrac{\pi}{3}$. $l$ cuts tangents in $A$ and $B$ to $\omega$ in $M$ and $N$ respectively. Suppose circumcircles of triangles $CAM$ and $CBN$, cut $\omega$ again in $Q$ and $R$ respectively and theirselves in $P$ (other than $C$). Prove that $OP\perp QR$.
[i]Proposed by Mehdi E'tesami Fard, Ali Khezeli[/i]
Let $ABCD$ be an isosceles trapezoid with bases $AD> BC$. Let $X$ be the intersection of the bisectors of $\angle BAC$ and $BC$. Let $E$ be the intersection of$ DB$ with the parallel to the bisector of $\angle CBD$ through $X$ and let $F$ be the intersection of $DC$ with the parallel to the bisector of $\angle DCB$ through $X$. Show that quadrilateral $AEFD$ is cyclic.