Found problems: 844
The area of the shaded region $\text{BEDC}$ in parallelogram $\text{ABCD}$ is
[asy]
unitsize(10);
pair A,B,C,D,E;
A=origin; B=(4,8); C=(14,8); D=(10,0); E=(4,0);
draw(A--B--C--D--cycle);
fill(B--E--D--C--cycle,gray);
label("A",A,SW); label("B",B,NW); label("C",C,NE); label("D",D,SE); label("E",E,S);
label("$10$",(9,8),N); label("$6$",(7,0),S); label("$8$",(4,4),W);
draw((3,0)--(3,1)--(4,1));
[/asy]
$\text{(A)}\ 24 \qquad \text{(B)}\ 48 \qquad \text{(C)}\ 60 \qquad \text{(D)}\ 64 \qquad \text{(E)}\ 80$
Let $ABCD$ be a trapezoid in which $AB \parallel CD$ and $AB = 2CD$. A line $\ell$ perpendicular to $CD$ was drawn through point $C$. A circle with center at point $D$ and radius $DA$ intersects line $\ell$ at points $P$ and $Q$. Prove that $AP \perp BQ$.
Let $ABC$ and $PQR$ be two triangles such that
[list]
[b](a)[/b] $P$ is the mid-point of $BC$ and $A$ is the midpoint of $QR$.
[b](b)[/b] $QR$ bisects $\angle BAC$ and $BC$ bisects $\angle QPR$
[/list]
Prove that $AB+AC=PQ+PR$.
Given an acute triangle $ PBC$ with $ PB\neq PC.$ Points $ A,D$ lie on $ PB,PC,$ respectively. $ AC$ intersects $ BD$ at point $ O.$ Let $ E,F$ be the feet of perpendiculars from $ O$ to $ AB,CD,$ respectively. Denote by $ M,N$ the midpoints of $ BC,AD.$
$ (1)$: If four points $ A,B,C,D$ lie on one circle, then $ EM\cdot FN \equal{} EN\cdot FM.$
$ (2)$: Determine whether the converse of $ (1)$ is true or not, justify your answer.
Let $ ABCD$ be a trapezoid with $ AB\parallel CD,AB>CD$ and $ \angle{A} \plus{} \angle{B} \equal{} 90^\circ$. Prove that the distance between the midpoints of the bases is equal to the semidifference of the bases.
Let $ABCD$ be a trapezoid with $AB // CD$, $2|AB| = |CD|$ and $BD \perp BC$. Let $M$ be the midpoint of $CD$ and let $E$ be the intersection $BC$ and $AD$. Let $O$ be the intersection of $AM$ and $BD$. Let $N$ be the intersection of $OE$ and $AB$.
(a) Prove that $ABMD$ is a rhombus.
(b) Prove that the line $DN$ passes through the midpoint of the line segment $BE$.
Let $A'$ be a point on one of the sides of the trapezoid $ABCD$ such that line $AA'$ divides the area of the trapezoid in half. Points $B'$, $C'$, $D'$ are defined similarly. Prove that the intersection points of the diagonals of quadrilaterals $ABCD$ and $A'B'C'D'$ are symmetrical wrt the midpoint of midline of trapezoid $ABCD$.
In triangle $ ABC$, $ AB \equal{} AC \equal{} 100$, and $ BC \equal{} 56$. Circle $ P$ has radius $ 16$ and is tangent to $ \overline{AC}$ and $ \overline{BC}$. Circle $ Q$ is externally tangent to $ P$ and is tangent to $ \overline{AB}$ and $ \overline{BC}$. No point of circle $ Q$ lies outside of $ \triangle ABC$. The radius of circle $ Q$ can be expressed in the form $ m \minus{} n\sqrt {k}$, where $ m$, $ n$, and $ k$ are positive integers and $ k$ is the product of distinct primes. Find $ m \plus{} nk$.
Quadrilateral $XABY$ is inscribed in the semicircle $\omega$ with diameter $XY$. Segments $AY$ and $BX$ meet at $P$. Point $Z$ is the foot of the perpendicular from $P$ to line $XY$. Point $C$ lies on $\omega$ such that line $XC$ is perpendicular to line $AZ$. Let $Q$ be the intersection of segments $AY$ and $XC$. Prove that \[\dfrac{BY}{XP}+\dfrac{CY}{XQ}=\dfrac{AY}{AX}.\]
Show how to construct a chord $BPC$ of a given angle $A$ through a given point $P$ such that $\tfrac{1}{BP}\plus{} \tfrac{1}{PC}$ is a maximum.
[asy]
size(200);
defaultpen(linewidth(0.7));
pair A = origin, B = (5,0), C = (4.2,3), P = waypoint(B--C,0.65);
pair Bp = 1.3 * B, Cp = 1.2 * C;
draw(A--B--C--A^^Bp--A--Cp);
dot(P);
label("$A$",A,W);
label("$B$",B,S);
label("$C$",C,dir(B--C));
label("$P$",P,dir(A--P));
[/asy]
Given trapezoid $ABCD$ with parallel sides $AB$ and $CD$, let $E$ be a point on line $BC$ outside segment $BC$, such that segment $AE$ intersects segment $CD$. Assume that there exists a point $F$ inside segment $AD$ such that $\angle EAD=\angle CBF$. 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 $EF$, and assume that $K$ is different from $I$ and $J$.
Prove that $K$ belongs to the circumcircle of $\triangle ABI$ if and only if $K$ belongs to the circumcircle of $\triangle CDJ$.
A small square is constructed inside a square of area 1 by dividing each side of the unit square into $n$ equal parts, and then connecting the vertices to the division points closest to the opposite vertices. Find the value of $n$ if the the area of the small square is exactly 1/1985.
[asy]
size(200);
pair A=(0,1), B=(1,1), C=(1,0), D=origin;
draw(A--B--C--D--A--(1,1/6));
draw(C--(0,5/6)^^B--(1/6,0)^^D--(5/6,1));
pair point=( 0.5 , 0.5 );
//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("$1/n$", (11/12,1), N, fontsize(9));[/asy]
A circle center $O$ is inscribed in the quadrilateral $ABCD$. $AB$ is parallel to and longer than $CD$ and has midpoint $M$. The line $OM$ meets $CD$ at $F$. $CD$ touches the circle at $E$. Show that $DE = CF$ iff $AB = 2CD$.
In a trapezoid $ABCD$ whose parallel sides $AB$ and $CD$ are in ratio $\frac{AB}{CD}=\frac32$, the points $ N$ and $M$ are marked on the sides $BC$ and $AB$ respectively, in such a way that $BN = 3NC$ and $AM = 2MB$ and segments $AN$ and $DM$ are drawn that intersect at point $P$, find the ratio between the areas of triangle $APM$ and trapezoid $ABCD$.
[img]https://cdn.artofproblemsolving.com/attachments/7/8/21d59ca995d638dfcb76f9508e439fd93a5468.png[/img]
In a trapezoid $ABCD$ with $AB // CD, E$ is the midpoint of $BC$. Prove that if the quadrilaterals $ABED$ and $AECD$ are tangent, then the sides $a = AB, b = BC, c =CD, d = DA$ of the trapezoid satisfy the equalities $a+c = \frac{b}{3} +d$ and $\frac1a +\frac1c = \frac3b$ .
Let $ ABCD $ be a convex quadrilateral with no pair of parallel sides, such that $ \angle ABC = \angle CDA $. Assume that the intersections of the pairs of neighbouring angle bisectors of $ ABCD $ form a convex quadrilateral $ EFGH $. Let $ K $ be the intersection of the diagonals of $ EFGH$. Prove that the lines $ AB $ and $ CD $ intersect on the circumcircle of the triangle $ BKD $.
Two disjoint circles $\omega_1$ and $\omega_2$ are inscribed into an angle. Consider all pairs of parallel lines $l_1$ and $l_2$ such that $l_1$ touches $\omega_1$ and $l_2$ touches $\omega_2$ ($\omega_1$, $\omega_2$ lie between $l_1$ and $l_2$). Prove that the medial lines of all trapezoids formed by $l_1$ and $l_2$ and the sides of the angle touch some fixed circle.
In trapezoid $ ABCD$ we have $ \overline{AB}$ parallel to $ \overline{DC}$, $ E$ as the midpoint of $ \overline{BC}$, and $ F$ as the midpoint of $ \overline{DA}$. The area of $ ABEF$ is twice the area of $ FECD$. What is $ AB/DC$?
$ \textbf{(A)}\ 2\qquad
\textbf{(B)}\ 3\qquad
\textbf{(C)}\ 5\qquad
\textbf{(D)}\ 6\qquad
\textbf{(E)}\ 8$
Let $n=3k$ be a positive integer (with $k\geq 2$). An equilateral triangle is divided in $n^2$ unit equilateral triangles with sides parallel to the initial, forming a grid. We will call "trapezoid" the trapezoid which is formed by three equilateral triangles (one base is equal to one and the other is equal to two). We colour the points of the grid with three colours (red, blue and green) such that each two neighboring points have different colour.
Finally, the colour of a "trapezoid" will be the colour of the midpoint of its big base.
Find the number of all "trapezoids" in the grid (not necessarily disjoint) and determine the number of red, blue and green "trapezoids".
A trapezoid $ABCD$ with $BC // AD$ is given. The points $B'$ and $C'$ are symmetrical to $B$ and $C$ with respect to $CD$ and $AB$, respectively. Prove that the midpoint of the segment joining the circumcentres of $ABC'$ and $B'CD$ is equidistant from $A$ and $D$.
The area of trapezoid $ ABCD$ is $ 164 \text{cm}^2$. The altitude is $ 8 \text{cm}$, $ AB$ is $ 10 \text{cm}$, and $ CD$ is $ 17 \text{cm}$. What is $ BC$, in centimeters?
[asy]/* AMC8 2003 #21 Problem */
size(4inch,2inch);
draw((0,0)--(31,0)--(16,8)--(6,8)--cycle);
draw((11,8)--(11,0), linetype("8 4"));
draw((11,1)--(12,1)--(12,0));
label("$A$", (0,0), SW);
label("$D$", (31,0), SE);
label("$B$", (6,8), NW);
label("$C$", (16,8), NE);
label("10", (3,5), W);
label("8", (11,4), E);
label("17", (22.5,5), E);[/asy]
$ \textbf{(A)}\ 9\qquad\textbf{(B)}\ 10\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 20$
Let $ABCD$ be an isosceles trapezoid such that $AB \parallel CD$ and $\overline{AD}=\overline{BC}, \overline{AB}>\overline{CD}$. Let $E$ be a point such that $\overline{EC}=\overline{AC}$ and $EC \perp BC$, and $\angle ACE<90^{\circ}$. Let $\Gamma$ be a circle with center $D$ and radius $DA$, and $\Omega$ be the circumcircle of triangle $AEB$. Suppose that $\Gamma$ meets $\Omega$ again at $F(\neq A)$, and let $G$ be a point on $\Gamma$ such that $\overline{BF}=\overline{BG}$.
Prove that the lines $EG, BD$ meet on $\Omega$.
Let $ ABCDE$ be a convex pentagon such that $ AB\plus{}CD\equal{}BC\plus{}DE$ and $ k$ a circle with center on side $ AE$ that touches the sides $ AB$, $ BC$, $ CD$ and $ DE$ at points $ P$, $ Q$, $ R$ and $ S$ (different from vertices of the pentagon) respectively. Prove that lines $ PS$ and $ AE$ are parallel.
In the isosceles trapezoid with the area of $28$, a circle of radius $2$ is inscribed. Find the length of the side of the trapezoid.
A square $ ABCD$ with sides of length 1 is divided into two congruent trapezoids and a pentagon, which have equal areas, by joining the center of the square with points $ E,F,G$ where $ E$ is the midpoint of $ BC$, $ F,G$ are on $ AB$ and $ CD$, respectively, and they're positioned that $ AF < FB, DG < GC$ and $ F$ is the directly opposite of $ G$. If $ FB \equal{} x$, the length of the longer parallel side of each trapezoid, find the value of $ x$.
[asy]unitsize(2.5cm);
defaultpen(linewidth(.8pt)+fontsize(10pt));
dotfactor=4;
pair[] dotted={(0,0),(0,1),(1,1),(1,0),(1/6,0),(1/6,1),(1/2,1/2),(1,1/2)};
draw(unitsquare);
draw((1/6,0)--(1/2,1/2)--(1/6,1));
draw((1/2,1/2)--(1,1/2));
dot(dotted);
label("$x$",midpoint((1/6,1)--(1,1)),N);[/asy]$ \displaystyle \textbf{(A)}\ \frac {3}{5} \qquad \textbf{(B)}\ \frac {2}{3} \qquad \textbf{(C)}\ \frac {3}{4} \qquad \textbf{(D)}\ \frac {5}{6} \qquad \textbf{(E)}\ \frac {7}{8}$