Found problems: 844
A rectangular piece of paper with side lengths 5 by 8 is folded along the dashed lines shown below, so that the folded flaps just touch at the corners as shown by the dotted lines. Find the area of the resulting trapezoid.
[asy]
size(150);
defaultpen(linewidth(0.8));
draw(origin--(8,0)--(8,5)--(0,5)--cycle,linewidth(1));
draw(origin--(8/3,5)^^(16/3,5)--(8,0),linetype("4 4"));
draw(origin--(4,3)--(8,0)^^(8/3,5)--(4,3)--(16/3,5),linetype("0 4"));
label("$5$",(0,5/2),W);
label("$8$",(4,0),S);
[/asy]
$ABCD$ 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.
Let $ A_1,A_2,...,A_{3n} $ be $ 3n\ge 3 $ planar points such that $ A_1A_2A_3 $ is an equilateral triangle and $ A_{3k+1} ,A_{3k+2} ,A_{3k+3} $ are the midpoints of the sides of $ A_{3k-2}A_{3k-1}A_{3k} , $ for all $ 1\le k<n. $ Of two different colors, each one of these points are colored, either with one, either with another.
[b]a)[/b] Prove that, if $ n\ge 7, $ then some of these points form a monochromatic (only one color) isosceles trapezoid.
[b]b)[/b] What about $ n=6? $
Let $ABCD$ be a cyclic quadrilateral which is not a trapezoid and whose diagonals meet at $E$. The midpoints of $AB$ and $CD$ are $F$ and $G$ respectively, and $\ell$ is the line through $G$ parallel to $AB$. The feet of the perpendiculars from E onto the lines $\ell$ and $CD$ are $H$ and $K$, respectively. Prove that the lines $EF$ and $HK$ are perpendicular.
Let $ ABCD$ be an isosceles trapezium with $ AB \parallel{} CD$ and $ \bar{BC} \equal{} \bar{AD}.$ The parallel to $ AD$ through $ B$ meets the perpendicular to $ AD$ through $ D$ in point $ X.$ The line through $ A$ drawn which is parallel to $ BD$ meets the perpendicular to $ BD$ through $ D$ in point $ Y.$ Prove that points $ C,X,D$ and $ Y$ lie on a common circle.
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]
A diagonal is drawn in an isosceles trapezoid. By the contour of each of the resulting two triangles creeps its own
beetle. The velocities of the beetles are constant and identical. Beetles don't change directions around their contours, and along the diagonal of the trapezoid they crawl in different directions.
Prove that for any starting positions of the beetles they will ever meet.
A trapezoid $ABCD$ with bases $BC$ and $AD$ is given. The points $K$ and $L$ are chosen on the sides $AB$ and $CD$, respectively, so that $KL \parallel AD$. It turned out that the areas of the quadrilaterals $AKLD$ and $KBCL$ are equal. Find the length $KL$ if $BC = 3, AD = 5$.
There is an Equilateral trapezoid $ ABCD. $
$ \bar{AB} =60, \bar{BC}=\bar{DA}= 36, \bar{CD}=108. $
$ M $ is the middle point of $ \bar {AB} $, and point $P$ on $ \bar{AM} $ follows that $ \bar {AP} $ =10.
The foot of perpendicular dropped from $P$ to $ \bar {BD} $ is $E$.
$ \bar{AC} \cap \bar{BD} $ is $ F $.
Point $X$ is on $ \bar {AF} $ which follows $ \bar{MX}=\bar{ME} $
Find $ \bar{AX} \times \bar{AF} $
Let $ABCD$ be a trapezoid such that $|AC|=8$, $|BD|=6$, and $AD \parallel BC$. Let $P$ and $S$ be the midpoints of $[AD]$ and $[BC]$, respectively. If $|PS|=5$, find the area of the trapezoid $ABCD$.
Let $ABC$ be an isosceles triangle with $AB=AC$ and let $\Gamma$ denote its circumcircle. A point $D$ is on arc $AB$ of $\Gamma$ not containing $C$. A point $E$ is on arc $AC$ of $\Gamma$ not containing $B$. If $AD=CE$ prove that $BE$ is parallel to $AD$.
Let $ABCD$ be a trapezoid with $AD=a, AB=2a, BC=3a$ and $\angle A=\angle B =90 ^o$. Let $E,Z$ be the midpoints of the sides $AB ,CD$ respectively and $I$ be the foot of the perpendicular from point $Z$ on $BC$. Prove that :
i) triangle $BDZ$ is isosceles
ii) midpoint $O$ of $EZ$ is the centroid of triangle $BDZ$
iii) lines $AZ$ and $DI$ intersect at a point lying on line $BO$
Two circles $O_1$ and $O_2$ intersect at $M,N$. The common tangent line nearer to $M$ of the two circles touches $O_1,O_2$ at $A,B$ respectively. Let $C,D$ be the symmetric points of $A,B$ with respect to $M$ respectively. The circumcircle of triangle $DCM$ intersects circles $O_1$ and $O_2$ at points $E,F$ respectively which are distinct from $M$. Prove that the circumradii of the triangles $MEF$ and $NEF$ are equal.
Consider a trapezoid $ABCD,AB\parallel CD,AB>CD.$ Let us denote intersections of lines as follows: $E=AC\cap BD, F=AD\cap BC.$ Let $GH$ be a line such that $G\in AD,H\in BC, E\in GH,GH\parallel AB.$ Moreover, denote $K,L$ midpoints of the bases $AB,CD$ respectively. Show that
(a) the points $K,L$ lie on the line $EF,$
(b) lines $AC,KH$ and $BD,KG$ are not parallel (denote $M=AC\cap KH,N=BD\cap KG$),
(c) the points $F,M,N$ are collinear.
An isosceles trapezoid is circumscribed around a circle. The longer base of the trapezoid is $16$, and one of the base angles is $\arcsin(.8)$. Find the area of the trapezoid.
$ \textbf{(A)}\ 72\qquad\textbf{(B)}\ 75\qquad\textbf{(C)}\ 80\qquad\textbf{(D)}\ 90\qquad\textbf{(E)}\ \text{not uniquely determined} $
Let $ABCD$ be a cyclic quadrilateral. Let $E$ and $F$ be variable points on the sides $AB$ and $CD$, respectively, such that $AE:EB=CF:FD$. Let $P$ be the point on the segment $EF$ such that $PE:PF=AB:CD$. Prove that the ratio between the areas of triangles $APD$ and $BPC$ does not depend on the choice of $E$ and $F$.
The diagram below shows a regular hexagon with an inscribed square where two sides of the square are parallel to two sides of the hexagon. There are positive integers $m$, $n$, and $p$ such that the ratio of the area of the hexagon to the area of the square can be written as $\tfrac{m+\sqrt{n}}{p}$ where $m$ and $p$ are relatively prime. Find $m + n + p$.
[asy]
import graph; size(4cm);
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps);
draw((0,1)--(1,1)--(1.5,1.87)--(1,2.73)--(0,2.73)--(-0.5,1.87)--cycle);
filldraw((1.13,2.5)--(-0.13,2.5)--(-0.13,1.23)--(1.13,1.23)--cycle,grey);
draw((0,1)--(1,1));
draw((1,1)--(1.5,1.87));
draw((1.5,1.87)--(1,2.73));
draw((1,2.73)--(0,2.73));
draw((0,2.73)--(-0.5,1.87));
draw((-0.5,1.87)--(0,1));
draw((1.13,2.5)--(-0.13,2.5));
draw((-0.13,2.5)--(-0.13,1.23));
draw((-0.13,1.23)--(1.13,1.23));
draw((1.13,1.23)--(1.13,2.5)); [/asy]
$ABCD$ is a trapezium where $AD\parallel BC$. $P$ is the point on the line $AB$ such that $\angle CPD$ is maximal. $Q$ is the point on the line $CD$ such that $\angle BQA$ is maximal. Given that $P$ lies on the segment $AB$, prove that $\angle CPD=\angle BQA$.
A circle $k$ is given with a non-diameter chord $AC$. On the tangent at point $A$ select point $X \ne A$ and mark $D$ the intersection of the circle $k$ with the interior of the line $XC$ (if any). Let $B$ a point in circle $k$ such that quadrilateral $ABCD$ is a trapezoid . Determine the set of intersections of lines $BC$ and $AD$ belonging to all such trapezoids.
If $P_1P_2P_3P_4P_5P_6$ is a regular hexagon whose apothem (distance from the center to midpoint of a side) is $2$, and $Q_i$ is the midpoint of side $P_iP_{i+1}$ for $i=1,2,3,4$, then the area of quadrilateral $Q_1Q_2Q_3Q_4$ is
$\textbf{(A) }6\qquad\textbf{(B) }2\sqrt{6}\qquad\textbf{(C) }\frac{8\sqrt{3}}{3}\qquad\textbf{(D) }3\sqrt{3}\qquad\textbf{(E) }4\sqrt{3}$
Two circles $O_1$ and $O_2$ intersect each other at $M$ and $N$. The common tangent to two circles nearer to $M$ touch $O_1$ and $O_2$ at $A$ and $B$ respectively. Let $C$ and $D$ be the reflection of $A$ and $B$ respectively with respect to $M$. The circumcircle of the triangle $DCM$ intersect circles $O_1$ and $O_2$ respectively at points $E$ and $F$ (both distinct from $M$). Show that the circumcircles of triangles $MEF$ and $NEF$ have same radius length.
The polygon(s) formed by $y=3x+2$, $y=-3x+2$, and $y=-2$, is (are):
$ \textbf{(A) }\text{An equilateral triangle}\qquad\textbf{(B) }\text{an isosceles triangle} \qquad\textbf{(C) }\text{a right triangle} \qquad$
$\textbf{(D) }\text{a triangle and a trapezoid}\qquad\textbf{(E) }\text{a quadrilateral} $
In this diagram the center of the circle is $O$, the radius is $a$ inches, chord $EF$ is parallel to chord $CD, O, G, H, J$ are collinear, and $G$ is the midpoint of $CD$. Let $K$ (sq. in.) represent the area of trapezoid $CDFE$ and let $R$ (sq. in.) represent the area of rectangle $ELMF$. Then, as $CD$ and $EF$ are translated upward so that $OG$ increases toward the value $a$, while $JH$ always equals $HG$, the ratio $K:R$ become arbitrarily close to:
[asy]
size((270));
draw((0,0)--(10,0)..(5,5)..(0,0));
draw((5,0)--(5,5));
draw((9,3)--(1,3)--(1,1)--(9,1)--cycle);
draw((9.9,1)--(.1,1));
label("O", (5,0), S);
label("a", (7.5,0), S);
label("G", (5,1), SE);
label("J", (5,5), N);
label("H", (5,3), NE);
label("E", (1,3), NW);
label("L", (1,1), S);
label("C", (.1,1), W);
label("F", (9,3), NE);
label("M", (9,1), S);
label("D", (9.9,1), E);
[/asy]
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ \sqrt{2} \qquad\textbf{(D)}\ \frac{1}{\sqrt{2}}+\frac{1}{2} \qquad\textbf{(E)}\ \frac{1}{\sqrt{2}}+1$
$ABC$ is a right triangle with $\angle C=90$,$CD$ is perpendicular to $AB$,and $D$ is the foot,$\omega$ is the circumcircle of triangle $BCD$,$\omega_{1}$ is a circle inside triangle $ACD$,tangent to $AD$ and $AC$ at $M$ and $N$ respectively,and $\omega_{1}$ is also tangent to $\omega$.prove that:
(1)$BD*CN+BC*DM=CD*BM$
(2)$BM=BC$
Let $ABCD$ be an isosceles trapezoid such that $AB \parallel CD$. Suppose that there exists a point $P$ in $ABCD$ such that $\angle APB > \angle ADC$ and $\angle DPC > \angle ABC$. Prove that $$AB+CD>DA+BC.$$