Found problems: 844
Let $ABCD$ be an isosceles trapezoid with $AB=AD=BC, AB//CD, AB>CD$. Let $E= AC \cap BD$ and $N$ symmetric to $B$ wrt $AC$. Prove that the quadrilateral $ANDE$ is cyclic.
Given a triangle $ABC$, a point $P$ is chosen on side $BC$. Points $M$ and $N$ lie on sides $AB$ and $AC$, respectively, such that $MP \parallel AC$ and $NP \parallel AB$. Point $P$ is reflected across $MN$ to point $Q$. Show that triangle $QMB$ is similar to triangle $CNQ$.
[i]Brian Hamrick.[/i]
Let be a natural number $ n\ge 2 $ and a real number $ r>1. $ Determine the natural numbers $ k $ having the property that the affixes of $ r^ke^{\pi ki/n} ,r^{k+1}e^{\pi (k+1)i/n} ,r^{k+n}e^{\pi (k+n)i/n} ,r^{k+n+1}e^{\pi (k+n+1) i/n} $ in the complex plane represent the vertices of a trapezoid.
Let $n$ be a positive integer. An equilateral triangle with side $n$ will be denoted by $T_n$ and 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).
Let also $m$ be a positive integer with $m<n$ and suppose that $T_n$ and $T_m$ can be tiled with "trapezoids".
Prove that, if from $T_n$ we remove a $T_m$ with the same orientation, then the rest can be tiled with "trapezoids".
A hexagon that is inscribed in a circle has side lengths $22$, $22$, $20$, $22$, $22$, and $20$ in that order. The radius of the circle can be written as $p+\sqrt{q}$, where $p$ and $q$ are positive integers. Find $p+q$.
Let $ABC$ be an acute triangle inscribed in a circle centered at $O$. Let $M$ be a point on the small arc $AB$ of the triangle's circumcircle. The perpendicular dropped from $M$ on the ray $OA$ intersects the sides $AB$ and $AC$ at the points $K$ and $L$, respectively. Similarly, the perpendicular dropped from $M$ on the ray $OB$ intersects the sides $AB$ and $BC$ at $N$ and $P$, respectively. Assume that $KL=MN$. Find the size of the angle $\angle{MLP}$ in terms of the angles of the triangle $ABC$.
Consider a regular heptagon ( polygon of $7$ equal sides and angles) $ABCDEFG$ as in the figure below:-
$(a).$ Prove $\frac{1}{\sin\frac{\pi}{7}}=\frac{1}{\sin\frac{2\pi}{7}}+\frac{1}{\sin\frac{3\pi}{7}}$
$(b).$ Using $(a)$ or otherwise, show that $\frac{1}{AG}=\frac{1}{AF}+\frac{1}{AE}$
[asy]
draw(dir(360/7)..dir(2*360/7),blue);
draw(dir(2*360/7)..dir(3*360/7),blue);
draw(dir(3*360/7)..dir(4*360/7),blue);
draw(dir(4*360/7)..dir(5*360/7),blue);
draw(dir(5*360/7)..dir(6*360/7),blue);
draw(dir(6*360/7)..dir(7*360/7),blue);
draw(dir(7*360/7)..dir(360/7),blue);
draw(dir(2*360/7)..dir(4*360/7),blue);
draw(dir(4*360/7)..dir(1*360/7),blue);
label("$A$",dir(4*360/7),W);
label("$B$",dir(5*360/7),S);
label("$C$",dir(6*360/7),S);
label("$D$",dir(7*360/7),E);
label("$E$",dir(1*360/7),E);
label("$F$",dir(2*360/7),N);
label("$G$",dir(3*360/7),W);
[/asy]
$ABCD$ is a trapezoid with $AB || CD$. There are two circles $\omega_1$ and $\omega_2$ is the trapezoid such that $\omega_1$ is tangent to $DA$, $AB$, $BC$ and $\omega_2$ is tangent to $BC$, $CD$, $DA$. Let $l_1$ be a line passing through $A$ and tangent to $\omega_2$(other than $AD$), Let $l_2$ be a line passing through $C$ and tangent to $\omega_1$ (other than $CB$).
Prove that $l_1 || l_2$.
A disk with radius $1$ is externally tangent to a disk with radius $5$. Let $A$ be the point where the disks are tangent, $C$ be the center of the smaller disk, and $E$ be the center of the larger disk. While the larger disk remains fixed, the smaller disk is allowed to roll along the outside of the larger disk until the smaller disk has turned through an angle of $360^\circ$. That is, if the center of the smaller disk has moved to the point $D$, and the point on the smaller disk that began at $A$ has now moved to point $B$, then $\overline{AC}$ is parallel to $\overline{BD}$. Then $\sin^2(\angle BEA)=\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
Find all isosceles trapezoids that are divided into $2$ isosceles triangles by a diagonal.
On square $ABCD,$ points $E,F,G,$ and $H$ lie on sides $\overline{AB},\overline{BC},\overline{CD},$ and $\overline{DA},$ respectively, so that $\overline{EG} \perp \overline{FH}$ and $EG=FH = 34.$ Segments $\overline{EG}$ and $\overline{FH}$ intersect at a point $P,$ and the areas of the quadrilaterals $AEPH, BFPE, CGPF,$ and $DHPG$ are in the ratio $269:275:405:411.$ Find the area of square $ABCD$.
[asy]
size(200);
defaultpen(linewidth(0.8)+fontsize(10.6));
pair A = (0,sqrt(850));
pair B = (0,0);
pair C = (sqrt(850),0);
pair D = (sqrt(850),sqrt(850));
draw(A--B--C--D--cycle);
dotfactor = 3;
dot("$A$",A,dir(135));
dot("$B$",B,dir(215));
dot("$C$",C,dir(305));
dot("$D$",D,dir(45));
pair H = ((2sqrt(850)-sqrt(120))/6,sqrt(850));
pair F = ((2sqrt(850)+sqrt(306)+7)/6,0);
dot("$H$",H,dir(90));
dot("$F$",F,dir(270));
draw(H--F);
pair E = (0,(sqrt(850)-6)/2);
pair G = (sqrt(850),(sqrt(850)+sqrt(100))/2);
dot("$E$",E,dir(180));
dot("$G$",G,dir(0));
draw(E--G);
pair P = extension(H,F,E,G);
dot("$P$",P,dir(60));
label("$w$", (H+E)/2,fontsize(15));
label("$x$", (E+F)/2,fontsize(15));
label("$y$", (G+F)/2,fontsize(15));
label("$z$", (H+G)/2,fontsize(15));
label("$w:x:y:z=269:275:405:411$",(sqrt(850)/2,-4.5),fontsize(11));
[/asy]
Given trapezoid $ABCD$ ($AD\parallel BC$) with $AD \perp AB$ and $T=AC\cap BD$. A circle centered at point $O$ is inscribed in the trapezoid and touches the side $CD$ at point $Q$. Let $P$ be the intersection point (different from $Q$) of the side $CD$ and the circle passing through $T,Q$ and $O$. Prove that $TP \parallel AD$.
I. Voronovich
On side \( AB \) of an isosceles trapezoid \( ABCD \) (\( AD \parallel BC \)), points \( E \) and \( F \) are chosen such that a circle can be inscribed in quadrilateral \( CDEF \). Prove that the circumcircles of triangles \( ADE \) and \( BCF \) are tangent to each other.
[i]Proposed by Matthew Kurskyi[/i]
All of the polygons in the figure below are regular. Prove that $ABCD$ is an isosceles trapezoid.
[img]https://cdn.artofproblemsolving.com/attachments/e/a/3f4de32becf4a90bf0f0b002fb4d8e724e8844.png[/img]
[i]Proposed by Mahdi Etesamifard - Iran[/i]
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]
Let $ABCDEF$ be a convex hexagon and it`s diagonals have one common point $M$. It is known that the circumcenters of triangles $MAB,MBC,MCD,MDE,MEF,MFA$ lie on a circle.
Show that the quadrilaterals $ABDE,BCEF,CDFA$ have equal areas.
Let $MAZN$ be an isosceles trapezium inscribed in a circle $(c)$ with centre $O$. Assume that $MN$ is a diameter of $(c)$ and let $ B$ be the midpoint of $AZ$. Let $(\epsilon)$ be the perpendicular line on $AZ$ passing through $ A$. Let $C$ be a point on $(\epsilon)$, let $E$ be the point of intersection of $CB$ with $(c)$ and assume that $AE$ is perpendicular to $CB$. Let $D$ be the point of intersection of $CZ$ with $(c)$ and let $F$ be the antidiametric point of $D$ on $(c)$. Let $ P$ be the point of intersection of $FE$ and $CZ$. Assume that the tangents of $(c)$ at the points $M$ and $Z$ meet the lines $AZ$ and $PA$ at the points $K$ and $T$ respectively. Prove that $OK$ is perpendicular to $TM$.
Theoklitos Parayiou, Cyprus
In the diagram below, angle $ABC$ is a right angle. Point $D$ is on $\overline{BC}$, and $\overline{AD}$ bisects angle $CAB$. Points $E$ and $F$ are on $\overline{AB}$ and $\overline{AC}$, respectively, so that $AE=3$ and $AF=10.$ Given that $EB=9$ and $FC=27$, find the integer closest to the area of quadrilateral $DCFG.$
[asy]
size(250);
pair A=(0,12), E=(0,8), B=origin, C=(24*sqrt(2),0), D=(6*sqrt(2),0), F=A+10*dir(A--C), G=intersectionpoint(E--F, A--D);
draw(A--B--C--A--D^^E--F);
pair point=G+1*dir(250);
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("$E$", E, dir(point--E));
label("$F$", F, dir(point--F));
label("$G$", G, dir(point--G));
markscalefactor=0.1;
draw(rightanglemark(A,B,C));
label("10", A--F, dir(90)*dir(A--F));
label("27", F--C, dir(90)*dir(F--C));
label("3", (0,10), W);
label("9", (0,4), W);[/asy]
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.
We consider a right trapezoid $ABCD$, in which $AB||CD,AB>CD,AD\perp AB$ and $AD>CD$. The diagonals $AC$ and $BD$ intersect at $O$. The parallel through $O$ to $AB$ intersects $AD$ in $E$ and $BE$ intersects $CD$ in $F$. Prove that $CE\perp AF$ if and only if $AB\cdot CD=AD^2-CD^2$ .
Suppose $ABCD$ is a trapezoid with $AB\parallel CD$ and $AB\perp BC$. Let $X$ be a point on segment $\overline{AD}$ such that $AD$ bisects $\angle BXC$ externally, and denote $Y$ as the intersection of $AC$ and $BD$. If $AB=10$ and $CD=15$, compute the maximum possible value of $XY$.
The longer base of trapezoid $ABCD$ is $AB$, while the shorter base is $CD$. Diagonal $AC$ bisects the interior angle at $A$. The interior bisector at $B$ meets diagonal $AC$ at $E$. Line $DE$ meets segment $AB$ at $F$. Suppose that $AD = FB$ and $BC = AF$. Find the interior angles of quadrilateral $ABCD$, if we know that $\angle BEC = 54^o$.
The point $D$ at the side $AB$ of triangle $ABC$ is given. Construct points $E,F$ at sides $BC, AC$ respectively such that the midpoints of $DE$ and $DF$ are collinear with $B$ and the midpoints of $DE$ and $EF$ are collinear with $C.$
Three segments $2$ cm, $5$ cm and $12$ cm long are constructed on the plane. Construct a trapezoid with bases of $2$ cm and $5$ cm, the sum of the sides of which is $12$ cm, and one of the angles is $60^o$.
(Bogdan Rublev)
In a given trapezium $ ABCD $ with $ AB$ parallel to $ CD $ and $ AB > CD $, the line $ BD $ bisects the angle $ \angle ADC $. The line through $ C $ parallel to $ AD $ meets the segments $ BD $ and $ AB $ in $ E $ and $ F $, respectively. Let $ O $ be the circumcenter of the triangle $ BEF $. Suppose that $ \angle ACO = 60^{\circ} $. Prove the equality
\[ CF = AF + FO .\]