Found problems: 844
Construct an equilateral trapezoid given the height and the midline, if it is known that the midline is divided by diagonals into three equal parts.
(Grigory Filippovsky)
$ABCD$ is an isosceles trapezoid with $BC || AD$. A circle $\omega$ passing through $B$ and $C$ intersects the side $AB$ and the diagonal $BD$ at points $X$ and $Y$ respectively. Tangent to $\omega$ at $C$ intersects the line $AD$ at $Z$. Prove that the points $X$, $Y$, and $Z$ are collinear.
Let $ABCD$ be a square of side length $4$. Points $E$ and $F$ are chosen on sides $BC$ and $DA$, respectively, such that $EF = 5$. Find the sum of the minimum and maximum possible areas of trapezoid $BEDF$.
[i]Proposed by Andrew Wu[/i]
Let $ABC$ be an equilateral triangle with the side of $20$ units. Amir divides this triangle into $400$ smaller equilateral triangles with the sides of $1$ unit. Reza then picks $4$ of the vertices of these smaller triangles. The vertices lie inside the triangle $ABC$ and form a parallelogram with sides parallel to the sides of the triangle $ABC.$ There are exactly $46$ smaller triangles that have at least one point in common with the sides of this parallelogram. Find all possible values for the area of this parallelogram.
[asy]
unitsize(150);
defaultpen(linewidth(0.7));
int n = 20; /* # of vertical lines, including BC */
pair A = (0,0), B = dir(-30), C = dir(30);
draw(A--B--C--cycle,linewidth(1)); dot(A,UnFill(0)); dot(B,UnFill(0)); dot(C,UnFill(0));
label("$A$",A,W); label("$C$",C,NE); label("$B$",B,SE);
for(int i = 1; i < n; ++i) {
draw((i*A+(n-i)*B)/n--(i*A+(n-i)*C)/n);
draw((i*B+(n-i)*A)/n--(i*B+(n-i)*C)/n);
draw((i*C+(n-i)*A)/n--(i*C+(n-i)*B)/n);
}[/asy]
[Thanks azjps for drawing the diagram.]
[hide="Note"][i]Note:[/i] Vid changed to Amir, and Eva change to Reza![/hide]
Let $ ABCD$ be an isosceles trapezoid with $ \overline{AD}\parallel{}\overline{BC}$ whose angle at the longer base $ \overline{AD}$ is $ \dfrac{\pi}{3}$. The diagonals have length $ 10\sqrt {21}$, and point $ E$ is at distances $ 10\sqrt {7}$ and $ 30\sqrt {7}$ from vertices $ A$ and $ D$, respectively. Let $ F$ be the foot of the altitude from $ C$ to $ \overline{AD}$. The distance $ EF$ can be expressed in the form $ m\sqrt {n}$, where $ m$ and $ n$ are positive integers and $ n$ is not divisible by the square of any prime. Find $ m \plus{} n$.
As shown in figure , $\odot O$ is the inscribed circle of trapezoid $ABCD$, and the tangent points are $E, F, G, H$, $AB \parallel CD$. The line passing through$ B$, parallel to $AD$ intersects extension of $DC$ at point $P$. The extension of $AO$ intersects $CP$ at point $Q$. If $AE=BE$ , prove that $\angle CBQ = \angle PBQ$.
[img]https://cdn.artofproblemsolving.com/attachments/d/2/7c3a04bb1c59bc6d448204fd78f553ea53cb9e.png[/img]
Is there exist four points on plane, such that distance between any two of them is integer odd number?
May be it is geometry or number theory or combinatoric, I don't know, so it here :blush:
Circles with centers $A ,~ B,$ and $C$ each have radius $r$, where $1 < r < 2$. The distance between each pair of centers is $2$. If $B'$ is the point of intersection of circle $A$ and circle $C$ which is outside circle $B$, and if $C'$ is the point of intersection of circle $A$ and circle $B$ which is outside circle $C$, then length $B'C'$ equals
$\textbf{(A) }3r-2\qquad\textbf{(B) }r^2\qquad\textbf{(C) }r+\sqrt{3(r-1)}\qquad$
$\textbf{(D) }1+\sqrt{3(r^2-1)}\qquad\textbf{(E) }\text{none of these}$
[asy]
//Holy crap, CSE5 is freaking amazing!
import cse5;
pathpen=black;
pointpen=black;
dotfactor=3;
size(200);
pair A=(1,2),B=(2,0),C=(0,0);
D(CR(A,1.5));
D(CR(B,1.5));
D(CR(C,1.5));
D(MP("$A$",A));
D(MP("$B$",B));
D(MP("$C$",C));
pair[] BB,CC;
CC=IPs(CR(A,1.5),CR(B,1.5));
BB=IPs(CR(A,1.5),CR(C,1.5));
D(BB[0]--CC[1]);
MP("$B'$",BB[0],NW);MP("$C'$",CC[1],NE);
//Credit to TheMaskedMagician for the diagram
[/asy]
Let $ABCD$ be a trapezoid with the longer base $AB$ such that its diagonals $AC$ and $BD$ are perpendicular. Let $O$ be the circumcenter of the triangle $ABC$ and $E$ be the intersection of the lines $OB$ and $CD$. Prove that $BC^2 = CD \cdot CE$.
A rectangle intersects a circle as shown: $AB=4$, $BC=5$, and $DE=3$. Then $EF$ equals:
[asy]size(200);
defaultpen(linewidth(0.7)+fontsize(10));
pair D=origin, E=(3,0), F=(10,0), G=(12,0), H=(12,1), A=(0,1), B=(4,1), C=(9,1), O=circumcenter(B,C,F);
draw(D--G--H--A--cycle);
draw(Circle(O, abs(O-C)));
label("$A$", A, NW);
label("$B$", B, NW);
label("$C$", C, NE);
label("$D$", D, SW);
label("$E$", E, SE);
label("$F$", F, SW);
label("4", (2,0.85), N);
label("3", D--E, S);
label("5", (6.5,0.85), N);
[/asy]
$\mathbf{(A)}\; 6\qquad \mathbf{(B)}\; 7\qquad \mathbf{(C)}\; \frac{20}3\qquad \mathbf{(D)}\; 8\qquad \mathbf{(E)}\; 9$
Let $ABCD$ be a trapezoid (quadrilateral with one pair of parallel sides) such that $AB < CD$. Suppose that $AC$ and $BD$ meet at $E$ and $AD$ and $BC$ meet at $F$. Construct the parallelograms $AEDK$ and $BECL$. Prove that $EF$ passes through the midpoint of the segment $KL$.
A trapezium has parallel sides of length equal to $a$ and $b$ ($a <b$), and the distance between the parallel sides is the altitude $h$. The extensions of the non-parallel lines intersect at a point that is a vertex of two triangles that have as sides the parallel sides of the trapezium. Express the areas of the triangles as functions of $a,b$ and $h$.
In the right-angled trapezoid $AB CD$, $AB \parallel CD$, $m( \angle A) = 90°$, $AD = DC = a$ and $AB =2a$. On the perpendiculars raised in $C$ and $D$ on the plane containing the trapezoid one considers points $E$ and $F$ (on the same side of the plane) such that $CE = 2a$ and $DF = a$. Find the distance from the point $B$ to the plane $(AEF)$ and the measure of the angle between the lines $AF$ and $BE$.
Let $ABC$ be an acute,non-isosceles triangle with $AB<AC<BC$.Let $D,E,Z$ be the midpoints of $BC,AC,AB$ respectively and segments $BK,CL$ are altitudes.In the extension of $DZ$ we take a point $M$ such that the parallel from $M$ to $KL$ crosses the extensions of $CA,BA,DE$ at $S,T,N$ respectively (we extend $CA$ to $A$-side and $BA$ to $A$-side and $DE$ to $E$-side).If the circumcirle $(c_{1})$ of $\triangle{MBD}$ crosses the line $DN$ at $R$ and the circumcirle $(c_{2})$ of $\triangle{NCD}$ crosses the line $DM$ at $P$ prove that $ST\parallel PR$.
An isosceles trapezoid $ABCD$, inscribed in $\omega$, satisfies $AB=CD, AD<BC, AD<CD$.
A circle with center $D$ and passing $A$ hits $BD, CD, \omega$ at $E, F, P(\not= A)$, respectively.
Let $AP \cap EF = Q$, and $\omega$ meet $CQ$ and the circumcircle of $\triangle BEQ$ at $R(\not= C), S(\not= B)$, respectively.
Prove that $\angle BER= \angle FSC$.
The incircle of a triangle $ABC$ touches the sides $BC$, $CA$, $AB$ at the points $D$, $E$, $F$ respectively. Let the line $AD$ intersect this incircle of triangle $ABC$ at a point $X$ (apart from $D$). Assume that this point $X$ is the midpoint of the segment $AD$, this means, $AX = XD$. Let the line $BX$ meet the incircle of triangle $ABC$ at a point $Y$ (apart from $X$), and let the line $CX$ meet the incircle of triangle $ABC$ at a point $Z$ (apart from $X$). Show that $EY = FZ$.
A checkered polygon $A$ is drawn on the checkered plane. We call a cell of $A$ [i]internal[/i] if all $8$ of its adjacent cells belong to $A$. All other (non-internal) cells of $A$ we call [i]boundary[/i]. It is known that $1)$ each boundary cell has exactly two common sides with no boundary cells; and 2) the union of all boundary cells can be divided into isosceles trapezoid of area $2$ with vertices at the grid nodes (and acute angles of the trapezoids are equal $45^\circ$).
Prove that the area of the polygon $A$ is congruent to $1$ modulo $4$.
All edges of the triangular pyramid $ABCD$ are equal in length. Let $M$ be the midpoint of $DB$, $N$ is the point on $\overline{AB}$, such that $2NA=NB$ and $N\not\in AB$ and $P$ is a point on the altitude through point $D$ in $\triangle BCD$. Find $\angle MPD$ if the intersection of the pyramid with the plane $(NMP)$ is a trapezoid.
The points $X, Y,Z$ are marked on the sides $AB, BC,AC$ of the triangle $ABC$, respectively. Points $A',B', C'$ are on the $XZ, XY, YZ$ sides of the triangle $XYZ$, respectively, so that $\frac{AB}{A'B'} = \frac{AB}{A'B'} =\frac{BC}{B'C'}= 2$ and $ABB'A',BCC'B',ACC'A'$ are trapezoids in which the sides of the triangle $ABC$ are bases.
a) Determine the ratio between the area of the trapezium $ABB'A'$ and the area of the triangle $A'B'X$.
b) Determine the ratio between the area of the triangle $XYZ$ and the area of the triangle $ABC$.
Let $\gamma$ be the circumcircle of the acute triangle $ABC$. Let $P$ be the midpoint of the minor arc $BC$. The parallel to $AB$ through $P$ cuts $BC, AC$ and $\gamma$ at points $R,S$ and $T$, respectively. Let $K \equiv AP \cap BT$ and $L \equiv BS \cap AR$. Show that $KL$ passes through the midpoint of $AB$ if and only if $CS = PR$.
If the larger base of an isosceles trapezoid equals a diagonal and the smaller base equals the altitude, then the ratio of the smaller base to the larger base is:
$ \textbf{(A)}\ \frac{1}{2} \qquad\textbf{(B)}\ \frac{2}{3} \qquad\textbf{(C)}\ \frac{3}{4} \qquad\textbf{(D)}\ \frac{3}{5} \qquad\textbf{(E)}\ \frac{2}{5}$
Let $ABCD$ be a trapezoid with $AB\parallel CD$ and inscribed in a circumference $\Gamma$. Let $P$ and $Q$ be two points on segment $AB$ ($A$, $P$, $Q$, $B$ appear in that order and are distinct) such that $AP=QB$. Let $E$ and $F$ be the second intersection points of lines $CP$ and $CQ$ with $\Gamma$, respectively. Lines $AB$ and $EF$ intersect at $G$. Prove that line $DG$ is tangent to $\Gamma$.
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.
Let $PQ$ be a line segment of constant length $\lambda$ taken on the side $BC$ of a triangle $ABC$ with the order $B,P,Q,C$, and let the lines through $P$ and $Q$ parallel to the lateral sides meet $AC$ at $P_1$ and $Q_1$ and $AB$ at $P_2$ and $Q_2$ respectively. Prove that the sum of the areas of the trapezoids $PQQ_1P_1$ and $PQQ_2P_2$ is independent of the position of $PQ$ on $BC.$
In trapezoid $ ABCD$ with bases $ AB$ and $ CD$, we have $ AB\equal{}52$, $ BC\equal{}12$, $ CD\equal{}39$, and $ DA\equal{}5$. The area of $ ABCD$ is
[asy]
pair A,B,C,D;
A=(0,0);
B=(52,0);
C=(38,20);
D=(5,20);
dot(A);
dot(B);
dot(C);
dot(D);
draw(A--B--C--D--cycle);
label("$A$",A,S);
label("$B$",B,S);
label("$C$",C,N);
label("$D$",D,N);
label("52",(A+B)/2,S);
label("39",(C+D)/2,N);
label("12",(B+C)/2,E);
label("5",(D+A)/2,W);[/asy]
$ \text{(A)}\ 182 \qquad
\text{(B)}\ 195 \qquad
\text{(C)}\ 210 \qquad
\text{(D)}\ 234 \qquad
\text{(E)}\ 260$