Found problems: 844
Rectangle $ABCD$ and right triangle $DCE$ have the same area. They are joined to form a trapezoid, as shown. What is $DE$?
[asy]
size(250);
defaultpen(linewidth(0.8));
pair A=(0,5),B=origin,C=(6,0),D=(6,5),E=(18,0);
draw(A--B--E--D--cycle^^C--D);
draw(rightanglemark(D,C,E,30));
label("$A$",A,NW);
label("$B$",B,SW);
label("$C$",C,S);
label("$D$",D,N);
label("$E$",E,S);
label("$5$",A/2,W);
label("$6$",(A+D)/2,N);
[/asy]
$\textbf{(A) }12\qquad\textbf{(B) }13\qquad\textbf{(C) }14\qquad\textbf{(D) }15\qquad \textbf{(E) }16$
Let $ABC$ be an isosceles triangle with $AB=AC$. Consider a variable point $P$ on the extension of the segment $BC$ beyound $B$ (in other words, $P$ lies on the line $BC$ such that the point $B$ lies inside the segment $PC$). Let $r_{1}$ be the radius of the incircle of the triangle $APB$, and let $r_{2}$ be the radius of the $P$-excircle of the triangle $APC$. Prove that the sum $r_{1}+r_{2}$ of these two radii remains constant when the point $P$ varies.
[i]Remark.[/i] The $P$-excircle of the triangle $APC$ is defined as the circle which touches the side $AC$ and the [i]extensions[/i] of the sides $AP$ and $CP$.
A circle is inscribed in the trapezoid [i]ABCD[/i]. Let [i]K, L, M, N[/i] be the points of tangency of this circle with the diagonals [i]AC[/i] and [i]BD[/i], respectively ([i]K[/i] is between [i]A[/i] and [i]L[/i], and [i]M[/i] is between [i]B[/i] and [i]N[/i]). Given that $AK\cdot LC=16$ and $BM\cdot ND=\frac94$, find the radius of the circle.
[color=red][Moderator edit: A solution of this problem can be found on http://www.ajorza.org/math/mathfiles/scans/belarus.pdf , page 20 (the statement of the problem is on page 6). The author of the problem is I. Voronovich.][/color]
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]
A trapezoid $ABCD$ is bicentral. The vertex $A$, the incenter $I$, the circumcircle $\omega$ and its center $O$ are given and the trapezoid is erased. Restore it using only a ruler.
Let $ABCD$ be a trapezoid of parallel bases $ BC$ and $AD$. If $\angle CAD = 2\angle CAB, BC = CD$ and $AC = AD$, determine all the possible values of the measure of the angle $\angle CAB$.
In the diagram, $ABDF$ is a trapezoid with $AF$ parallel to $BD$ and $AB$ perpendicular to $BD.$ The circle with center $B$ and radius $AB$ meets $BD$ at $C$ and is tangent to $DF$ at $E.$ Suppose that $x$ is equal to the area of the region inside quadrilateral $ABEF$ but outside the circle, that y is equal to the area of the region inside $\triangle EBD$ but outside the circle, and that $\alpha = \angle EBC.$ Prove that there is exactly one measure $\alpha,$ with $0^\circ \leq \alpha \leq 90^\circ,$ for which $x = y$ and that this value of $\frac 12 < \sin \alpha < \frac{1}{\sqrt 2}.$
[asy]
import graph; size(150); real lsf = 0.5; pen dp = linewidth(0.7) + fontsize(10); defaultpen(dp); pen ds = black; pen qqttff = rgb(0,0.2,1); pen fftttt = rgb(1,0.2,0.2);
draw(circle((6.04,2.8),1.78),qqttff); draw((6.02,4.58)--(6.04,2.8),fftttt); draw((6.02,4.58)--(6.98,4.56),fftttt); draw((6.04,2.8)--(8.13,2.88),fftttt); draw((6.98,4.56)--(8.13,2.88),fftttt);
dot((6.04,2.8),ds); label("$B$", (5.74,2.46), NE*lsf); dot((6.02,4.58),ds); label("$A$", (5.88,4.7), NE*lsf); dot((6.98,4.56),ds); label("$F$", (7.06,4.6), NE*lsf); dot((7.39,3.96),ds); label("$E$", (7.6,3.88), NE*lsf); dot((8.13,2.88),ds); label("$D$", (8.34,2.56), NE*lsf); dot((7.82,2.86),ds); label("$C$", (7.5,2.46), NE*lsf); clip((-4.3,-10.94)--(-4.3,6.3)--(16.18,6.3)--(16.18,-10.94)--cycle);
[/asy]
Circles $ C_1,C_2$ intersect at $ P$. A line $ \Delta$ is drawn arbitrarily from $ P$ and intersects with $ C_1,C_2$ at $ B,C$. What is locus of $ A$ such that the median of $ AM$ of triangle $ ABC$ has fixed length $ k$.
In trapezoid $ABCD$: $BC <AD, AB = CD, K$ is midpoint of $AD, M$ is midpoint of $CD, CH$ is height. Prove that lines $AM, CK$ and $BH$ intersect at one point.
A regular octagon $ ABCDEFGH$ has sides of length two. Find the area of $ \triangle{ADG}$.
$ \textbf{(A)}\ 4 \plus{} 2 \sqrt{2} \qquad
\textbf{(B)}\ 6 \plus{} \sqrt{2} \qquad
\textbf{(C)}\ 4 \plus{} 3 \sqrt{2} \qquad
\textbf{(D)}\ 3 \plus{} 4 \sqrt{2} \qquad
\textbf{(E)}\ 8 \plus{} \sqrt{2}$
Let $ABC$ be a triangle, let the $A$-altitude meet $BC$ at $D$, let the $B$-altitude meet $AC$ at $E$, and let $T\neq A$ be the point on the circumcircle of $ABC$ such that $AT || BC$. Given that $D,E,T$ are collinear, if $BD=3$ and $AD=4$, then the area of $ABC$ can be written as $a+\sqrt{b}$, where $a$ and $b$ are positive integers. What is $a+b$?
[i]2021 CCA Math Bonanza Individual Round #12[/i]
Let $ ABCD$ be a trapezoid with $ AB$ and $ CD$ parallel, $ \angle D \equal{} 2 \angle B, AD \equal{} 5,$ and $ CD \equal{} 2.$ Then $ AB$ equals
A. 7
B. 8
C. 13/2
D. 27/4
E. $ 5 \plus{} \frac{3 \sqrt{2}}{2}$
On each side of a unit square, an equilateral triangle of side length 1 is constructed. On each new side of each equilateral triangle, another equilateral triangle of side length 1 is constructed. The interiors of the square and the 12 triangles have no points in common. Let $ R$ be the region formed by the union of the square and all the triangles, and $ S$ be the smallest convex polygon that contains $ R$. What is the area of the region that is inside $ S$ but outside $ R$?
$ \textbf{(A)} \; \frac{1}{4} \qquad \textbf{(B)} \; \frac{\sqrt{2}}{4} \qquad \textbf{(C)} \; 1 \qquad \textbf{(D)} \; \sqrt{3} \qquad \textbf{(E)} \; 2 \sqrt{3}$
In acute triangle $ABC$, $AB > AC$.
$D$ and $E$ are the midpoints of $AB$, $AC$ respectively.
The circumcircle of $ADE$ intersects the circumcircle of $BCE$ again at $P$.
The circumcircle of $ADE$ intersects the circumcircle $BCD$ again at $Q$.
Prove that $AP = AQ$.
$ABCD$ is a trapezoid with $BC\parallel AD$ and $BC = 2AD$. Point $M$ is chosen on the side $CD$ such that $AB = AM$. Prove that $BM \perp CD$.
[i]Proposed by Bogdan Rublov[/i]
Let $ABCD$ be a trapezoid with bases $ AB$ and $CD$ $(AB>CD)$. Diagonals $AC$ and $BD$ intersect in point $ N$ and lines $AD$ and $BC$ intersect in point $ M$. The circumscribed circles of $ADN$ and $BCN$ intersect in point $ P$, different from point $ N$. Prove that the angles $AMP$ and $BMN$ are equal.
Vertex $ E$ of equilateral $ \triangle{ABE}$ is in the interior of unit square $ ABCD$. Let $ R$ be the region consisting of all points inside $ ABCD$ and outside $ \triangle{ABE}$ whose distance from $ \overline{AD}$ is between $ \frac{1}{3}$ and $ \frac{2}{3}$. What is the area of $ R$?
$ \textbf{(A)}\ \frac{12\minus{}5\sqrt3}{72} \qquad
\textbf{(B)}\ \frac{12\minus{}5\sqrt3}{36} \qquad
\textbf{(C)}\ \frac{\sqrt3}{18} \qquad
\textbf{(D)}\ \frac{3\minus{}\sqrt3}{9} \qquad
\textbf{(E)}\ \frac{\sqrt3}{12}$
$O$ is the circumcenter of acute $\Delta ABC$, $H$ is the Orthocenter. $AD \bot BC$, $EF$ is the perpendicular bisector of $AO$,$D,E$ on the $BC$. Prove that the circumcircle of $\Delta ADE$ through the midpoint of $OH$.
A convex quadrilateral $ABCD$ is inscribed in a circle with center $O$. The diagonals $AC$, $BD$ of $ABCD$ meet at $P$. Circumcircles of $\triangle ABP$ and $\triangle CDP$ meet at $P$ and $Q$ ($O,P,Q$ are pairwise distinct). Show that $\angle OQP=90^{\circ}$.
Two planes, $P$ and $Q$, intersect along the line $p$. The point $A$ is given in the plane $P$, and the point $C$ in the plane $Q$; neither of these points lies on the straight line $p$. Construct an isosceles trapezoid $ABCD$ (with $AB \parallel CD$) in which a circle can be inscribed, and with vertices $B$ and $D$ lying in planes $P$ and $Q$ respectively.
Let $A_1A_2 \cdots A_{10}$ be a regular decagon such that $[A_1A_4]=b$ and the length of the circumradius is $R$. What is the length of a side of the decagon?
$
\textbf{a)}\ b-R
\qquad\textbf{b)}\ b^2-R^2
\qquad\textbf{c)}\ R+\dfrac b2
\qquad\textbf{d)}\ b-2R
\qquad\textbf{e)}\ 2b-3R
$
Prove that in an isosceles trapezoid circumscibed around a circle, the segments connecting the points of tangency of opposite sides with the circle pass through the point of intersection of the diagonals.
Let $ABCS$ be an isosceles trapezoid. Denote $A',B',C',D'$ the incenters of triangles $BCD,CDA,$ $DAB,ABC,$ respectively. Show that $A',B',C',D'$ are vertices of a rectangle.
Let $ABCD$ be a trapezoid with $AD\parallel BC$. A point $M $ is chosen inside the trapezoid, and a point $N$ is chosen inside the triangle $BMC$ such that $AM\parallel CN$, $BM\parallel DN$. Prove that triangles $ABN$ and $CDM$ have equal areas.
Quadrilateral $ABCD$ is circumscribed around the circle with centre $I$. Let points $M$ and $N$ be the midpoints of sides $AB$ and $CD$ respectively and let $\frac{IM}{AB} = \frac{IN}{CD}$. Prove that $ABCD$ is either a trapezoid or a parallelogram.