Found problems: 107
Point $ P$ lies on side $ AB$ of a convex quadrilateral $ ABCD$. Let $ \omega$ be the incircle of triangle $ CPD$, and let $ I$ be its incenter. Suppose that $ \omega$ is tangent to the incircles of triangles $ APD$ and $ BPC$ at points $ K$ and $ L$, respectively. Let lines $ AC$ and $ BD$ meet at $ E$, and let lines $ AK$ and $ BL$ meet at $ F$. Prove that points $ E$, $ I$, and $ F$ are collinear.
[i]Author: Waldemar Pompe, Poland[/i]
Given a trapezoid, divide it by a line into two quadrilaterals in such a way that both of them are cyclic with the same circumradius. Discuss conditions of solvability.
The vertices of a quadrilateral lie on the graph of $y = \ln x$, and the $x$-coordinates of these vertices are consecutive positive integers. The area of the quadrilateral is $\ln \frac{91}{90}$. What is the $x$-coordinate of the leftmost vertex?
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 13$
The diagonals of a quadrilateral $ ABCD$ are perpendicular: $ AC \perp BD.$ Four squares, $ ABEF,BCGH,CDIJ,DAKL,$ are erected externally on its sides. The intersection points of the pairs of straight lines $ CL, DF, AH, BJ$ are denoted by $ P_1,Q_1,R_1, S_1,$ respectively (left figure), and the intersection points of the pairs of straight lines $ AI, BK, CE DG$ are denoted by $ P_2,Q_2,R_2, S_2,$ respectively (right figure). Prove that $ P_1Q_1R_1S_1 \cong P_2Q_2R_2S_2$ where $ P_1,Q_1,R_1, S_1$ and $ P_2,Q_2,R_2, S_2$ are the two quadrilaterals.
[i]Alternative formulation:[/i] Outside a convex quadrilateral $ ABCD$ with perpendicular diagonals, four squares $ AEFB, BGHC, CIJD, DKLA,$ are constructed (vertices are given in counterclockwise order). Prove that the quadrilaterals $ Q_1$ and $ Q_2$ formed by the lines $ AG, BI, CK, DE$ and $ AJ, BL, CF, DH,$ respectively, are congruent.
Let $ABCD$ be a convex quadrilateral with non perpendicular diagonals and with the sides $AB$ and $CD$ non parallel . Denote by $O$ the intersection of the diagonals , $H_1$ the orthocenter of the triangle $AOB$ and $H_2$ the orthocenter of the triangle $COD$ . Also denote with $M$ the midpoint of the side $AB$ and with $N$ the midpoint of the side $CD$ . Prove that $H_1H_2$ and $MN$ are parallel if and only if $AC=BD$
Convex quadrilateral $ABCD$ has $\angle ABC = \angle CDA = 90^{\circ}$. Point $H$ is the foot of the perpendicular from $A$ to $BD$. Points $S$ and $T$ lie on sides $AB$ and $AD$, respectively, such that $H$ lies inside triangle $SCT$ and \[
\angle CHS - \angle CSB = 90^{\circ}, \quad \angle THC - \angle DTC = 90^{\circ}. \] Prove that line $BD$ is tangent to the circumcircle of triangle $TSH$.
$(HUN 3)$ In the plane $4000$ points are given such that each line passes through at most $2$ of these points. Prove that there exist $1000$ disjoint quadrilaterals in the plane with vertices at these points.
Let $ABCM$ be a quadrilateral and $D$ be an interior point such that $ABCD$ is a parallelogram. It is known that $\angle AMB =\angle CMD$. Prove that $\angle MAD =\angle MCD$.
Let $ABCD$ be a quadrilateral inscribed in circle $\omega$. Define $E = AA \cap CD$, $F = AA \cap BC$, $G = BE \cap \omega$, $H = BE \cap AD$, $I = DF \cap \omega$, and $J = DF \cap AB$. Prove that $GI$, $HJ$, and the $B$-symmedian are concurrent.
[i]Proposed by Robin Park[/i]
In a convex quadrilateral $ABCD, AB = 10, BC = 14, CD = 11$ and $DA = 5$. Determine the angle between its diagonals.
On a sheet of transparent paper, draw a quadrilateral with Chinese ink, which is illuminated with a lamp. Show that it is always possible to locate the sheet in such a way that the shadow projected on the desk is a parallelogram.
$ ABCD$ is a quadrilateral with $ BC$ parallel to $ AD$. $ M$ is the midpoint of $ CD$, $ P$ is the midpoint of $ MA$ and $ Q$ is the midpoint of $ MB$. The lines $ DP$ and $ CQ$ meet at $ N$. Prove that $ N$ is inside the quadrilateral $ ABCD$.
Let $ABCD$ be a cyclic quadrilateral. Let $P$, $Q$, $R$ be the feet of the perpendiculars from $D$ to the lines $BC$, $CA$, $AB$, respectively. Show that $PQ=QR$ if and only if the bisectors of $\angle ABC$ and $\angle ADC$ are concurrent with $AC$.
A bicentric quadrilateral is one that is both inscribable in and circumscribable about a circle, i.e. both the incircle and circumcircle exists. Show that for such a quadrilateral, the centers of the two associated circles are collinear with the point of intersection of the diagonals.
In a regular 100-gon, 41 vertices are colored black and the remaining 59 vertices are colored white. Prove that there exist 24 convex quadrilaterals $Q_{1}, \ldots, Q_{24}$ whose corners are vertices of the 100-gon, so that
[list]
[*] the quadrilaterals $Q_{1}, \ldots, Q_{24}$ are pairwise disjoint, and
[*] every quadrilateral $Q_{i}$ has three corners of one color and one corner of the other color.
[/list]
There is given a convex quadrilateral $ ABCD$. Prove that there exists a point $ P$ inside the quadrilateral such that
\[
\angle PAB \plus{} \angle PDC \equal{} \angle PBC \plus{} \angle PAD \equal{} \angle PCD \plus{} \angle PBA \equal{} \angle PDA \plus{} \angle PCB = 90^{\circ}
\]
if and only if the diagonals $ AC$ and $ BD$ are perpendicular.
[i]Proposed by Dusan Djukic, Serbia[/i]
Let be two quadrilaterals $ ABCD,A'B'C'D' $ with $ AB,BC,CD,AC,BD $ being perpendicular to $ A'B',B'C',C'D',A'C',B'D', $ respectively. Show that $ AD $ is perpendicular to $ A'D'. $
The quadrilateral $ABCD$ satisfies $\angle ACD = 2\angle CAB, \angle ACB = 2\angle CAD $ and $CB = CD.$ Show that $$\angle CAB=\angle CAD.$$
There is given a convex quadrilateral $ ABCD$. Prove that there exists a point $ P$ inside the quadrilateral such that
\[
\angle PAB \plus{} \angle PDC \equal{} \angle PBC \plus{} \angle PAD \equal{} \angle PCD \plus{} \angle PBA \equal{} \angle PDA \plus{} \angle PCB = 90^{\circ}
\]
if and only if the diagonals $ AC$ and $ BD$ are perpendicular.
[i]Proposed by Dusan Djukic, Serbia[/i]
In convex quadrilateral $KLMN$ side $\overline{MN}$ is perpendicular to diagonal $\overline{KM}$, side $\overline{KL}$ is perpendicular to diagonal $\overline{LN}$, $MN = 65$, and $KL = 28$. The line through $L$ perpendicular to side $\overline{KN}$ intersects diagonal $\overline{KM}$ at $O$ with $KO = 8$. Find $MO$.
Find whether among all quadrilaterals, whose interiors lie inside a semi-circle of radius $r$, there exist one (or more) with maximum area. If so, determine their shape and area.
Let $ ABCD$ be a convex quadrilateral and let $ P$ and $ Q$ be points in $ ABCD$ such that $ PQDA$ and $ QPBC$ are cyclic quadrilaterals. Suppose that there exists a point $ E$ on the line segment $ PQ$ such that $ \angle PAE \equal{} \angle QDE$ and $ \angle PBE \equal{} \angle QCE$. Show that the quadrilateral $ ABCD$ is cyclic.
[i]Proposed by John Cuya, Peru[/i]
Let $ ABCD$ be a convex quadrilateral and let $ P$ and $ Q$ be points in $ ABCD$ such that $ PQDA$ and $ QPBC$ are cyclic quadrilaterals. Suppose that there exists a point $ E$ on the line segment $ PQ$ such that $ \angle PAE \equal{} \angle QDE$ and $ \angle PBE \equal{} \angle QCE$. Show that the quadrilateral $ ABCD$ is cyclic.
[i]Proposed by John Cuya, Peru[/i]
Let $ ABCD$ be a convex quadrilateral and let $ P$ and $ Q$ be points in $ ABCD$ such that $ PQDA$ and $ QPBC$ are cyclic quadrilaterals. Suppose that there exists a point $ E$ on the line segment $ PQ$ such that $ \angle PAE \equal{} \angle QDE$ and $ \angle PBE \equal{} \angle QCE$. Show that the quadrilateral $ ABCD$ is cyclic.
[i]Proposed by John Cuya, Peru[/i]
Convex quadrilateral $ABCD$ has $\angle ABC = \angle CDA = 90^{\circ}$. Point $H$ is the foot of the perpendicular from $A$ to $BD$. Points $S$ and $T$ lie on sides $AB$ and $AD$, respectively, such that $H$ lies inside triangle $SCT$ and \[
\angle CHS - \angle CSB = 90^{\circ}, \quad \angle THC - \angle DTC = 90^{\circ}. \] Prove that line $BD$ is tangent to the circumcircle of triangle $TSH$.