Found problems: 107
Let $ABCD$ be a convex quadrilateral such that the circle with diameter $AB$ is tangent to the line $CD$, and the circle with diameter $CD$ is tangent to the line $AB$. Prove that the two intersection points of these circles and the point $AC \cap BD$ are collinear.
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.
Let $ABCD$ be a convex quadrilateral with these properties: $\angle ADC = 135^o$ and $\angle ADB - \angle ABD = 2\angle DAB = 4\angle CBD$.
If $BC = \sqrt2 CD$ , prove that $AB = BC + AD$.
by Mahdi Etesami Fard
Let $ABCD$ a convex quadrilateral with $[BC]$ and $[CD]$'s midpoint is $P$ and $N$ respectively. If
$$[AP]+[AN]=d$$
Show that, area of the $ABCD$ is less then $\frac{1}{2}d^2$
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 $CL,DF; DF,AH; AH,BJ; BJ,CL$ are denoted by $P_1,Q_1,R_1, S_1$, respectively, and the intersection points of the pairs of straight lines $AI,BK; BK,CE;$ $ CE,DG; DG,AI$ are denoted by $P_2,Q_2,R_2, S_2$, respectively. Prove that $P_1Q_1R_1S_1 \cong P_2Q_2R_2S_2.$
Five different points $O,A,B,C,D$ are given in plane such that \[OA\le OB\le OC\le OD.\] Show that for area $P$ of any convex quadrilateral with vertices $A,B,C,D$ (not necessarily in this order) the inequality \[P\le \frac12(OA+OD)(OB+OC)\] holds and determine when equality occurs.
Circle $\omega$ is inscribed in quadrilateral $ABCD$ such that $AB$ and $CD$ are not parallel and
intersect at point $O.$ Circle $\omega_1$ touches the side $BC$ at $K$ and touches line $AB$ and $CD$ at
points which are located outside quadrilateral $ABCD;$ circle $\omega_2$ touches side $AD$ at $L$ and
touches line $AB$ and $CD$ at points which are located outside quadrilateral $ABCD.$ If $O,K,$
and $L$ are collinear$,$ then show that the midpoint of side $BC,AD,$ and the center of circle
$\omega$ are also collinear.
A point $P$ lies in the interior of the triangle $ABC$. The lines $AP, BP$, and $CP$ intersect $BC, CA$, and $AB$ at points $D, E$, and $F$, respectively. Prove that if two of the quadrilaterals $ABDE, BCEF, CAFD, AEPF, BFPD$, and $CDPE$ are concyclic, then all six are concyclic.
In the quadrilateral $ABCD$, the length of the sides $AB$ and $BC$ is equal to $1, \angle B= 100^o , \angle D= 130^o$ . Find the length of $BD$.
The point $M$ is inside the convex quadrilateral $ABCD$, such that
\[ MA = MC, \hspace{0,2cm} \widehat{AMB} = \widehat{MAD} + \widehat{MCD} \quad \textnormal{and} \quad \widehat{CMD} = \widehat{MCB} + \widehat{MAB}. \]
Prove that $AB \cdot CM = BC \cdot MD$ and $BM \cdot AD = MA \cdot CD.$
Quadrilateral $ABCD$ satisfies $\angle ABC = \angle ACD = 90^{\circ}, AC = 20$, and $CD = 30$. Diagonals $\overline{AC}$ and $\overline{BD}$ intersect at point $E$, and $AE = 5$. What is the area of quadrilateral $ABCD$?
$\textbf{(A) } 330 \qquad\textbf{(B) } 340 \qquad\textbf{(C) } 350 \qquad\textbf{(D) } 360 \qquad\textbf{(E) } 370$
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 $ABCD$ be a convex quadrilateral such that the circle with diameter $AB$ is tangent to the line $CD$, and the circle with diameter $CD$ is tangent to the line $AB$. Prove that the two intersection points of these circles and the point $AC \cap BD$ are collinear.
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]
Let $\mathcal{P}$ be a convex quadrilateral. Consider a point $X$ inside $\mathcal{P}.$ Let $M,N,P,Q$ be the projections of $X$ on the sides of $\mathcal{P}.$ We know that $M,N,P,Q$ all sit on a circle of center $L.$ Let $J$ and $K$ be the midpoints of the diagonals of $\mathcal{P}.$ Prove that $J,K$ and $L$ lie on a line.
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 $TSH$.
Show that an arbitrary quadrilateral can be divided into nine isosceles triangles.
Let ABCD be a convex quadrilateral and O a point inside it. Let the parallels to the lines BC, AB, DA, CD through the point O meet the sides AB, BC, CD, DA of the quadrilateral ABCD at the points E, F, G, H, respectively. Then, prove that $ \sqrt {\left|AHOE\right|} \plus{} \sqrt {\left|CFOG\right|}\leq\sqrt {\left|ABCD\right|}$, where $ \left|P_1P_2...P_n\right|$ is an abbreviation for the non-directed area of an arbitrary polygon $ P_1P_2...P_n$.
Let $ABCD$ be a convex quadrilateral with $\angle DAB=\angle BCD=90^{\circ}$ and $\angle ABC> \angle CDA$. Let $Q$ and $R$ be points on segments $BC$ and $CD$, respectively, such that line $QR$ intersects lines $AB$ and $AD$ at points $P$ and $S$, respectively. It is given that $PQ=RS$.Let the midpoint of $BD$ be $M$ and the midpoint of $QR$ be $N$.Prove that the points $M,N,A$ and $C$ lie on a circle.
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]
Let $ABCD$ be a convex quadrilateral such that $\angle{BAD} = 90^{\circ}$ and its diagonals $AC$ and $BD$ are perpendicular. Let $M$ be the midpoint of side $CD$, and $E$ be the intersection of $BM$ and $AC$. Let $F$ be a point on side $AD$ such that $BM$ and $EF$ are perpendicular. If $CE = AF\sqrt{2}$ and $FD = CE\sqrt{2}$, show that $ABCD$ is a square.
Let $A$, $B$, $C$, $D$ be four points in the plane, with $C$ and $D$ on the same side of the line $AB$, such that $AC \cdot BD = AD \cdot BC$ and $\angle ADB = 90^{\circ}+\angle ACB$. Find the ratio
\[\frac{AB \cdot CD}{AC \cdot BD}, \]
and prove that the circumcircles of the triangles $ACD$ and $BCD$ are orthogonal. (Intersecting circles are said to be orthogonal if at either common point their tangents are perpendicuar. Thus, proving that the circumcircles of the triangles $ACD$ and $BCD$ are orthogonal is equivalent to proving that the tangents to the circumcircles of the triangles $ACD$ and $BCD$ at the point $C$ are perpendicular.)
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]
Let $ABCD$ be a convex quadrilateral. We assume that there exists a point $P$ inside the quadrilateral such that
the areas of the triangles $ABP, BCP, CDP$, and $DAP$ are equal. Show that at least one of the diagonals of the
quadrilateral bisects the other diagonal.
Circle $\omega$ is inscribed in quadrilateral $ABCD$ such that $AB$ and $CD$ are not parallel and
intersect at point $O.$ Circle $\omega_1$ touches the side $BC$ at $K$ and touches line $AB$ and $CD$ at
points which are located outside quadrilateral $ABCD;$ circle $\omega_2$ touches side $AD$ at $L$ and
touches line $AB$ and $CD$ at points which are located outside quadrilateral $ABCD.$ If $O,K,$
and $L$ are collinear$,$ then show that the midpoint of side $BC,AD,$ and the center of circle
$\omega$ are also collinear.