Found problems: 1049
$ABCD$ is convex quadrilateral. Circumcircle of $ABC$ intersect $AD$ and $DC$ at points $P$ and $Q$. Circumcircle of $ADC$ intersect $AB$ and $BC$ at points $S$ and $R$. Prove that if $PQRS$ is parallelogram then $ABCD$ is parallelogram
On the sides $AB$ and $BC$ arbitrarily mark points $M$ and $N$, respectively. Let $P$ be the point of intersection of segments $AN$ and $BM$. In addition, we note the points $Q$ and $R$ such that quadrilaterals $MCNQ$ and $ACBR$ are parallelograms. Prove that the points $P,Q$ and $R$ lie on one line.
Two circles intersect at two points $A$ and $B$. A line $\ell$ which passes through the point $A$ meets the two circles again at the points $C$ and $D$, respectively. Let $M$ and $N$ be the midpoints of the arcs $BC$ and $BD$ (which do not contain the point $A$) on the respective circles. Let $K$ be the midpoint of the segment $CD$. Prove that $\measuredangle MKN = 90^{\circ}$.
Let $ABCD$ be a parallelogram. $P \in (CD), Q \in (AB)$, $M= AP \cap DQ$, $N=BP \cap CQ$, $ K=MN \cap AD$, $L= MN \cap BC$. Prove that $BL=DK$.
Let $\triangle ABC$ be a triangle with $BC = 4, CA= 5, AB= 6$, and let $O$ be the circumcenter of $\triangle ABC$. Let $O_b$ and $O_c$ be the reflections of $O$ about lines $CA$ and $AB$ respectively. Suppose $BO_b$ and $CO_c$ intersect at $T$, and let $M$ be the midpoint of $BC$. Given that $MT^2 = \frac{p}{q}$ for some coprime positive integers $p$ and $q$, find $p+q$.
[i]Proposed by Sreejato Bhattacharya[/i]
(A.Myakishev) Let triangle $ A_1B_1C_1$ be symmetric to $ ABC$ wrt the incenter of its medial triangle. Prove that the orthocenter of $ A_1B_1C_1$ coincides with the circumcenter of the triangle formed by the excenters of $ ABC$.
Let $ABXC$ be a parallelogram. Points $K,P,Q$ lie on $\overline{BC}$ in this order such that $BK = \frac{1}{3} KC$ and $BP = PQ = QC = \frac{1}{3} BC$. Rays $XP$ and $XQ$ meet $\overline{AB}$ and $\overline{AC}$ at $D$ and $E$, respectively. Suppose that $\overline{AK} \perp \overline{BC}$, $EK-DK=9$ and $BC=60$. Find $AB+AC$.
[i]Proposed by Evan Chen[/i]
On side, $BC, AB$ of a parallelogram $ABCD$ lie points $M,N$ respectively such that $|AM| =|CN|$. Let $P$ be the intersection of $AM$ and $CN$. Prove that the angle bisector of $\angle APC$ passes through $D$.
Given is a quadrilateral $ABCD$ in which we can inscribe circle. The segments $AB, BC, CD$ and $DA$ are the diameters of the circles $o1, o2, o3$ and $o4$, respectively. Prove that there exists a circle tangent to all of the circles $o1, o2, o3$ and $o4$.
Let $ABC$ be a triangle such that it's circumcircle radius is equal to the radius of outer inscribed circle with respect to $A$.
Suppose that the outer inscribed circle with respect to $A$ touches $BC,AC,AB$ at $M,N,L$.
Prove that $O$ (Center of circumcircle) is the orthocenter of $MNL$.
Let $ABCD$ be a tetragonal.
[b](a)[/b] If the plane $(P)$ cuts $ABCD,$ find the necessary and sufficient condition such that the area formed from the intersection of the plane $(P)$ and the tetragonal be a parallelogram. Prove that the problem has three solutions in this case.
[b](b)[/b] Consider one of the solutions of [b](a)[/b]. Find the situation of the plane $(P)$ for which the parallelogram has maximum area.
[b](c)[/b] Find a plane $(P)$ for which the parallelogram be a lozenge and then find the length side of his lozenge in terms of the length of the edges of $ABCD.$
$ABCD$ is a parallelogram. $K$ is the circumcircle of $ABD$. The lines $BC$ and $CD$ meet $K$ again at $E$ and $F$. Show that the circumcenter of $CEF$ lies on $K$.
Let $B$ and $C$ be two fixed points on a circle centered at $O$ that are not diametrically opposed. Let $A$ be a variable point on the circle distinct from $B$ and $C$ and not belonging to the perpendicular bisector of $BC$. Let $H$ be the orthocenter of $\triangle ABC$, and $M$ and $N$ be the midpoints of the segments $BC$ and $AH$, respectively. The line $AM$ intersects the circle again at $D$, and finally, $NM$ and $OD$ intersect at $P$. Determine the locus of points $P$ as $A$ moves around the circle.
The area of the shaded region $\text{BEDC}$ in parallelogram $\text{ABCD}$ is
[asy]
unitsize(10);
pair A,B,C,D,E;
A=origin; B=(4,8); C=(14,8); D=(10,0); E=(4,0);
draw(A--B--C--D--cycle);
fill(B--E--D--C--cycle,gray);
label("A",A,SW); label("B",B,NW); label("C",C,NE); label("D",D,SE); label("E",E,S);
label("$10$",(9,8),N); label("$6$",(7,0),S); label("$8$",(4,4),W);
draw((3,0)--(3,1)--(4,1));
[/asy]
$\text{(A)}\ 24 \qquad \text{(B)}\ 48 \qquad \text{(C)}\ 60 \qquad \text{(D)}\ 64 \qquad \text{(E)}\ 80$
$C_0:x^2+y^2=1,C_1:\frac{x^2}{a^2}+\frac{y^2}{b^2}(a>b>0)$. Find all $(a,b)$ such that for any point $P$ on $C_1$, we can find a parallelogram with an apex $P$, and it is externally tangent to $C_0$, inscribed to $C_1$.
Laura won the local math olympiad and was awarded a "magical" ruler. With it, she can draw (as usual) lines in the plane, and she can also measure segments and replicate them anywhere in the plane; but she can also divide a segment into as many equal parts as she wishes; for instance, she can divide any segment into $17$ equal parts. Laura drew a parallelogram $ABCD$ and decided to try out her magical ruler; with it, she found the midpoint $M$ of side $CD$, and she extended $CB$ beyond $B$ to point $N$ so that segments $CB$ and $BN$ were equal in length. Unfortunately, her mischievous little brother came along and erased everything on Laura's picture except for points $A, M$, and $N$. Using Laura's magical ruler, help her reconstruct the original parallelogram $ABCD$: write down the steps that she needs to follow and prove why this will lead to reconstructing the original parallelogram $ABCD$.
In a parallelogram $ABCD$ , $BD$ is the largest diagonal. By matching $B$ with $D$ by a bend, a regular pentagon is formed. Calculate the measures of the angles formed by the diagonal $BD$ with each of the sides of the parallelogram.
Circles $ \omega_{1}$ and $ \omega_{2}$ intersect at points $ A$ and $ B.$ Points $ C$ and $ D$ are on circles $ \omega_{1}$ and $ \omega_{2},$ respectively,
such that lines $ AC$ and $ AD$ are tangent to circles $ \omega_{2}$ and $ \omega_{1},$ respectively.
Let $ I_{1}$ and $ I_{2}$ be the incenters of triangles $ ABC$ and $ ABD,$ respectively. Segments $ I_{1}I_{2}$ and $ AB$ intersect at $ E$.
Prove that: $ \frac {1}{AE} \equal{} \frac {1}{AC} \plus{} \frac {1}{AD}$
Let $ABCD$ be a parallelogram. Suppose a line passing through $C$ and lying outside the parallelogram meets $AB$ and $AD$ produced at $E$ and $F$ respectively. Show that \[ AC^2 + CE \cdot CF = AB \cdot AE + AD \cdot AF . \]
On the sides $AB,AD$ of the rhombus $ABCD$ are the points $E,F$ such that $AE=DF$. The lines $BC,DE$ intersect at $P$ and $CD,BF$ intersect at $Q$. Prove that:
(a) $\frac{PE}{PD} + \frac{QF}{QB} = 1$;
(b) $P,A,Q$ are collinear.
[i]Virginia Tica, Vasile Tica[/i]
Let $B$ and $C$ be two fixed points on a circle centered at $O$ that are not diametrically opposed. Let $A$ be a variable point on the circle distinct from $B$ and $C$ and not belonging to the perpendicular bisector of $BC$. Let $H$ be the orthocenter of $\triangle ABC$, and $M$ and $N$ be the midpoints of the segments $BC$ and $AH$, respectively. The line $AM$ intersects the circle again at $D$, and finally, $NM$ and $OD$ intersect at $P$. Determine the locus of points $P$ as $A$ moves around the circle.
Let $ ABC $ be an acute triangle. Denote by $ D $ the foot of the perpendicular line drawn from the point $ A $ to the side $ BC $, by $M$ the midpoint of $ BC $, and by $ H $ the orthocenter of $ ABC $. Let $ E $ be the point of intersection of the circumcircle $ \Gamma $ of the triangle $ ABC $ and the half line $ MH $, and $ F $ be the point of intersection (other than $E$) of the line $ ED $ and the circle $ \Gamma $. Prove that $ \tfrac{BF}{CF} = \tfrac{AB}{AC} $ must hold.
(Here we denote $XY$ the length of the line segment $XY$.)
In an acute scalene triangle $ABC$ the bisector of the acute angle between the altitudes $AA_1$ and $CC_1$ meets the sides $AB$ and $BC$ at $P$ and $Q$ respectively. The bisector of the angle $B$ intersects the segment joining the orthocenter of $ABC$ and the midpoint of $AC$ at point $R$. Prove that $P$, $B$, $Q$, $R$ lie on a circle.
Let $ABCD$ be a parallelogram in the plane. We draw two circles of radius $R$, one through the points $A$ and
$B$, the other through $B$ and $C$. Let $E$ be the other intersection point of the circles. We assume that $E$ is not a vertex of the parallelogram. Show that the circle passing through $A, D$, and $E$ also has radius $R$.
Let $ABC$ be a triangle and let $M$ be the midpoint $BC$. On the exterior of the triangle, consider the parallelogram $BCDE$ such that $BE//AM$ and $BE=AM/2$ . Prove that line $EM$ passes through the midpoint of segment $AD$.