Found problems: 1049
The area of a convex quadrilateral is $S$, and the angle between the diagonals is $a$. On the sides of this quadrilateral, as on the bases, isosceles triangles with vertex angle equal to $\phi$, wherein two opposite triangles are located on the other side of the corresponding side of the quadrilateral than the quadrilateral itself, and the other two are located on the other side. Prove that the vertices of the constructed triangles, different from the vertices of the quadrilateral, serve as the vertices of a parallelogram. Find the area of this parallelogram.
In trapezoid $ABCD$, $AD$ is perpendicular to $DC$, $AD=AB=3$, and $DC=6$. In addition, E is on $DC$, and $BE$ is parallel to $AD$. Find the area of $\Delta BEC$.
[asy]
defaultpen(linewidth(0.7));
pair A=(0,3), B=(3,3), C=(6,0), D=origin, E=(3,0);
draw(E--B--C--D--A--B);
draw(rightanglemark(A, D, C));
label("$A$", A, NW);
label("$B$", B, NW);
label("$C$", C, SE);
label("$D$", D, SW);
label("$E$", E, NW);
label("$3$", A--D, W);
label("$3$", A--B, N);
label("$6$", E, S);[/asy]
$\textbf{(A)} \: 3\qquad \textbf{(B)} \: 4.5\qquad \textbf{(C)} \: 6\qquad \textbf{(D)} \: 9\qquad \textbf{(E)} \: 18\qquad $
Let $M$ be the midpoint of the side $BC$ of a acute triangle $ABC$. Incircle of the triangle $ABM$ is tangent to the side $AB$ at the point $D$. Incircle of the triangle $ACM$ is tangent to the side $AC$ at the point $E$. Let $F$ be the such point, that the quadrilateral $DMEF$ is a parallelogram. Prove that $F$ lies on the bisector of $\angle BAC$.
$ \Delta ABC$ is a triangle such that $ AB \neq AC$. The incircle of $ \Delta ABC$ touches $ BC, CA, AB$ at $ D, E, F$ respectively. $ H$ is a point on the segment $ EF$ such that $ DH \bot EF$. Suppose $ AH \bot BC$, prove that $ H$ is the orthocentre of $ \Delta ABC$.
Remark: the original question has missed the condition $ AB \neq AC$
A parallelogram $ABCD$ is given ($AB \neq BC$). Points $E$ and $G$ are chosen on the line $\overline{CD}$ such that $\overline{AC}$ is the angle bisector of both angles $\angle EAD$ and $\angle BAG$. The line $\overline{BC}$ intersects $\overline{AE}$ and $\overline{AG}$ at $F$ and $H$, respectively. Prove that the line $\overline{FG}$ passes through the midpoint of $HE$.
[i]Proposed by Mahdi Etesamifard[/i]
Let $ABCD$ be a parallelogram, and let $P$ be a point on the side $AB$. Let the line through $P$ parallel to $BC$ intersect the diagonal $AC$ at point $Q$. Prove that $$|DAQ|^2 = |PAQ| \times |BCD| ,$$ where $|XY Z|$ denotes the area of triangle $XY Z$.
The diagonals of a convex quadrilateral divide it into four triangles. The radii of the circles circumscribed around these triangles are equal. Can such a property have a quadrilateral other than:
a) parallelogram,
b) rhombus?
(Sharygin Igor)
Given a quadrangle $ABCD$ and a point $M$ inside it such that $ABMD$ is a parallelogram. $ \angle CBM = \angle CDM$. Prove that the $ \angle ACD = \angle BCM$.
Let $ABCD$ be a quadrilateral such that $\angle ABC = \angle ADC = 90º$ and $\angle BCD$ > $90º$. Let $P$ be a point inside of the $ABCD$ such that $BCDP$ is parallelogram, the line $AP$ intersects $BC$ in $M$. If $BM = 2, MC = 5, CD = 3$. Find the length of $AM$.
The points $P$ and $Q$ lie on the sides $BC$ and $CD$ of the parallelogram $ABCD$ so that $BP = QD$. Show that the intersection point between the lines $BQ$ and $DP$ lies on the line bisecting $\angle BAD$.
Let $ ABCD $ be a convex quadrilateral, and $ P $ be a point that isn't found on any of the lines formed by the sides of the quadrilateral. Prove that the centers of mass of the triangles $ PAB, PBC, PCD $ and $ PDA, $ form a parallelogram, and calculate the legths of its sides in terms of its diagonals.
in an acute triangle $ABC$,$D$ is a point on $BC$,let $Q$ be the intersection of $AD$ and the median of $ABC$from $C$,$P$ is a point on $AD$,distinct from $Q$.the circumcircle of $CPD$ intersects $CQ$ at $C$ and $K$.prove that the circumcircle of $AKP$ passes through a fixed point differ from $A$.
Let $P$ be a variable point on arc $BC$ of the circumcircle of triangle $ABC$ not containing $A$. Let $I_1$ and $I_2$ be the incenters of the triangles $PAB$ and $PAC$, respectively. Prove that:
[b](a)[/b] The circumcircle of $?PI_1I_2$ passes through a fixed point.
[b](b)[/b] The circle with diameter $I_1I_2$ passes through a fixed point.
[b](c)[/b] The midpoint of $I_1I_2$ lies on a fixed circle.
Let $ABCD$ be a parallelogram ($AB < BC$). The bisector of the angle $BAD$ intersects the side $BC$ at the point K; and the bisector of the angle $ADC$ intersects the diagonal $AC$ at the point $F$. Suppose that $KD \perp BC$. Prove that $KF \perp BD$.
Let two chords $AC$ and $BD$ of a circle $k$ meet at the point $K$, and let $O$ be the center of $k$. Let $M$ and $N$ be the circumcenters of triangles $AKB$ and $CKD$. Show that the quadrilateral $OMKN$ is a parallelogram.
$ABC$ is an isosceles triangle with $AB=AC$ and the angle in $A$ is less than $60^{\circ}$. Let $D$ be a point on $AC$ such that $\angle{DBC}=\angle{BAC}$. $E$ is the intersection between the perpendicular bisector of $BD$ and the line parallel to $BC$ passing through $A$. $F$ is a point on the line $AC$ such that $FA=2AC$ ($A$ is between $F$ and $C$).
Show that $EB$ and $AC$ are parallel and that the perpendicular from $F$ to $AB$, the perpendicular from $E$ to $AC$ and $BD$ are concurrent.
Let $P$ be an interior point of the parallelogram $ABCD$. Prove that $\angle APB+ \angle CPD = 180^\circ$ if and only if $\angle PDC = \angle PBC$.
$(BEL 2) (a)$ Find the equations of regular hyperbolas passing through the points $A(\alpha, 0), B(\beta, 0),$ and $C(0, \gamma).$
$(b)$ Prove that all such hyperbolas pass through the orthocenter $H$ of the triangle $ABC.$
$(c)$ Find the locus of the centers of these hyperbolas.
$(d)$ Check whether this locus coincides with the nine-point circle of the triangle $ABC.$
A convex quadrangle without parallel sides is given. For each triple of its vertices, a point is constructed that supplements this triple to a parallelogram, one of the diagonals of which coincides with the diagonal of the quadrangle. Prove that of the four points constructed, exactly one lies inside the original quadrangle.
$ABCD$ is parallelogram, $M$ is the midpoint of side $CD$ and $H$ is the foot of the perpendicular from $B$ to line $AM$. Prove that $BCH$ is an isosceles triangle.
(M Volchkevich)
Suppose that $ABCD$ is a parallelogram such that $DAB>90$. Let the point $H$ to be on $AD$ such that $BH$ is perpendicular to $AD$. Let the point $M$ to be the midpoint of $AB$. Let the point $K$ to be the intersecting point of the line $DM$ with the circumcircle of $ADB$. Prove that $HKCD$ is concyclic.
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]
Consider $\triangle OAB$ with acute angle $AOB$. Thorugh a point $M \neq O$ perpendiculars are drawn to $OA$ and $OB$, the feet of which are $P$ and $Q$ respectively. The point of intersection of the altitudes of $\triangle OPQ$ is $H$. What is the locus of $H$ if $M$ is permitted to range over
a) the side $AB$;
b) the interior of $\triangle OAB$.
The parallelogram $ABCD$ has $AB=a,AD=1,$ $\angle BAD=A$, and the triangle $ABD$ has all angles acute. Prove that circles radius $1$ and center $A,B,C,D$ cover the parallelogram if and only
\[a\le\cos A+\sqrt3\sin A.\]
In parallelogram $ABCD$ holds $AB=BD$. Let $K$ be a point on $AB$, different from $A$, such that $KD=AD$. Let $M$ be a point symmetric to $C$ with respect to $K$, and $N$ be a point symmetric to point $B$ with respect to $A$. Prove that $DM=DN$