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)

The quadrilateral $ABCD$ is a rhombus with acute angle at $A.$ Points $M$ and $N$ are on segments $\overline{AC}$ and $\overline{BC}$ such that $|DM| = |MN|.$ Let $P$ be the intersection of $AC$ and $DN$ and let $R$ be the intersection of $AB$ and $DM.$ Prove that $|RP| = |PD|.$

In quadrilateral $ABCD$, $AC = BD$ and $\measuredangle B = 60^\circ$. Denote by $M$ and $N$ the midpoints of $\overline{AB}$ and $\overline{CD}$, respectively. If $MN = 12$ and the area of quadrilateral $ABCD$ is 420, then compute $AC$. [i]Proposed by Aaron Lin[/i]

Let $AB\varGamma\varDelta$ be a rhombus. (a) Prove that you can construct a circle $(c)$ which is inscribed in the rhombus and is tangent to its sides. (b) The points $\varTheta,H,K,I$ are on the sides $\varDelta\varGamma,B\varGamma,AB,A\varDelta$ of the rhombus respectively, such that the line segments $KH$ and $I\varTheta$ are tangent on the circle $(c)$. Prove that the quadrilateral defined from the points $\varTheta,H,K,I$ is a trapezium.

Rhombus $PQRS$ is inscribed in rectangle $ABCD$ so that vertices $P$, $Q$, $R$, and $S$ are interior points on sides $\overline{AB}$, $\overline{BC}$, $\overline{CD}$, and $\overline{DA}$, respectively. It is given that $PB=15$, $BQ=20$, $PR=30$, and $QS=40$. Let $m/n$, in lowest terms, denote the perimeter of $ABCD$. Find $m+n$.

Let $ABCD$ be a cyclic quadrilateral. Points $K, L, M, N$ are chosen on $AB, BC, CD, DA$ such that $KLMN$ is a rhombus with $KL \parallel AC$ and $LM \parallel BD$. Let $\omega_A, \omega_B, \omega_C, \omega_D$ be the incircles of $\triangle ANK, \triangle BKL, \triangle CLM, \triangle DMN$. Prove that the common internal tangents to $\omega_A$, and $\omega_C$ and the common internal tangents to $\omega_B$ and $\omega_D$ are concurrent.

On the sides $[AB]$ and $[BC]$ of the parallelogram $ABCD$ are constructed the equilateral triangles $ABE$ and $BCF,$ so that the points $D$ and $E$ are on the same side of the line $AB$, and $F$ and $D$ on different sides of the line $BC$. If the points $E,D$ and $F$ are collinear, then prove that $ABCD$ is rhombus.

The rhombus $ABCD$ is given. Let $E$ be one of the points of intersection of the circles $\Gamma_B$ and $\Gamma_C$, where $\Gamma_B$ is the circle centered at $B$ and passing through $C$, and $\Gamma_C$ is the circle centered at $C$ and passing through $B$. The line $ED$ intersects $\Gamma_B$ at point $F$. Find the value of angle $\angle AFB$. [i](S. Mazanik)[/i]

A rhombus is formed by two radii and two chords of a circle whose radius is $ 16$ feet. The area of the rhombus in square feet is: $ \textbf{(A)}\ 128 \qquad\textbf{(B)}\ 128\sqrt {3} \qquad\textbf{(C)}\ 256 \qquad\textbf{(D)}\ 512 \qquad\textbf{(E)}\ 512\sqrt {3}$

Points $M,\ N,\ K,\ L$ are taken on the sides $AB,\ BC,\ CD,\ DA$ of a rhombus $ABCD,$ respectively, in such a way that $MN\parallel LK$ and the distance between $MN$ and $KL$ is equal to the height of $ABCD.$ Show that the circumcircles of the triangles $ALM$ and $NCK$ intersect each other, while those of $LDK$ and $MBN$ do not.

In a triangle $ABC$, let $D$ and $E$ be the feet of the angle bisectors of angles $A$ and $B$, respectively. A rhombus is inscribed into the quadrilateral $AEDB$ (all vertices of the rhombus lie on different sides of $AEDB$). Let $\varphi$ be the non-obtuse angle of the rhombus. Prove that $\varphi \le \max \{ \angle BAC, \angle ABC \}$.

Let $\Delta ABC$ be a triangle with $AB=AC$ and $\angle A = \alpha$, and let $O,H$ be its circumcenter and orthocenter, respectively. If $P,Q$ are points on $AB$ and $AC$, respectively, such that $APHQ$ forms a rhombus, determine $\angle POQ$ in terms of $\alpha$.

Let $ABCD$ be a square with $2cm$ side length and with center $T$. A rhombus $ARTE$ is drawn where point $E$ lies on line $DC$. What is the area of $ARTE$?

The points $A,B,C$ are in this order on line $D$, and $AB = 4BC$. Let $M$ be a variable point on the perpendicular to $D$ through $C$. Let $MT_1$ and $MT_2$ be tangents to the circle with center $A$ and radius $AB$. Determine the locus of the orthocenter of the triangle $MT_1T_2.$

Each angle of a rectangle is trisected. The intersections of the pairs of trisectors adjacent to the same side always form: $ \textbf{(A)}\ \text{a square} \qquad\textbf{(B)}\ \text{a rectangle} \qquad\textbf{(C)}\ \text{a parallelogram with unequal sides} \\ \textbf{(D)}\ \text{a rhombus} \qquad\textbf{(E)}\ \text{a quadrilateral with no special properties}$

In a ABCD cyclic quadrilateral 4 points K, L ,M, N are taken on AB , BC , CD and DA , respectively such that KLMN is a parallelogram. Lines AD, BC and KM have a common point. And also lines AB, DC and NL have a common point. Prove that KLMN is rhombus.

Rhombus $ ABCD$ is similar to rhombus $ BFDE$. The area of rhombus $ ABCD$ is 24, and $ \angle BAD \equal{} 60^\circ$. What is the area of rhombus $ BFDE$? [asy] size(180); defaultpen(linewidth(0.7)+fontsize(11)); pair A=origin, B=(2,0), C=(3, sqrt(3)), D=(1, sqrt(3)), E=(1, 1/sqrt(3)), F=(2, 2/sqrt(3)); pair point=(3/2, sqrt(3)/2); draw(B--C--D--A--B--F--D--E--B); label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$D$", D, dir(point--D)); label("$E$", E, dir(point--E)); label("$F$", F, dir(point--F));[/asy] $ \textbf{(A) } 6 \qquad \textbf{(B) } 4\sqrt {3} \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 6\sqrt {3}$

$ABC$ is an acute angle triangle such that $AB>AC$ and $\hat{BAC}=60^{\circ}$. Let's denote by $O$ the center of the circumscribed circle of the triangle and $H$ the intersection of altitudes of this triangle. Line $OH$ intersects $AB$ in point $P$ and $AC$ in point $Q$. Find the value of the ration $\frac{PO}{HQ}$.

Olga and Sasha play a game on an infinite hexagonal grid. They take turns in placing a stone on a free hexagon of their choice. Olga starts the game. Just before the $2018$th stone is placed, a new rule comes into play. A stone may now be placed only on those free hexagons having at least two occupied neighbors. A player loses when she or he either is unable to make a move, or makes a move such that a pattern of the rhomboid shape as shown (rotated in any possible way) appears. Determine which player, if any, possesses a winning strategy.

A rhombus $ABCD$ with $\angle BAD=60^{\circ}$ is given. Points $E$ on side $AB$ and $F$ on side $AD$ are such that $\angle ECF=\angle ABD$. Lines $CE$ and $CF$ respectively meet line $BD$ at $P$ and $Q$. Prove that $\frac{PQ}{EF}=\frac{AB}{BD}$.

Let $M$ be the intersection of two diagonal, $AC$ and $BD$, of a rhombus $ABCD$, where angle $A<90^\circ$. Construct $O$ on segment $MC$ so that $OB<OC$ and let $t=\frac{MA}{MO}$, provided that $O \neq M$. Construct a circle that has $O$ as centre and goes through $B$ and $D$. Let the intersections between the circle and $AB$ be $B$ and $X$. Let the intersections between the circle and $BC$ be $B$ and $Y$. Let the intersections of $AC$ with $DX$ and $DY$ be $P$ and $Q$, respectively. Express $\frac{OQ}{OP}$ in terms of $t$.

The vertex $ A$ of the acute triangle $ ABC$ is equidistant from the circumcenter $ O$ and the orthocenter $ H.$ Determine all possible values for the measure of angle $ A.$

Let $E$ be the point of intersection of the diagonals of convex quadrilateral $ABCD$, and let $P,Q,R,$ and $S$ be the centers of the circles circumscribing triangles $ABE,$ $BCE$, $CDE$, and $ADE$, respectively. Then $\textbf{(A) }PQRS\text{ is a parallelogram}$ $\textbf{(B) }PQRS\text{ is a parallelogram if an only if }ABCD\text{ is a rhombus}$ $\textbf{(C) }PQRS\text{ is a parallelogram if an only if }ABCD\text{ is a rectangle}$ $\textbf{(D) }PQRS\text{ is a parallelogram if an only if }ABCD\text{ is a parallelogram}$ $\textbf{(E) }\text{none of the above are true}$

In a triangle $ABC$, $\angle C = 60^o$, $D, E, F$ are points on the sides $BC, AB, AC$ respectively, and $M$ is the intersection point of $AD$ and $BF$. Suppose that $CDEF$ is a rhombus. Prove that $DF^2 = DM \cdot DA$

Let $ABCD$ be rhombus $ABCD$ where the triangles $ABD$ and $BCD$ are equilateral. Let $M$ and $N$ be points on the sides $BC$ and $CD$, respectively, such that $\angle MAN = 30^o$. Let $X$ be the intersection point of the diagonals $AC$ and $BD$. Prove that $\angle XMN = \angle\ DAM$ and $\angle XNM = \angle BAN$.