Found problems: 43
The point $M$ inside a convex quadrilateral $ABCD$ is equidistant from the lines $AB$ and $CD$ and is equidistant from the lines $BC$ and $AD$. The area of $ABCD$ occurred to be equal to $MA\cdot MC +MB \cdot MD$. Prove that the quadrilateral $ABCD$ is
a) tangential (circumscribed),
b) cyclic (inscribed).
(Nairi Sedrakyan)
Which of the following four statements are true and which are false?
a) If a polygon inscribed in a circle is equilateral, then it is also equiangular.
b) If a polygon inscribed in a circle is equiangular, then it is also equilateral.
c) If a polygon circumscribed to a circle is equilateral, then it is also equiangular.
d) If a polygon circumscribed to a circle is equiangular, then it is also equilateral.
Quadrilateral $ABCD$ is circumscribed around circle $k$. Gind the smallest possible value of
$$\frac{AB + BC + CD + DA}{AC + BD}$$, as well as all quadrilaterals with the above property where it is reached.
We are given trapezoid $ABCD$ and point $M$ on the intersection of its diagonals. The parallel sides are $AD$ and $BC$ and it is known that $AB$ is perpendicular to $AD$ and that the trapezoid can have an inscribed circle. If the radius of this inscribed circle is $R$ find the area of triangle $DCM$ .
a) The triangle $ABC$ was turned around the centre of the circumscribed circle by the angle less than $180$ degrees and thus was obtained the triangle $A_1B_1C_1$. The corresponding segments $[AB]$ and $[A_1B_1]$ intersect in the point $C_2, [BC]$ and $[B_1C_1]$ -- $A_2, [AC]$ and $[A_1C_1]$ -- $B_2$. Prove that the triangle $A_2B_2C_2$ is similar to the triangle $ABC$.
b) The quadrangle $ABCD$ was turned around the centre of the circumscribed circle by the angle less than $180$ degrees and thus was obtained the quadrangle $A_1B_1C_1D_1$. Prove that the points of intersection of the corresponding lines ( $(AB$) and $(A_1B_1), (BC)$ and $(B_1C_1), (CD)$ and $(C_1D_1), (DA)$ and $(D_1A_1)$ ) are the vertices of the parallelogram.
In the convex quadrilateral $ABCD$, the bisectors of angles $A$ and $C$ intersect in $I$. Prove that $ABCD$ is circumscriptible if and only if
$$S[AIB] + S[CID] =S[AID]+S[BIC]$$
( $S[XYZ]$ denotes the area of the triangle $XYZ$)
$ABCD$ is inscribed in unit circle $\Gamma$. Let $\Omega_1$, $\Omega_2$ be the circumcircles of $\vartriangle ABD$, $\vartriangle CBD$ respectively. Circles $\Omega_1$, $\Omega_2$ are tangent to segment $BD$ at $M$,$N$ respectively. Line A$M$ intersects $\Gamma$, $\Omega_1$ again at points $X_1$,$X_2$ respectively (different from $A$, $M$). Let $\omega_1$ be the circle passing through $X_1$, $X_2$ and tangent to $\Omega_1$. Line $CN$ intersects $\Gamma$, $\Omega_2$ again at points $Y_1$, $Y_2$ respectively (different from $C$, $N$). Let $\omega_2$ be the circle passing through $Y_1$, $Y_2$ and tangent to $\Omega_2$. Circles $\Omega_1$,$\Omega_2$, $\omega_1$, $\omega_2$ have radii $R_1$, $R_2$, $r_1$, $r_2$ respectively. Prove that $$r_1+r_2-R_1-R_2=1.$$
[img]https://cdn.artofproblemsolving.com/attachments/1/5/70471f2419fadc4b2183f5fe74f0c7a2e69ed4.png[/img]
[url=https://www.geogebra.org/m/vxx8ghww]geogebra file[/url]
The quadrangle $ABCD$ is circumscribed around the circle with the centre $O$. Prove that $$\angle AOB+ \angle COD=180^o. $$
Nine numbers are arranged in a square table:
$a_1 \,\,\, a_2 \,\,\,a_3$
$b_1 \,\,\,b_2 \,\,\,b_3$
$c_1\,\,\, c_2 \,\,\,c_3$ .
It is known that the six numbers obtained by summing the rows and columns of the table are equal:
$a_1 + a_2 + a_3 = b_1 + b_2 + b_3 = c_1 + c_2 + c_3 = a_1 + b_1 + c_1 = a_2 + b_2 + c_2 = a_3 + b_3 + c_3$ .
Prove that the sum of products of numbers in the rows is equal to the sum of products of numbers in the columns:
$a_1 b_1 c_1 + a_2 b_2c_2 + a_3b_3c_3 = a_1a_2a_3 + b_1 b_2 b_3 + c_1 c_2c_3$ .
(V Proizvolov)
Is there a $1987$-gon with consecutive sides lengths $1, 2, 3,..., 1986, 1987$, in which you can fit a circle?
The diagonals of the square $ABCD$ intersect at $P$ and the midpoint of the side $AB$ is $E$. Segment $ED$ intersects the diagonal $AC$ at point $F$ and segment $EC$ intersects the diagonal $BD$ at $G$. Inside the quadrilateral $EFPG$, draw a circle of radius $r$ tangent to all the sides of this quadrilateral. Prove that $r = | EF | - | FP |$.
It is known that a circle can be inscribed in a trapezium $ABCD$.
Prove that the two circles, constructed on its oblique sides as diameters, touch each other.
(D. Fomin, Leningrad)
Let $ABCD$ be a quadrilateral and let $\Gamma$ be a circle of center $O$ that is internally tangent to its four sides. If $M$ is the midpoint of $AC$ and $N$ is the midpoint of $BD$, prove that $M,O, N$ are collinear.
Prove that if a convex quadrilateral has the property that there exists a circle tangent to its sides (i.e. an inscribed circle), and also a circle tangent to the extensions of its sides (an excircle), then the diagonals of the quadrilateral are perpendicular to each other.
Considers a quadrilateral $[ABCD]$ that has an inscribed circle and a circumscribed circle. The sides $[AD]$ and $[BC]$ are tangent to the circle inscribed at points $E$ and $F$, respectively. Prove that $AE \cdot F C = BF \cdot ED$.
[img]https://1.bp.blogspot.com/-6o1fFTdZ69E/X4XMo98ndAI/AAAAAAAAMno/7FXiJnWzJgcfSn-qSRoEAFyE8VgxmeBjwCLcBGAsYHQ/s0/2005%2BPortugal%2Bp5.png[/img]
The point $M$ inside a convex quadrilateral $ABCD$ is equidistant from the lines $AB$ and $CD$ and is equidistant from the lines $BC$ and $AD$. The area of $ABCD$ occurred to be equal to $MA\cdot MC +MB \cdot MD$. Prove that the quadrilateral $ABCD$ is
a) tangential (circumscribed),
b) cyclic (inscribed).
(Nairi Sedrakyan)
In the convex quadrilateral $ ABCD $, the corresponding points $ M $ and $ N $ are chosen on the adjacent sides $ \overline{AB} $ and $ \overline{BC} $ and the intersection point of the segments $ AN $ and $ GM $ is marked by 0. Prove that if circles can be inscribed in the quadrilaterals $ AOCD $ and $ BMON $, then a circle can also be inscribed in the quadrilateral $ ABCD $.
Let $O$ be the center of a circle inscribed in a convex quadrilateral $ABCD$ and $|AB|= a$, $|CD|=$c.
Prove that $$\frac{a}{c}=\frac{AO\cdot BO}{CO\cdot DO}.$$