Find the equation of the circle in the complex plane determined by the roots of the equation $z^3 +(-1+i)z^2+(1-i)z+i= 0$.

Find the smallest positive number $\lambda$, such that for any $12$ points on the plane $P_1,P_2,\ldots,P_{12}$(can overlap), if the distance between any two of them does not exceed $1$, then $\sum_{1\le i<j\le 12} |P_iP_j|^2\le \lambda$.

On the side $ CD $ of the square $ ABCD, $ consider $ E $ for which $ \angle ABE =60^{\circ } . $ On the line $ AB, $ take the point $ F $ distinct from $ B $ such that $ BE=BF $ and such that it is on the segment $ AB, $ or $ A $ is on $ BF. $ Moreover, $ M $ is the intersection of $ EF,AD. $ [b]a)[/b] Show that $ \angle BME =75^{\circ } . $ [b]b)[/b] If the bisector of $ \angle CBE $ intersects $ CD $ in $ N, $ show that $ BMN $ is equilateral.

$ 5 $ points are situated in the plane so that any three of them form a triangle of area at most $ 1. $ Prove that there is a trapezoid of area at most $ 3 $ which contains all these points ('including' here means that the points can also be on the sides of the trapezoid).

Prove that there is no square with its four vertices on four concentric circles whose radii form an arithmetic progression.

Let $ABC$ be an acute triangle with circumcircle $\omega$ such that $AB<AC$. Let $M$ be the midpoint of the arc $BC$ of~$\omega$ containing the point~$A$, and let $X\neq M$ be the other point on $\omega$ such that $AX=AM$. Points $E$ and $F$ are chosen on sides $AC$ and $AB$ of the triangle $ABC$ such that $EX=EC$ and $FX=FB$. Prove that $AE=AF$.

Let $ABC$ be an acute triangle with altitudes $AD$, $BE$, $CF$ where $D$, $E$, $F$ lie on $BC$, $AC$, $AB$, respectively. Let $M$ be the midpoint of $BC$. The circumcircle of triangle $AEF$ cuts the line $AM$ at $A$ and $X$. The line $AM$ cuts the line $CF$ at $Y$. Let $Z$ be the point of intersection of $AD$ and $BX$. Show that the lines $YZ$ and $BC$ are parallel.

Triangle $\vartriangle ABC$ has side lengths $AB = 26$, $BC = 34$, and $CA = 24\sqrt2$. A fourth point $D$ makes a right angle $\angle BDC$. What is the smallest possible length of $\overline{AD}$?

A trapezoid $ABCD$ and a line $\ell$ perpendicular to its bases $AD$ and $BC$ are given. A point $X$ moves along $\ell$. The perpendiculars from $A$ to $BX$ and from $D$ to $CX$ meet at point $Y$ . Find the locus of $Y$ . by D.Prokopenko

Let $M$ and $N$ be touching points of incircle with sides $AB$ and $AC$ of triangle $ABC$, and $P$ intersection point of line $MN$ and angle bisector of $\angle ABC$. Prove that $\angle BPC =90 ^{\circ}$

In the plane of a triangle $ABC,$ in its exterior$,$ we draw the triangles $ABR, BCP, CAQ$ so that $\angle PBC = \angle CAQ = 45^{\circ}$, $\angle BCP = \angle QCA = 30^{\circ}$, $\angle ABR = \angle RAB = 15^{\circ}$. Prove that [b]a.)[/b] $\angle QRP = 90\,^{\circ},$ and [b]b.)[/b] $QR = RP.$

With a compass and a ruler, split a triangle into two smaller triangles with the same sum of squares of sides.

A circle touches the sides $AB,BC, CD,DA$ of a square at points $K,L,M,N$ respectively, and $BU, KV$ are parallel lines such that $U$ is on $DM$ and $V$ on $DN$. Prove that $UV$ touches the circle.

In a trapezoid $ABCD$ with $AB\parallel CD$, the diagonals $AC$ and $BD$ meet at $O$. Let $M$ and $N$ be the centers of the regular hexagons constructed on the sides $AB$ and $CD$ in the exterior of the trapezoid. Prove that $M,O$ and $N$ are collinear.

Let $M$ be the midpoint of side $BC$ of a triangle $ABC$. Let the median $AM$ intersect the incircle of $ABC$ at $K$ and $L,K$ being nearer to $A$ than $L$. If $AK = KL = LM$, prove that the sides of triangle $ABC$ are in the ratio $5 : 10 : 13$ in some order.

Points $D$, $E$, $F$ are taken respectively on sides $AB$, $BC$, and $CA$ of triangle $ABC$ so that $AD:DB=BE:CE=CF:FA=1:n$. The ratio of the area of triangle $DEF$ to that of triangle $ABC$ is: $\textbf{(A)}\ \frac{n^2-n+1}{(n+1)^2}\qquad \textbf{(B)}\ \frac{1}{(n+1)^2}\qquad \textbf{(C)}\ \frac{2n^2}{(n+1)^2}\qquad \textbf{(D)}\ \frac{n^2}{(n+1)^2}\qquad \textbf{(E)}\ \frac{n(n-1)}{n+1}$

Triangle $ABC$ is isosceles with $AB + AC$ and $BC = 65$ cm. $P$ is a point on $\overline{BC}$ such that the perpendicular distances from $P$ to $\overline{AB}$ and $\overline{AC}$ are $24$ cm and $36$ cm, respectively. The area of $\triangle ABC$, in cm$^2$, is $ \mathrm{(A) \ } 1254 \qquad \mathrm{(B) \ } 1640 \qquad \mathrm {(C) \ } 1950 \qquad \mathrm{(D) \ } 2535 \qquad \mathrm{(E) \ } 2942$

Let $\vartriangle ABC$ be a triangle with $AB = 5$, $BC = 8$, and, $CA = 7$. Let the center of the $A$-excircle be $O$, and let the $A$-excircle touch lines $BC$, $CA$, and,$ AB$ at points $X, Y$ , and, $Z$, respectively. Let $h_1$, $h_2$, and, $h_3$ denote the distances from $O$ to lines $XY$ , $Y Z$, and, ZX, respectively. If $h^2_1+ h^2_2+ h^2_3$ can be written as $\frac{m}{n}$ for relatively prime positive integers $m, n$, find $m + n$.

The figure shown is the union of a circle and two semicircles of diameters of $ a$ and $ b$, all of whose centers are collinear. The ratio of the area of the shaded region to that of the unshaded region is $ \displaystyle \textbf{(A)}\ \sqrt {\frac {a}{b}} \qquad \textbf{(B)}\ \ \frac {a}{b} \qquad \textbf{(C)}\ \ \frac {a^2}{b^2} \qquad \textbf{(D)}\ \ \frac {a \plus{} b}{2b} \qquad \textbf{(E)}\ \ \frac {a^2 \plus{} 2ab}{b^2 \plus{} 2ab}$ [asy]unitsize(2cm); defaultpen(fontsize(10pt)+linewidth(.8pt)); fill(Arc((1/3,0),2/3,0,180)--reverse(Arc((-2/3,0),1/3,180,360))--reverse(Arc((0,0),1,0,180))--cycle,mediumgray); draw(unitcircle); draw(Arc((-2/3,0),1/3,360,180)); draw(Arc((1/3,0),2/3,0,180)); label("$a$",(-2/3,0)); label("$b$",(1/3,0)); draw((-2/3+1/15,0)--(-1/3,0),EndArrow(4)); draw((-2/3-1/15,0)--(-1,0),EndArrow(4)); draw((1/3+1/15,0)--(1,0),EndArrow(4)); draw((1/3-1/15,0)--(-1/3,0),EndArrow(4));[/asy]

Coordinate axes (without any marks, with the same scale) and the graph of a quadratic trinomial $y = x^2 + ax + b$ are drawn in the plane. The numbers $a$ and $b$ are not known. How to draw a unit segment using only ruler and compass?

The area of a convex quadrilateral $ABCD$ is $18$. If $|AB|+|BD|+|DC|=12$, then what is $|AC|$? $ \textbf{(A)}\ 9 \qquad\textbf{(B)}\ 6\sqrt 3 \qquad\textbf{(C)}\ 8 \qquad\textbf{(D)}\ 6 \qquad\textbf{(E)}\ 6\sqrt 2 $

Let $ABCDEF$ be a convex hexagon and denote by $A',B',C',D',E',F'$ the middle points of the sides $AB$, $BC$, $CD$, $DE$, $EF$ and $FA$ respectively. Given are the areas of the triangles $ABC'$, $BCD'$, $CDE'$, $DEF'$, $EFA'$ and $FAB'$. Find the area of the hexagon. [i]Kvant Magazine[/i]

Let $ABCD$ be a parallelogram with an acute angle at $A$. Let $G$ be a point on the line $AB$, distinct from $B$, such that $|CG| = |CB|$. Let $H$ be a point on the line $BC$, distinct from $B$, such that $|AB| =|AH|$. Prove that triangle $DGH$ is isosceles. [asy] unitsize(1.5 cm); pair A, B, C, D, G, H; A = (0,0); B = (2,0); D = (0.5,1.5); C = B + D - A; G = reflect(A,B)*(C) + C - B; H = reflect(B,C)*(H) + A - B; draw(H--A--D--C--G); draw(interp(A,G,-0.1)--interp(A,G,1.1)); draw(interp(C,H,-0.1)--interp(C,H,1.1)); draw(D--G--H--cycle, dashed); dot("$A$", A, SW); dot("$B$", B, SE); dot("$C$", C, E); dot("$D$", D, NW); dot("$G$", G, NE); dot("$H$", H, SE); [/asy]

Let $ ABC $ be a triangle, $ M_A $ be the midpoint of the side $ BC, $ and $ P_A $ be the orthogonal projection of $ A $ on $ BC. $ Similarly, define $ M_B,M_C,P_B,P_C. M_BM_C $ intersects $ P_BP_C $ at $ S_A, $ and the tangent of the circumcircle of $ ABC $ at $ A $ meets $ BC $ at $ T_A. $ Similarly, define $ S_B,S_C,T_B,T_C. $ Show that the perpendiculars through $ A,B,C, $ to $ S_AT_A,S_BT_B, $ respectively, $ S_CT_C, $ are concurent. [i]Flavian Georgescu[/i]

In the plane we are given $3 \cdot n$ points ($n>$1) no three collinear, and the distance between any two of them is $\leq 1$. Prove that we can construct $n$ pairwise disjoint triangles such that: The vertex set of these triangles are exactly the given 3n points and the sum of the area of these triangles $< 1/2$.