$ABC$ is acute-angled. $D$ s a variable point on the side BC. $O_1$ is the circumcenter of $ABD$, $O_2$ is the circumcenter of $ACD$, and $O$ is the circumcenter of $AO_1O_2$. Find the locus of $O$.

Let $D$ be an arbitrary point on side $AB$ of triangle $ABC$. Circumcircles of triangles $BCD$ and $ACD$ intersect sides $AC$ and $BC$ at points $E$ and $F$, respectively. Perpendicular bisector of $EF$ cuts $AB$ at point $M$, and line perpendicular to $AB$ at $D$ at point $N$. Lines $AB$ and $EF$ intersect at point $T$, and the second point of intersection of circumcircle of triangle $CMD$ and line $TC$ is $U$. Prove that $NC=NU$

Let $ K $ be the midpoint of the side $ AB $ of a given triangle $ ABC $. Let $ L $ and $ M$ be points on the sides $ AC $ and $ BC$, respectively, such that $ \angle CLK = \angle KMC $. Prove that the perpendiculars to the sides $ AB, AC, $ and $ BC $ passing through $ K,L, $ and $M$, respectively, are concurrent.

Let ${\cal C}_1$ and ${\cal C}_2$ be concentric circles, with ${\cal C}_2$ in the interior of ${\cal C}_1$. From a point $A$ on ${\cal C}_1$ one draws the tangent $AB$ to ${\cal C}_2$ ($B\in {\cal C}_2$). Let $C$ be the second point of intersection of $AB$ and ${\cal C}_1$, and let $D$ be the midpoint of $AB$. A line passing through $A$ intersects ${\cal C}_2$ at $E$ and $F$ in such a way that the perpendicular bisectors of $DE$ and $CF$ intersect at a point $M$ on $AB$. Find, with proof, the ratio $AM/MC$.

In a plane are given a straight line $\ell$ and a point $P$ not located on $\ell$. Is there a circle in this plane such that there exist more than three different points $S$ on $\ell$ with the property that the perpendicular bisector of $PS$ is tangent to the circle ? Explain the answer.

Let $ A, B$ and $ C$ be non-collinear points. Prove that there is a unique point $ X$ in the plane of $ ABC$ such that \[ XA^2 \plus{} XB^2 \plus{} AB^2 \equal{} XB^2 \plus{} XC^2 \plus{} BC^2 \equal{} XC^2 \plus{} XA^2 \plus{} CA^2.\]

Let $ABCDEF$ be a convex hexagon satisfying $AC=DF$, $CE=FB$ and $EA=BD$. Prove that the lines connecting the midpoints of opposite sides of the hexagon $ABCDEF$ intersect in one point.

A line segment of length 1 is given on the plane. Show that a line segment of length $\sqrt{2010}$ can be constructed using only a straightedge and a compass.

Let $\angle A$ in a triangle $ABC$ be $60^\circ$. Let point $N$ be the intersection of $AC$ and perpendicular bisector to the side $AB$ while point $M$ be the intersection of $AB$ and perpendicular bisector to the side $AC$. Prove that $CB = MN$. [i](3 points)[/i]

In equilateral triangle $ABC$, the midpoint of $\overline{BC}$ is $M$. If the circumcircle of triangle $MAB$ has area $36\pi$, then find the perimeter of the triangle. [i]Proposed by Isabella Grabski [/i]

In the figure of [url]http://www.artofproblemsolving.com/Forum/download/file.php?id=50643&mode=view[/url] $\odot O_1$ and $\odot O_2$ intersect at two points $A$, $B$. The extension of $O_1A$ meets $\odot O_2$ at $C$, and the extension of $O_2A$ meets $\odot O_1$ at $D$, and through $B$ draw $BE \parallel O_2A$ intersecting $\odot O_1$ again at $E$. If $DE \parallel O_1A$, prove that $DC \perp CO_2$.

A circle is tangential to sides $AB$ and $AD$ of convex quadrilateral $ABCD$ at $G$ and $H$ respectively, and cuts diagonal $AC$ at $E$ and $F$. What are the necessary and sufficient conditions such that there exists another circle which passes through $E$ and $F$, and is tangential to $DA$ and $DC$ extended?

In triangle $ ABC$, one has marked the incenter, the foot of altitude from vertex $ C$ and the center of the excircle tangent to side $ AB$. After this, the triangle was erased. Restore it.

Let $ABC$ be a triangle inscribed in circle $\Gamma$ with center $O$. Let $H$ be the orthocenter of triangle $ABC$ and let $K$ be the midpoint of $OH$. Tangent of $\Gamma$ at $B$ intersects the perpendicular bisector of $AC$ at $L$. Tangent of $\Gamma$ at $C$ intersects the perpendicular bisector of $AB$ at $M$. Prove that $AK$ and $LM$ are perpendicular. by Michael Sarantis, Greece

Point $P$ is taken interior to a square with side-length $a$ and such that is it equally distant from two consecutive vertices and from the side opposite these vertices. If $d$ represents the common distance, then $d$ equals: $\textbf{(A)}\ \dfrac{3a}{5} \qquad \textbf{(B)}\ \dfrac{5a}{8} \qquad \textbf{(C)}\ \dfrac{3a}{8} \qquad \textbf{(D)}\ \dfrac{a\sqrt{2}}{2} \qquad \textbf{(E)}\ \dfrac{a}{2}$

As shown in the figure below, the in-circle of $ABC$ is tangent to sides $AB$ and $AC$ at $D$ and $E$ respectively, and $O$ is the circumcenter of $BCI$. Prove that $\angle ODB = \angle OEC$. [asy]import graph; size(5.55cm); pathpen=linewidth(0.7); pointpen=black; pen fp=fontsize(10); pointfontpen=fp; real xmin=-5.76,xmax=4.8,ymin=-3.69,ymax=3.71; pen zzttqq=rgb(0.6,0.2,0), wwwwqq=rgb(0.4,0.4,0), qqwuqq=rgb(0,0.39,0); pair A=(-2,2.5), B=(-3,-1.5), C=(2,-1.5), I=(-1.27,-0.15), D=(-2.58,0.18), O=(-0.5,-2.92); D(A--B--C--cycle,zzttqq); D(arc(D,0.25,-104.04,-56.12)--(-2.58,0.18)--cycle,qqwuqq); D(arc((-0.31,0.81),0.25,-92.92,-45)--(-0.31,0.81)--cycle,qqwuqq); D(A--B,zzttqq); D(B--C,zzttqq); D(C--A,zzttqq); D(CR(I,1.35),linewidth(1.2)+dotted+wwwwqq); D(CR(O,2.87),linetype("2 2")+blue); D(D--O); D((-0.31,0.81)--O); D(A); D(B); D(C); D(I); D(D); D((-0.31,0.81)); D(O); MP( "A", A, dir(110)); MP("B", B, dir(140)); D("C", C, dir(20)); D("D", D, dir(150)); D("E", (-0.31, 0.81), dir(60)); D("O", O, dir(290)); D("I", I, dir(100)); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]

Chord $EF$ is the perpendicular bisector of chord $BC$, intersecting it in $M$. Between $B$ and $M$ point $U$ is taken, and $EU$ extended meets the circle in $A$. Then, for any selection of $U$, as described, triangle $EUM$ is similar to triangle: [asy] pair B = (-0.866, -0.5); pair C = (0.866, -0.5); pair E = (0, -1); pair F = (0, 1); pair M = midpoint(B--C); pair A = (-0.99, -0.141); pair U = intersectionpoints(A--E, B--C)[0]; draw(B--C); draw(F--E--A); draw(unitcircle); label("$B$", B, SW); label("$C$", C, SE); label("$A$", A, W); label("$E$", E, S); label("$U$", U, NE); label("$M$", M, NE); label("$F$", F, N); //Credit to MSTang for the asymptote [/asy] $\textbf{(A)}\ EFA \qquad \textbf{(B)}\ EFC \qquad \textbf{(C)}\ ABM \qquad \textbf{(D)}\ ABU \qquad \textbf{(E)}\ FMC$

Let convex quadrilateral $ ABCD$ be inscribed in a circle centers at $ O.$ The opposite sides $ BA,CD$ meet at $ H$, the diagonals $ AC,BD$ meet at $ G.$ Let $ O_{1},O_{2}$ be the circumcenters of triangles $ AGD,BGC.$ $ O_{1}O_{2}$ intersects $ OG$ at $ N.$ The line $ HG$ cuts the circumcircles of triangles $ AGD,BGC$ at $ P,Q$, respectively. Denote by $ M$ the midpoint of $ PQ.$ Prove that $ NO \equal{} NM.$

Let $ ABCD$ be an isosceles trapezium. Determine the geometric location of all points $ P$ such that \[ |PA| \cdot |PC| \equal{} |PB| \cdot |PD|.\]

In a plane a circle with radius $R$ and center $w$ and a line $\Lambda$ are given. The distance between $w$ and $\Lambda$ is $d, d > R$. The points $M$ and $N$ are chosen on $\Lambda$ in such a way that the circle with diameter $MN$ is externally tangent to the given circle. Show that there exists a point $A$ in the plane such that all the segments $MN$ are seen in a constant angle from $A.$

Aaron takes a square sheet of paper, with one corner labeled $A$. Point $P$ is chosen at random inside of the square and Aaron folds the paper so that points $A$ and $P$ coincide. He cuts the sheet along the crease and discards the piece containing $A$. Let $p$ be the probability that the remaining piece is a pentagon. Find the integer nearest to $100p$. [i]Proposed by Aaron Lin[/i]

In an acute triangle $ABC$ with $\overline{AB} < \overline{AC}$, angle bisector of $A$ and perpendicular bisector of $\overline{BC}$ intersect at $D$. Let $P$ be an interior point of triangle $ABC$. Line $CP$ meets the circumcircle of triangle $ABP$ again at $K$. Prove that $B, D, K$ are collinear if and only if $AD$ and $BC$ meet on the circumcircle of triangle $APC$.

Let $\triangle ABC$ which $\angle ABC$ are right angle, Let $D$ be point on $AB$ ( $D \neq A , B$ ), Let $E$ be point on line $AB$ which $B$ is the midpoint of $DE$, Let $I$ be incenter of $\triangle ACE$ and $J$ be $A$-excenter of $\triangle ACD$. Prove that perpendicular bisector of $BC$ bisects $IJ$

Line passes the focal point $F$ of parabola $y^2=8(x+2)$ with bank angle of $60^{\circ}$ intersects the parabola at $A,B$. Perpendicular bisector of $AB$ intersects $x$-axis at $P$, then the length of $PF$ is $\text{(A)}\frac{16}{3}\qquad\text{(B)}\frac{8}{3}\qquad\text{(C)}\frac{16}{3}\sqrt3\qquad\text{(D)}8\sqrt3$

The vertices $ A$ and $ B$ of an equilateral triangle $ ABC$ lie on a circle $k$ of radius $1$, and the vertex $ C$ is in the interior of the circle $ k$. A point $ D$, different from $ B$, lies on $ k$ so that $ AD\equal{}AB$. The line $ DC$ intersects $ k$ for the second time at point $ E$. Find the length of the line segment $ CE$.