Let $ABCD$ be a convex quadrilateral whose sides $AD$ and $BC$ are not parallel. Suppose that the circles with diameters $AB$ and $CD$ meet at points $E$ and $F$ inside the quadrilateral. Let $\omega_E$ be the circle through the feet of the perpendiculars from $E$ to the lines $AB,BC$ and $CD$. Let $\omega_F$ be the circle through the feet of the perpendiculars from $F$ to the lines $CD,DA$ and $AB$. Prove that the midpoint of the segment $EF$ lies on the line through the two intersections of $\omega_E$ and $\omega_F$. [i]Proposed by Carlos Yuzo Shine, Brazil[/i]

Let $ABCC_1B_1A_1$ be a convex hexagon such that $AB=BC$, and suppose that the line segments $AA_1, BB_1$, and $CC_1$ have the same perpendicular bisector. Let the diagonals $AC_1$ and $A_1C$ meet at $D$, and denote by $\omega$ the circle $ABC$. Let $\omega$ intersect the circle $A_1BC_1$ again at $E \neq B$. Prove that the lines $BB_1$ and $DE$ intersect on $\omega$.

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 $A$, $B$ and $C$ be three points on a line (in this order). For each circle $k$ through the points $B$ and $C$, let $D$ be one point of intersection of the perpendicular bisector of $BC$ with the circle $k$. Further, let $E$ be the second point of intersection of the line $AD$ with $k$. Show that for each circle $k$, the ratio of lengths $\overline{BE}:\overline{CE}$ is the same.

Let $ABCD$ by a cyclic quadrilateral with circumcenter $O$. Let $P$ be the point of intersection of the diagonals $AC$ and $BD$, and $K, L, M, N$ the circumcenters of triangles $AOP, BOP$, $COP, DOP$, respectively. Prove that $KL = MN$.

Let $P,Q,R$ be points on the same side of a line $g$ in the plane. Let $M$ and $N$ be the feet of the perpendiculars from $P$ and $Q$ to $g$ respectively. Point $S$ lies between the lines $PM$ and $QN$ and satisfies and satisfies $PM = PS$ and $QN = QS$. The perpendicular bisectors of $SM$ and $SN$ meet in a point $R$. If the line $RS$ intersects the circumcircle of triangle $PQR$ again at $T$, prove that $S$ is the midpoint of $RT$.

Let $ABC$ be an isosceles triangle with $AB=AC$. Consider a variable point $P$ on the extension of the segment $BC$ beyound $B$ (in other words, $P$ lies on the line $BC$ such that the point $B$ lies inside the segment $PC$). Let $r_{1}$ be the radius of the incircle of the triangle $APB$, and let $r_{2}$ be the radius of the $P$-excircle of the triangle $APC$. Prove that the sum $r_{1}+r_{2}$ of these two radii remains constant when the point $P$ varies. [i]Remark.[/i] The $P$-excircle of the triangle $APC$ is defined as the circle which touches the side $AC$ and the [i]extensions[/i] of the sides $AP$ and $CP$.

Given an acute triangle $ ABC$. The incircle of triangle $ ABC$ touches $ BC,CA,AB$ respectively at $ D,E,F$. The angle bisector of $ \angle A$ cuts $ DE$ and $ DF$ respectively at $ K$ and $ L$. Suppose $ AA_1$ is one of the altitudes of triangle $ ABC$, and $ M$ be the midpoint of $ BC$. (a) Prove that $ BK$ and $ CL$ are perpendicular with the angle bisector of $ \angle BAC$. (b) Show that $ A_1KML$ is a cyclic quadrilateral.

In convex quadrilateral $ ABCD$, $ CB,DA$ are external angle bisectors of $ \angle DCA,\angle CDB$, respectively. Points $ E,F$ lie on the rays $ AC,BD$ respectively such that $ CEFD$ is cyclic quadrilateral. Point $ P$ lie in the plane of quadrilateral $ ABCD$ such that $ DA,CB$ are external angle bisectors of $ \angle PDE,\angle PCF$ respectively. $ AD$ intersects $ BC$ at $ Q.$ Prove that $ P$ lies on $ AB$ if and only if $ Q$ lies on segment $ EF$.

Let $ABCD$ be a quadrilateral inscribed in a circle $k$. $AC$ and $BD$ meet at $E$. The rays $\overrightarrow{CB}, \overrightarrow{DA}$ meet at $F$. Prove that the line through the incenters of $\triangle ABE\,,\, \triangle ABF$ and the line through the incenters of $\triangle CDE\,,\, \triangle CDF$ meet at a point lying on the circle $k$. [i]Proposed by N. Beluhov[/i]

Let $ABC$ be an acute triangle. The circumference with diameter $AB$ intersects sides $AC$ and $BC$ at $E$ and $F$ respectively. The tangent lines to the circumference at the points $E$ and $F$ meet at $P$. Show that $P$ belongs to the altitude from $C$ of triangle $ABC$.

In a triangle $ABC$, the perpendicular bisectors of sides $AB$ and $AC$ intersect $BC$ at $X$ and $Y$. Prove that $BC = XY$ if and only if $\tan B\tan C = 3$ or $\tan B\tan C = -1$.

Consider a rectangle $ABCD$ with $AB = 2$ and $BC = \sqrt3$. The point $M$ lies on the side $AD$ so that $MD = 2 AM$ and the point $N$ is the midpoint of the segment $AB$. On the plane of the rectangle rises the perpendicular MP and we choose the point $Q$ on the segment $MP$ such that the measure of the angle between the planes $(MPC)$ and $(NPC)$ shall be $45^o$, and the measure of the angle between the planes $(MPC)$ and $(QNC)$ shall be $60^o$. a) Show that the lines $DN$ and $CM$ are perpendicular. b) Show that the point $Q$ is the midpoint of the segment $MP$.

Consider a circle $S$, and a point $P$ outside it. The tangent lines from $P$ meet $S$ at $A$ and $B$, respectively. Let $M$ be the midpoint of $AB$. The perpendicular bisector of $AM$ meets $S$ in a point $C$ lying inside the triangle $ABP$. $AC$ intersects $PM$ at $G$, and $PM$ meets $S$ in a point $D$ lying outside the triangle $ABP$. If $BD$ is parallel to $AC$, show that $G$ is the centroid of the triangle $ABP$. [i]Arnoldo Aguilar (El Salvador)[/i]

Let $ABC$ be a triangle such that $AB<AC$. The perpendicular bisector of the side $BC$ meets the side $AC$ at the point $D$, and the (interior) bisectrix of the angle $ADB$ meets the circumcircle $ABC$ at the point $E$. Prove that the (interior) bisectrix of the angle $AEB$ and the line through the incentres of the triangles $ADE$ and $BDE$ are perpendicular.

As shown in the figure, quadrilateral $ ABCD$ is inscribed in a circle with $ AC$ as its diameter, $ BD \perp AC,$ and $ E$ the intersection of $ AC$ and $ BD.$ Extend line segment $ DA$ and $ BA$ through $ A$ to $ F$ and $ G$ respectively, such that $ DG \parallel{} BF.$ Extend $ GF$ to $ H$ such that $ CH \perp GH.$ Prove that points $ B, E, F$ and $ H$ lie on one circle. [asy] defaultpen(linewidth(0.8)+fontsize(10));size(150); real a=4, b=6.5, c=9, d=a*c/b, g=14, f=sqrt(a^2+b^2)*sqrt(a^2+d^2)/g; pair E=origin, A=(0,a), B=(-b,0), C=(0,-c), D=(d,0), G=A+g*dir(B--A), F=A+f*dir(D--A), M=midpoint(G--C); path c1=circumcircle(A,B,C), c2=Circle(M, abs(M-G)); pair Hf=F+10*dir(G--F), H=intersectionpoint(F--Hf, c2); dot(A^^B^^C^^D^^E^^F^^G^^H); draw(c1^^c2^^G--D--C--A--G--F--D--B--A^^F--H--C--B--F); draw(H--B^^F--E^^G--C, linetype("2 2")); pair point= E; 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("$F$", F, dir(point--F)); label("$G$", G, dir(point--G)); label("$H$", H, dir(point--H)); label("$E$", E, NE);[/asy]

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 $M$ 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 $t$.

Let $H$ be the orthocenter of a triangle $ABC$, and let $D$ be the midpoint of the segment $AH$. The altitude $BH$ of triangle $ABC$ intersects the perpendicular to the line $AB$ through the point $A$ at the point $M$. The altitude $CH$ of triangle $ABC$ intersects the perpendicular to the line $CA$ through the point $A$ at the point $N$. The perpendicular bisector of the segment $AB$ intersects the perpendicular to the line $BC$ through the point $B$ at the point $U$. The perpendicular bisector of the segment $CA$ intersects the perpendicular to the line $BC$ through the point $C$ at the point $V$. Finally, let $E$ be the midpoint of the side $BC$ of triangle $ABC$. Prove that the points $D$, $M$, $N$, $U$, $V$ all lie on one and the same perpendicular to the line $AE$. [i]Extensions.[/i] In other words, we have to show that the points $M$, $N$, $U$, $V$ lie on the perpendicular to the line $AE$ through the point $D$. Additionally, one can find two more points on this perpendicular: [b](a)[/b] The nine-point circle of triangle $ABC$ is known to pass through the midpoint $E$ of its side $BC$. Let $D^{\prime}$ be the point where this nine-point circle intersects the line $AE$ apart from $E$. Then, the point $D^{\prime}$ lies on the perpendicular to the line $AE$ through the point $D$. [b](b)[/b] Let the tangent to the circumcircle of triangle $ABC$ at the point $A$ intersect the line $BC$ at a point $X$. Then, the point $X$ lies on the perpendicular to the line $AE$ through the point $D$. [i]Comment.[/i] The actual problem was created by Victor Thébault around 1950 (cf. Hyacinthos messages #1102 and #1551). The extension [b](a)[/b] initially was a (pretty trivial) lemma in Thébault's solution of the problem. Extension [b](b)[/b] is rather new; in the form "prove that $X\in UV$", it was [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=3659]proposed by Valentin Vornicu for the Balkan MO 2003[/url], however it circulated in the Hyacinthos newsgroup before (Hyacinthos messages #7240 and #7242), where different solutions of the problem were discussed as well. Hereby, "Hyacinthos" always refers to the triangle geometry newsgroup "Hyacinthos", which can be found at http://groups.yahoo.com/group/Hyacinthos . I proposed the problem for the QEDMO math fight wishing to draw some attention to it. It has a rather short and elementary solution, by the way (without using radical axes or inversion like the standard solutions). Darij

Let $ABC$ be a scalene (i.e. non-isosceles) triangle. Let $U$ be the center of the circumcircle of this triangle and $I$ the center of the incircle. Assume that the second point of intersection different from $C$ of the angle bisector of $\gamma = \angle ACB$ with the circumcircle of $ABC$ lies on the perpendicular bisector of $UI$. Show that $\gamma$ is the second-largest angle in the triangle $ABC$.

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 an acute-angled triangle $ABC, \angle ABC$ is the largest angle. The perpendicular bisectors of $BC$ and $BA$ intersect AC at $X$ and $Y$ respectively. Prove that circumcentre of triangle $ABC$ is incentre of triangle $BXY$ .

Let $ABC$ be a triangle with $AB=AC$. Also, let $D\in[BC]$ be a point such that $BC>BD>DC>0$, and let $\mathcal{C}_1,\mathcal{C}_2$ be the circumcircles of the triangles $ABD$ and $ADC$ respectively. Let $BB'$ and $CC'$ be diameters in the two circles, and let $M$ be the midpoint of $B'C'$. Prove that the area of the triangle $MBC$ is constant (i.e. it does not depend on the choice of the point $D$). [i]Greece[/i]

Let $ ABC $ be a isosceles triangle with $ AC=BC$. Let $ D $ a point on a line $ BA $ such that $ A $ lies between $ B, D $. Let $O_1 $ be the circumcircle of triangle $ DAC $. $ O_1 $ meets $ BC $ at point $ E $. Let $ F $ be the point on $ BC $ such that $ FD $ is tangent to circle $O_1 $, and let $O_2 $ be the circumcircle of $ DBF$. Two circles $O_1 , O_2 $ meet at point $ G ( \ne D) $. Let $ O $ be the circumcenter of triangle $ BEG$. Prove that the line $FG$ is tangent to circle $O$ if and only if $ DG \bot FO$.

In an acute triangle $ABC$ with $AC<BC$, lines $m_a$ and $m_b$ are the perpendicular bisectors of sides $BC$ and $AC$, respectively. Further, let $M_c$ be the midpoint of side $AB$. The Median $CM_c$ intersects $m_a$ in point $S_a$ and $m_b$ in point $S_b$; the lines $AS_b$ und $BS_a$ intersect in point $K$. Prove: $\angle ACM_c = \angle KCB$.