Two circles $K_1$ and $K_2$ of different radii intersect at two points $A$ and $B$, let $C$ and $D$ be two points on $K_1$ and $K_2$, respectively, such that $A$ is the midpoint of the segment $CD$. The extension of $DB$ meets $K_1$ at another point $E$, the extension of $CB$ meets $K_2$ at another point $F$. Let $l_1$ and $l_2$ be the perpendicular bisectors of $CD$ and $EF$, respectively. i) Show that $l_1$ and $l_2$ have a unique common point (denoted by $P$). ii) Prove that the lengths of $CA$, $AP$ and $PE$ are the side lengths of a right triangle.

Let $ ABCD$ be a triangular pyramid such that no face of the pyramid is a right triangle and the orthocenters of triangles $ ABC$, $ ABD$, and $ ACD$ are collinear. Prove that the center of the sphere circumscribed to the pyramid lies on the plane passing through the midpoints of $ AB$, $ AC$ and $ AD$.

Let $ABC$ be a triangle satisfying $AB<AC$. Let $M$ be the midpoint of $BC$. A point $P$ lies on the segment $AB$ with $AP>PB$. A point $Q$ lies on the segment $AC$ with $\angle MPA = \angle AQM$. The perpendicular bisectors of $BC$ and $PQ$ intersect at $S$. Prove that $\angle BAC + \angle QSP = \angle QMP$.

In a triangle $ABC$, let $D$ be the point on the segment $BC$ such that $AB+BD=AC+CD$. Suppose that the points $B$, $C$ and the centroids of triangles $ABD$ and $ACD$ lie on a circle. Prove that $AB=AC$.

Let $ABC$ be an acute-angled triangle with $AC>AB$, and $\Omega$ is your circumcircle. Let $P$ be the midpoint of the arc $BC$ of $\Omega$ (not containing $A$) and $Q$ be the midpoint of the arc $BC$ of $\Omega$(containing the point $A$). Let $M$ be the foot of perpendicular of $Q$ on the line $AC$. Prove that the circumcircle of $\triangle AMB$ cut the segment $AP$ in your midpoint.

A circle center $O$ meets a circle center $O'$ at $A$ and $B.$ The line $TT'$ touches the first circle at $T$ and the second at $T'$. The perpendiculars from $T$ and $T'$ meet the line $OO'$ at $S$ and $S'$. The ray $AS$ meets the first circle again at $R$, and the ray $AS'$ meets the second circle again at $R'$. Show that $R, B$ and $R'$ are collinear.

Let $ ABC $ be a triangle. The perpendicular in $ B $ of the bisector of the angle $ \angle ABC $ intersects the bisector of the angle $ \angle BAC $ in $ M. $ Show that $ MC $ is perpendicular to the bisector of $ \angle BCA. $

Let $ABCD$ be a rhombus. Let $K$ be a point on the line $CD$, other than $C$ or $D$, such that $AD = BK$. Let $P$ be the point of intersection of $BD$ with the perpendicular bisector of $BC$. Prove that $A, K$ and $P$ are collinear.

Find the least positive integer $n$ ($n\geq 3$), such that among any $n$ points (no three are collinear) in the plane, there exist three points which are the vertices of a non-isoscele triangle.

Given a triangle $ABC$ and three circles $x$, $y$ and $z$ such that $A \in y \cap z$, $B \in z \cap x$ and $C \in x \cap y$. The circle $x$ intersects the line $AC$ at the points $X_b$ and $C$, and intersects the line $AB$ at the points $X_c$ and $B$. The circle $y$ intersects the line $BA$ at the points $Y_c$ and $A$, and intersects the line $BC$ at the points $Y_a$ and $C$. The circle $z$ intersects the line $CB$ at the points $Z_a$ and $B$, and intersects the line $CA$ at the points $Z_b$ and $A$. (Yes, these definitions have the symmetries you would expect.) Prove that the perpendicular bisectors of the segments $Y_a Z_a$, $Z_b X_b$ and $X_c Y_c$ concur.

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 $AA_1$ and $BB_1$ be the altitudes of an acute-angled, non-isosceles triangle $ABC$. Also, let $A_0$ and $B_0$ be the midpoints of its sides $BC$ and $CA$, respectively. The line $A_1B_1$ intersects the line $A_0B_0$ at a point $C'$. Prove that the line $CC'$ is perpendicular to the Euler line of the triangle $ABC$ (this is the line that joins the orthocenter and the circumcenter of the triangle $ABC$).

A scalene triangle $ABC$ is inscribed in a circle $\Gamma$. The bisector of angle $A$ meets $BC$ at $E$. Let $M$ be the midpoint of the arc $BAC$. The line $ME$ intersects $\Gamma$ again at $D$. Show that the circumcentre of triangle $AED$ coincides with the intersection point of the tangent to $\Gamma$ at $D$ and the line $BC$.

The perpendicular bisector of the bisector $BL$ of the triangle $ABC$ intersects the bisectors of its external angles $A$ and $C$ at points $P$ and $Q$, respectively. Prove that the circle circumscribed around the triangle $PBQ$ is tangent to the circle circumscribed around the triangle $ABC$.

An arbitrary point $M$ is selected in the interior of the segment $AB$. The square $AMCD$ and $MBEF$ are constructed on the same side of $AB$, with segments $AM$ and $MB$ as their respective bases. The circles circumscribed about these squares, with centers $P$ and $Q$, intersect at $M$ and also at another point $N$. Let $N'$ denote the point of intersection of the straight lines $AF$ and $BC$. a) Prove that $N$ and $N'$ coincide; b) Prove that the straight lines $MN$ pass through a fixed point $S$ independent of the choice of $M$; c) Find the locus of the midpoints of the segments $PQ$ as $M$ varies between $A$ and $B$.

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.

Two cirlces $ C_1$ and $ C_2$, with center $ O_1$ and $ O_2$ respectively, intersect at $ A$ and $ B$. Let $ O_1$ lies on $ C_2$. A line $ l$ passes through $ O_1$ but does not pass through $ O_2$. Let $ P$ and $ Q$ be the projection of $ A$ and $ B$ respectively on the line $ l$ and let $ M$ be the midpoint of $ \overline{AB}$. Prove that $ MPQ$ is an isoceles triangle.

Let $ABC$ be an acute non isosceles triangle. The angle bisector of angle $A$ meets again the circumcircle of the triangle $ABC$ in $D$. Let $O$ be the circumcenter of the triangle $ABC$. The angle bisectors of $\angle AOB$, and $\angle AOC$ meet the circle $\gamma$ of diameter $AD$ in $P$ and $Q$ respectively. The line $PQ$ meets the perpendicular bisector of $AD$ in $R$. Prove that $AR // BC$.

Segments $AC$ and $BD$ intersect at point $P$ such that $PA = PD$ and $PB = PC$. Let $E$ be the foot of the perpendicular from $P$ to the line $CD$. Prove that the line $PE$ and the perpendicular bisectors of the segments $PA$ and $PB$ are concurrent.

Let $ABC$ be a triangle and $AB<AC<BC$. Let $D,E$ be points on the side $BC$ and the line $AB$, respectively ($A$ is between $B,E$) such that $BD=BE=AC$. The circumcircle of $\Delta BED$ meets the side $AC$ at $P$ and $BP$ meets the circumcircle of $\Delta ABC$ at $Q$. Prove that: \[ AQ+CQ=BP. \]

Let $ABC$ be a triangle, and let $D$ be the foot of the $A-$altitude. Points $P, Q$ are chosen on $BC$ such that $DP = DQ = DA$. Suppose $AP$ and $AQ$ intersect the circumcircle of $ABC$ again at $X$ and $Y$. Prove that the perpendicular bisectors of the lines $PX$, $QY$, and $BC$ are concurrent. [i]Proposed by Pranjal Srivastava[/i]

Let $M, N$ be the midpoints of sides $AB, AC$ respectively of a triangle $ABC$. The perpendicular bisector to the bisectrix $AL$ meets the bisectrixes of angles $B$ and $C$ at points $P$ and $Q$ respectively. Prove that the common point of lines $PM$ and $QN$ lies on the tangent to the circumcircle of $ABC$ at $A$.

Ben "One Hunna Dolla" Franklin is flying a kite $KITE$ such that $IE$ is the perpendicular bisector of $KT$. Let $IE$ meet $KT$ at $R$. The midpoints of $KI,IT,TE,EK$ are $A,N,M,D,$ respectively. Given that $[MAKE]=18,IT=10,[RAIN]=4,$ find $[DIME]$. Note: $[X]$ denotes the area of the figure $X$.

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]

In triangle $ABC$, $\angle B=2\angle C$ and $\angle A>90^\circ$. Let $D$ be the point on the line $AB$ such that $CD$ is perpendicular to $AC$, and let $M$ be the midpoint of $BC$. Prove that $\angle AMB=\angle DMC$.