If $A$ and $B$ are fixed points on a given circle not collinear with centre $O$ of the circle, and if $XY$ is a variable diameter, find the locus of $P$ (the intersection of the line through $A$ and $X$ and the line through $B$ and $Y$).

Let $O$ be the circumcenter and $H$ the orthocenter of an acute triangle $ABC$. Show that there exist points $D$, $E$, and $F$ on sides $BC$, $CA$, and $AB$ respectively such that \[ OD + DH = OE + EH = OF + FH\] and the lines $AD$, $BE$, and $CF$ are concurrent.

In triangle $ABC$(with incenter $I$) let the line parallel to $BC$ from $A$ intersect circumcircle of $\triangle ABC$ at $A_1$ let $AI\cap BC=D$ and $E$ is tangency point of incircle with $BC$ let $ EA_1\cap \odot (\triangle ADE)=T$ prove that $AI=TI$.

Let $ABC$ be a triangle in which $\angle BAC = 60^{\circ}$ . Let $P$ (similarly $Q$) be the point of intersection of the bisector of $\angle ABC$(similarly of $\angle ACB$) and the side $AC$(similarly $AB$). Let $r_1$ and $r_2$ be the in-radii of the triangles $ABC$ and $AP Q$, respectively. Determine the circum-radius of $APQ$ in terms of $r_1$ and $r_2$.

Point $X$ is chosen inside a convex $ABCD$ so that $\angle XBC = \angle XAD, \angle XCB = \angle XDA$. Rays $AB, DC$ intersect at point $O$, circumcircles of triangles $BCO, ADO$ intersect at point $T$. Prove that line $TX$ and the line through $O$ perpendicular to $BC$ intersect on the circumcircle of $\triangle AOD$. [i]Proposed by Anton Trygub[/i]

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.

Given a point $P$ in the circumcircle $\omega$ of an equilateral triangle $ABC$, prove that the segments $PA$, $PB$, and $PC$ form a triangle $T$. Let $R$ be the radius of the circumcircle $\omega$ and let $d$ be the distance between $P$ and the circumcenter. Find the area of $T$.

On the extensions of sides $CA$ and $AB$ of triangle $ABC$ beyond points $A$ and $B$, respectively, the segments $AE = BC$ and $BF = AC$ are drawn. A circle is tangent to segment $BF$ at point $N$, side $BC$ and the extension of side $AC$ beyond point $C$. Point $M$ is the midpoint of segment $EF$. Prove that the line $MN$ is parallel to the bisector of angle $A$.

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.

Let $ABC$ be an acute triangle with $AC > AB$ and $O$ its circumcenter. Let $D$ be a point on segment $BC$ such that $O$ lies inside triangle $ADC$ and $\angle DAO + \angle ADB = \angle ADC$. Let $P$ and $Q$ be the circumcenters of triangles $ABD$ and $ACD$ respectively, and let $M$ be the intersection of lines $BP$ and $CQ$. Show that lines $AM, PQ$ and $BC$ are concurrent. [i]Pablo Jaén, Panama[/i]

The three squares $ACC_{1}A''$, $ABB_{1}'A'$, $BCDE$ are constructed externally on the sides of a triangle $ABC$. Let $P$ be the center of the square $BCDE$. Prove that the lines $A'C$, $A''B$, $PA$ are concurrent.

Let $ABCD$ a convex quadrilateral with $AB=BC=CD$, with $AC$ not equal to $BD$ and $E$ be the intersection point of it's diagonals. Prove that $AE=DE$ if and only if $\angle BAD+\angle ADC = 120$.

Let $\triangle ABC$ be an acute angled triangle with orthocenter $H$ and let $M$ be a midpoint of $BC$. Circle with diameter $AH$ is $\omega_1$ and circle with center $M$ is $\omega_2$. If $\omega_2$ is tangent to circumcircle of $\triangle ABC$, then prove that circles $\omega_1$ and $\omega_2$ are tangent to each other.

The vertices of the convex quadrilateral $ABCD$ lie on the parabola $y=x^2$. It is known that $ABCD$ is cyclic and $AC$ is a diameter of its circumcircle. Let $M$ and $N$ be the midpoints of the diagonals of $AC$ and $BD$ respectively. Find the length of the projection of the segment $MN$ on the axis $Oy$.

On the side $BC$ of acute triangle $ABC$ point $P$ was chosen. Point $E$ is symmetric to point $B$ onto line $AP$. Segment $PE$ meets circumcircle of triangle $ABP$ in point $D$. $M$ is midpoint of side $AC$. Prove that $DE+AC>2BM$.

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

The median $AM$ is drawn in the acute-angled triangle $ABC$ with different sides. Its extension intersects the circumscribed circle $w$ of this triangle at the point $P$. Let $A {{H} _ {1}}$ be the altitude $\Delta ABC$, $H$ be the point of intersection of its altitudes. The rays $MH$ and $P {{H} _ {1}}$ intersect the circle $w$ at the points $K$ and $T$, respectively. Prove that the circumscribed circle of $\Delta KT {{H} _ {1}}$ touches the segment $BC$. (Hilko Danilo)

Given triangle $ ABC$. Point $ O$ is the center of the excircle touching the side $ BC$. Point $ O_1$ is the reflection of $ O$ in $ BC$. Determine angle $ A$ if $ O_1$ lies on the circumcircle of $ ABC$.

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 $ABC$).

Circles $ \omega_{1}$ and $ \omega_{2}$ intersect at points $ A$ and $ B.$ Points $ C$ and $ D$ are on circles $ \omega_{1}$ and $ \omega_{2},$ respectively, such that lines $ AC$ and $ AD$ are tangent to circles $ \omega_{2}$ and $ \omega_{1},$ respectively. Let $ I_{1}$ and $ I_{2}$ be the incenters of triangles $ ABC$ and $ ABD,$ respectively. Segments $ I_{1}I_{2}$ and $ AB$ intersect at $ E$. Prove that: $ \frac {1}{AE} \equal{} \frac {1}{AC} \plus{} \frac {1}{AD}$

Let $ABC$ be a triangle with $ \widehat{BAC}\neq 90^\circ $. Let $M$ be the midpoint of $BC$. We choose a variable point $D$ on $AM$. Let $(O_1)$ and $(O_2)$ be two circle pass through $ D$ and tangent to $BC$ at $B$ and $C$. The line $BA$ and $CA$ intersect $(O_1),(O_2)$ at $ P,Q$ respectively. [b]a)[/b] Prove that tangent line at $P$ on $(O_1)$ and $Q$ on $(O_2)$ must intersect at $S$. [b]b)[/b] Prove that $S$ lies on a fix line.

Points $A$, $B$, $C$, and $D$ lie on a circle $\Gamma$, in that order, with $AB=5$ and $AD=3$. The angle bisector of $\angle ABC$ intersects $\Gamma$ at point $E$ on the opposite side of $\overleftrightarrow{CD}$ as $A$ and $B$. Assume that $\overline{BE}$ is a diameter of $\Gamma$ and $AC=AE$. Compute $DE$.

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 $BC$.

Given acute angle triangle $ ABC$. Let $ CD$be the altitude , $ H$ be the orthocenter and $ O$ be the circumcenter of $ \triangle ABC$ The line through point $ D$ and perpendicular with $ OD$ , is intersect $ BC$ at $ E$. Prove that $ \angle DHE \equal{} \angle ABC$.

Let $ABC$ be an acute-angled triangle with circumcircle $\Gamma$ and orthocentre $H$. Let $K$ be a point of $\Gamma$ on the other side of $BC$ from $A$. Let $L$ be the reflection of $K$ in the line $AB$, and let $M$ be the reflection of $K$ in the line $BC$. Let $E$ be the second point of intersection of $\Gamma $ with the circumcircle of triangle $BLM$. Show that the lines $KH$, $EM$ and $BC$ are concurrent. (The orthocentre of a triangle is the point on all three of its altitudes.) [i]Luxembourg (Pierre Haas)[/i]