Point $D$ is chosen on the Euler line of triangle $ABC$ and it is inside of the triangle. Points $E,F$ are were the line $BD,CD$ intersect with $AC,AB$ respectively. Point $X$ is on the line $AD$ such that $\angle EXF =180 - \angle A$, also $A,X$ are on the same side of $EF$. If $P$ is the second intersection of circumcircles of $CXF,BXE$ then prove the lines $XP,EF$ meet on the altitude of $A$ Proposed by [i]Alireza Danaie[/i]

Point $D$ is chosen on the Euler line of triangle $ABC$ and it is inside of the triangle. Points $E,F$ are were the line $BD,CD$ intersect with $AC,AB$ respectively. Point $X$ is on the line $AD$ such that $\angle EXF =180 - \angle A$, also $A,X$ are on the same side of $EF$. If $P$ is the second intersection of circumcircles of $CXF,BXE$ then prove the lines $XP,EF$ meet on the altitude of $A$ Proposed by [i]Alireza Danaie[/i]

Let $ABC$ be an acute triangle, not being isoceles. Let $\ell_a$ be the line passing through the points of tangency of the escribed circles in the angle $A$ with the lines $AB, AC$ produced. Let $d_a$ be the line through $A$ parallel to the line that joins the incenter $I$ of the triangle $ABC$ and the midpoint of $BC$. Lines $\ell_b, d_b, \ell_c, d_c$ are defined in the same manner. Three lines $\ell_a, \ell_b, \ell_c$ intersect each other and these intersections make a triangle called $MNP$. Prove that the lines $d_a, d_b$ and $d_c$ are concurrent and their point of concurrency lies on the Euler line of the triangle $MNP$. Lê Phúc Lữ

Let $ABC$ be a triangle with orthocenter $H$ and circumcenter $O$. Let $P$ be a point in the plane such that $AP \perp BC$. Let $Q$ and $R$ be the reflections of $P$ in the lines $CA$ and $AB$, respectively. Let $Y$ be the orthogonal projection of $R$ onto $CA$. Let $Z$ be the orthogonal projection of $Q$ onto $AB$. Assume that $H \neq O$ and $Y \neq Z$. Prove that $YZ \perp HO$. [asy] import olympiad; unitsize(30); pair A,B,C,H,O,P,Q,R,Y,Z,Q2,R2,P2; A = (-14.8, -6.6); B = (-10.9, 0.3); C = (-3.1, -7.1); O = circumcenter(A,B,C); H = orthocenter(A,B,C); P = 1.2 * H - 0.2 * A; Q = reflect(A, C) * P; R = reflect(A, B) * P; Y = foot(R, C, A); Z = foot(Q, A, B); P2 = foot(A, B, C); Q2 = foot(P, C, A); R2 = foot(P, A, B); draw(B--(1.6*A-0.6*B)); draw(B--C--A); draw(P--R, blue); draw(R--Y, red); draw(P--Q, blue); draw(Q--Z, red); draw(A--P2, blue); draw(O--H, darkgreen+linewidth(1.2)); draw((1.4*Z-0.4*Y)--(4.6*Y-3.6*Z), red+linewidth(1.2)); draw(rightanglemark(R,Y,A,10), red); draw(rightanglemark(Q,Z,B,10), red); draw(rightanglemark(C,Q2,P,10), blue); draw(rightanglemark(A,R2,P,10), blue); draw(rightanglemark(B,P2,H,10), blue); label("$\textcolor{blue}{H}$",H,NW); label("$\textcolor{blue}{P}$",P,N); label("$A$",A,W); label("$B$",B,N); label("$C$",C,S); label("$O$",O,S); label("$\textcolor{blue}{Q}$",Q,E); label("$\textcolor{blue}{R}$",R,W); label("$\textcolor{red}{Y}$",Y,S); label("$\textcolor{red}{Z}$",Z,NW); dot(A, filltype=FillDraw(black)); dot(B, filltype=FillDraw(black)); dot(C, filltype=FillDraw(black)); dot(H, filltype=FillDraw(blue)); dot(P, filltype=FillDraw(blue)); dot(Q, filltype=FillDraw(blue)); dot(R, filltype=FillDraw(blue)); dot(Y, filltype=FillDraw(red)); dot(Z, filltype=FillDraw(red)); dot(O, filltype=FillDraw(black)); [/asy]

Let $ABC$ be an acute scalene triangle, and let $A_1, B_1, C_1$ be the feet of the altitudes from $A, B, C$. Let $A_2$ be the intersection of the tangents to the circle $ABC$ at $B, C$ and define $B_2, C_2$ similarly. Let $A_2A_1$ intersect the circle $A_2B_2C_2$ again at $A_3$ and define $B_3, C_3$ similarly. Show that the circles $AA_1A_3, BB_1B_3$, and $CC_1C_3$ all have two common points, $X_1$ and $X_2$ which both lie on the Euler line of the triangle $ABC$. [i]United Kingdom, Joe Benton[/i]

Let $ABC$ be an acute scalene triangle with circumcenter $O$, and let $T$ be on line $BC$ such that $\angle TAO = 90^{\circ}$. The circle with diameter $\overline{AT}$ intersects the circumcircle of $\triangle BOC$ at two points $A_1$ and $A_2$, where $OA_1 < OA_2$. Points $B_1$, $B_2$, $C_1$, $C_2$ are defined analogously. [list=a][*] Prove that $\overline{AA_1}$, $\overline{BB_1}$, $\overline{CC_1}$ are concurrent. [*] Prove that $\overline{AA_2}$, $\overline{BB_2}$, $\overline{CC_2}$ are concurrent on the Euler line of triangle $ABC$. [/list][i]Evan Chen[/i]

Let $D$ be the foot of perpendicular from $A$ to the Euler line (the line passing through the circumcentre and the orthocentre) of an acute scalene triangle $ABC$. A circle $\omega$ with centre $S$ passes through $A$ and $D$, and it intersects sides $AB$ and $AC$ at $X$ and $Y$ respectively. Let $P$ be the foot of altitude from $A$ to $BC$, and let $M$ be the midpoint of $BC$. Prove that the circumcentre of triangle $XSY$ is equidistant from $P$ and $M$.

Let $ABCD$ be a cyclic quadrilateral with two diagonals intersect at $E$. Let $ M$, $N$, $P$, $Q$ be the reflections of $ E $ in midpoints of $AB$, $BC$, $CD$, $DA$ respectively. Prove that the Euler lines of $ \triangle MAB$, $\triangle NBC$, $\triangle PCD,$ $\triangle QDA$ are concurrent. Trần Quang Hùng

The medians $AA_0, BB_0$, and $CC_0$ of the acute-angled triangle $ABC$ intersect at the point $M$, and heights $AA_1, BB_1$ and $CC_1$ at point $H$. Tangent to the circumscribed circle of triangle $A_1B_1C_1$ at $C_1$ intersects the line $A_0B_0$ at the point $C'$. Points $A'$ and $B'$ are defined similarly. Prove that $A', B'$ and $C'$ lie on one line perpendicular to the line $MH$.

The line $l$ passing through the orthocenter $H$ of a triangle $ABC$ $(BC>AB)$ and parallel to $AC$ meets $AB$ and $BC$ at points $D$ and $E$ respectively. The line passing through the circumcenter of the triangle and parallel to the median $BM$ meets $l$ at point $F$. Prove that the length of segment $HF$ is three times greater than the difference of $FE$ and $DH$ Proposed by: A.Mardanov, K.Mardanova

Let $G, H$ be the centroid and orthocentre of $\vartriangle ABC$ which has an obtuse angle at $\angle B$. Let $\omega$ be the circle with diameter $AG$. $\omega$ intersects $\odot(ABC)$ again at $L \ne A$. The tangent to $\omega$ at $L$ intersects $\odot(ABC)$ at $K \ne L$. Given that $AG = GH$, prove $\angle HKG = 90^o$ . [i]Sam Bealing, United Kingdom[/i]