Given the segment $ PQ$ and a circle . A chord $AB$ moves around the circle, equal to $PQ$. Let $T$ be the intersection point of the perpendicular bisectors of the segments $AP$ and $BQ$. Prove that all points of $T$ thus obtained lie on one line.

The inscribed circle $\omega$ of the triangle $ABC$ touches its sides $BC, CA$ and $AB$ at points $A_1, B_1$ and $C_1$, respectively. Let $S$ be the intersection point of lines passing through points $B$ and $C$ and parallel to $A_1C_1$ and $A_1B_1$ respectively, $A_0$ be the foot of the perpendicular drawn from point $A_1$ on $B_1C_1$, $G_1$ be the centroid of triangle $A_1B_1C_1$, $P$ be the intersection point of the ray $G_1A_0$ with $\omega$. Prove that points $S, A_1$, and $P$ lie on a straight line.

In triangle $ABC$, the altitude through $B$ intersects $AC$ at $E$ and the altitude through $C$ intersects $AB$ at $F$. Point $T$ is such that $AETF$ is a parallelogram and points $ A$ ,$T$ lie on different half-planes wrt the line $EF$. Point $D$ is such that $ABDC$ is a parallelogram and points $ A$ ,$D$ lie in different half-planes wrt line $BC$. Prove that $T, D$ and the orthocenter of $ABC$ are collinear.

Let $ABC$ be an acute triangle, not isoceles triangle and $(O), (I)$ be its circumcircle and incircle respectively. Let $A_1$ be the the intersection of the radical axis of $(O), (I)$ and the line $BC$. Let $A_2$ be the point of tangency (not on $BC$) of the tangent from $A_1$ to $(I)$. Points $B_1,B_2,C_1,C_2$ are defined in the same manner. Prove that (a) the lines $AA_2,BB_2,CC_2$ are concurrent. (b) the radical centers circles through triangles $BCA_2, CAB_2$ and $ABC_2$ are all on the line $OI$. Lê Phúc Lữ

Let $N$ be the set of natural numbers, and let $f: N \to N$ be a function satisfying $f(x) + f(x + 2) < 2 f(x + 1)$ for any $x \in N$. Prove that there exists a straight line in the $xy$-plane which contains infinitely many points with coordinates $(n,f(n))$.

Let $\vartriangle ABC$ be an acute triangle with altitudes $AA_1, BB_1, CC_1$ and orthocenter $H$. Let $K, L$ be the midpoints of $BC_1, CB_1$. Let $\ell_A$ be the external angle bisector of $\angle BAC$. Let $\ell_B, \ell_C$ be the lines through $B, C$ perpendicular to $\ell_A$. Let $\ell_H$ be the line through $H$ parallel to $\ell_A$. Prove that the centers of the circumcircles of $\vartriangle A_1B_1C_1, \vartriangle AKL$ and the rectangle formed by $\ell_A, \ell_B, \ell_C, \ell_H$ lie on the same line.

Given an acute $\vartriangle ABC$ with orthocenter $H$. Let $M_a$ be the midpoint of $BC. M_aH$ intersects the circumcircle of $\vartriangle ABC$ at $X_a$ and $AX_a$ intersects $BC$ at $Y_a$. Define $Y_b, Y_c$ in a similar way. Prove that $Y_a, Y_b,Y_c$ are collinear. [img]https://2.bp.blogspot.com/-yjISBHtRa0s/XnSKTrhhczI/AAAAAAAALds/e_rvs9glp60L1DastlvT0pRFyP7GnJnCwCK4BGAYYCw/s320/imoc2017%2Bg4.png[/img]

Two congruent equilateral triangles $ABC$ and $A'B'C'$ in the plane are given. Show that the midpoints of the segments $AA',BB', CC'$ either are collinear or form an equilateral triangle.

Let $ABC$ be an isosceles triangle with base $AB$. We choose a point $P$ inside the triangle on altitude through $C$. The circle with diameter $CP$ intersects the straight line through $B$ and $P$ again at the point $D_P$ and the Straight through $A$ and $C$ one more time at point $E_P$. Prove that there is a point $F$ such that for any choice of $P$ the points $D_P , E_P$ and $F$ lie on a straight line. [i](Walther Janous)[/i]

Let $P$ be an interior point of a circle $\Gamma$ centered at $O$ where $P \ne O$. Let $A$ and $B$ be distinct points on $\Gamma$. Lines $AP$ and $BP$ meet $\Gamma$ again at $C$ and $D$, respectively. Let $S$ be any interior point on line segment $PC$. The circumcircle of $\vartriangle ABS$ intersects line segment $PD$ at $T$. The line through $S$ perpendicular to $AC$ intersects $\Gamma$ at $U$ and $V$ . The line through $T$ perpendicular to $BD$ intersects $\Gamma$ at $X$ and $Y$ . Let $M$ and $N$ be the midpoints of $UV$ and $XY$ , respectively. Let $AM$ and $BN$ meet at $Q$. Suppose that $AB$ is not parallel to $CD$. Show that $P, Q$, and $O$ are collinear if and only if $S$ is the midpoint of $PC$.

Let $O$ be the centre of a parallelogram $ABCD$ and $P$ be any point in the plane. Let $M, N$ be the midpoints of $AP, BP$, respectively and $Q$ be the intersection of $MC$ and $ND$. Prove that $O, P$ and $Q$ are collinear.

Let $\Gamma$ be the circumcircle of a triangle $ABC$ with $AB <BC$, and let $M$ be the midpoint from the side $AC$ . The median of side $AC$ cuts $\Gamma$ at points $X$ and $Y$ ($X$ in the arc $ABC$). The circumcircle of the triangle $BMY$ cuts the line $AB$ at $P$ ($P \ne B$) and the line $BC$ in $Q$ ($Q \ne B$). The circumcircles of the triangles $PBC$ and $QBA$ are cut in $R$ ($R \ne B$). Prove that points $X, B$ and $R$ are collinear.

Triangle $ABC$ is inscribed in circle $\omega$. Point $P$ lies inside triangle $ABC$.Lines $AP,BP$ and $CP$ intersect $\omega$ again at points $A_1$, $B_1$ and $C_1$ (other than $A, B, C$), respectively. The tangent lines to $\omega$ at $A_1$ and $B_1$ intersect at $C_2$.The tangent lines to $\omega$ at $B_1$ and $C_1$ intersect at $A_2$. The tangent lines to $\omega$ at $C_1$ and $A_1$ intersect at $B_2$. Prove that the lines $AA_2,BB_2$ and $CC_2$ are concurrent.

Let $ABC$ be an acute triangle and $G$ be the intersection of the meadians of triangle $ABC$. Let $D $be the foot of the altitude drawn from $A$ to $BC$. Draw a parallel line such that it is parallel to $BC$ and one of the points of it is $A$.Donate the point $S$ as the intersection of the parallel line and circumcircle $ABC$. Prove that $S,G,D$ are co-linear [asy] size(6cm); defaultpen(fontsize(10pt)); pair A = dir(50), S = dir(130), B = dir(200), C = dir(-20), G = (A+B+C)/3, D = foot(A, B, C); draw(A--B--C--cycle, black+linewidth(1)); draw(A--S^^A--D, magenta); draw(S--D, red+dashed); draw(circumcircle(A, B, C), heavymagenta); string[] names = {"$A$", "$B$", "$C$","$D$", "$G$","$S$"}; pair[] points = {A, B, C,D,G,S}; pair[] ll = {A, B, C,D, G,S}; int pt = names.length; for (int i=0; i<pt; ++i) dot(names[i], points[i], dir(ll[i])); [/asy]

Given triangle $ABC$. Suppose $P$ and $P_1$ are points on $BC, Q$ lies on $CA, R$ lies on $AB$, such that $$\frac{AR}{RB}=\frac{BP}{PC}=\frac{CQ}{QA}=\frac{CP_1}{P_1B}$$ Let $G$ be the centroid of triangle $ABC$ and $K = AP_1 \cap RQ$. Prove that points $P,G$, and $K$ are collinear.

The quadrilateral $ABCD$ is inscribed in the circle $\omega$ with the center $O$. Suppose that the angles $B$ and $C$ are obtuse and lines $AD$ and $BC$ are not parallel. Lines $AB$ and $CD$ intersect at point $E$. Let $P$ and $R$ be the feet of the perpendiculars from the point $E$ on the lines $BC$ and $AD$ respectively. $Q$ is the intersection point of $EP$ and $AD, S$ is the intersection point of $ER$ and $BC$. Let K be the midpoint of the segment $QS$ . Prove that the points $E, K$, and $O$ are collinear.

Let $ABCD$ be a cyclic quadrilateral inscirbed in circle $O$. Let the tangent to $O$ at $A$ meet $BC$ at $S$, and the tangent to $O$ at $B$ meet $CD$ at $T$. Circle with $S$ as its center and passing $A$ meets $BC$ at $E$, and $AE$ meets $O$ again at $F(\ne A)$. The circle with $T$ as its center and passing $B$ meets $CD$ at $K$. Let $P = BK \cap AC$. Prove that $P,F,D$ are collinear if and only if $AB = AP$.

Let $ABC$ be an acute-angled triangle. $M ,N$ are any two points on the sides $AB , AC$ respectively. The circles with the diameters $BN$ and $CM$ intersect at points $P$ and $Q$. Show that the points $P, Q$ and the orthocenter of the triangle $ABC$ lie on a straight line.

Let $AD, BE, CF$ be segments whose midpoints are on the same line $\ell$. The points $X, Y, Z$ lie on the lines $EF, FD, DE$ respectively such that $AX \parallel BY \parallel CZ \parallel \ell$. Prove that $X, Y, Z$ are collinear. Tran Quang Hung, High School of Natural Sciences, Hanoi National University

In triangle $ABC$ with centroid $G$, let $M \in (AB)$ and $N \in (AC)$ be points on two of its sides. Prove that points $M, G, N$ are collinear if and only if $\frac{MB}{MA}+\frac{NC}{NA}=1$.

Given trapezoid $ABCD$ with bases $AB \parallel CD$ ($AB < CD$). Let $O$ be the intersection of $AC$ and $BD$. Two straight lines from $D$ and $C$ are perpendicular to $AC$ and $BD$ intersect at $E$ , i.e. $CE \perp BD$ and $DE \perp AC$ . By analogy, $AF \perp BD$ and $BF \perp AC$ . Are three points $E , O, F$ located on the same line?

In triangle $ABC$ let $I_a$ be the $A$-excenter. Let $\omega$ be an arbitrary circle that passes through $A,I_a$ and intersects the extensions of sides $AB,AC$ (extended from $B,C$) at $X,Y$ respectively. Let $S,T$ be points on segments $I_aB,I_aC$ respectively such that $\angle AXI_a=\angle BTI_a$ and $\angle AYI_a=\angle CSI_a$.Lines $BT,CS$ intersect at $K$. Lines $KI_a,TS$ intersect at $Z$. Prove that $X,Y,Z$ are collinear. [i]Proposed by Hooman Fattahi[/i]

The inscribed circle of the non-isosceles triangle $ABC$ touches sides $AB, BC$ and $AC$ at points $C_1, A_1$ and $B_1$, respectively. The circumscribed circle of the triangle $A_1BC_1$ intersects the lines $B_1A_1$ and $B_1C_1$ at the points $A_0$ and $C_0$, respectively. Prove that the orthocenter of triangle $A_0BC_0$, the center of the inscribed circle of triangle $ABC$ and the midpoint of the $AC$ lie on one straight line.

Consider the right isosceles $\vartriangle ABC$ at $ A$. Let $L$ be the intersection of the bisector of $\angle ACB$ with $AB$ and $K$ the intersection point of $CL$ with the bisector of $BC$. Let $X$ be the point on line $AK$ such that $\angle KCX = 90^o$ and let $Y$ be the point of intersection of $CX$ with the circumcircle of $\vartriangle ABC$. Let $Y'$ the reflection of point $Y$ wrt $BC$. Prove that $B - K -Y'$. Notation: $A-B-C$ means than points $A,B,C$ are collinear in that order i.e. $ B$ lies between $ A$ and $C$.

[b]Q.[/b] Let $ABC$ be a triangle, where $L_A, L_B, L_C$ denote the internal angle bisectors of $\angle BAC, \angle ABC, \angle ACB$ respectively and $\ell_A, \ell_B, \ell_C$, the altitudes from the corresponding vertices. Suppose $ L_A\cap \overline{BC} = \{A_1\}$, $\ell_A \cap \overline{BC} = \{A_2\}$ and the circumcircle of $\triangle AA_1A_2$ meets $AB$ and $AC$ at $S$ and $T$ respectively. If $\overline{ST} \cap \overline{BC} = \{A'\}$, prove that $A',B',C'$ are collinear, where $B'$ and $C'$ are defined in a similar manner. [i]Proposed by Functional_equation[/i]