Let $ABC$ be an equiangular triangle with circumcircle $\omega$. Let point $F\in AB$ and point $E\in AC$ so that $\angle ABE+\angle ACF=60^{\circ}$. The circumcircle of triangle $AFE$ intersects the circle $\omega$ in the point $D$. The halflines $DE$ and $DF$ intersect the line through $B$ and $C$ in the points $X$ and $Y$. Prove that the incenter of the triangle $DXY$ is independent of the choice of $E$ and $F$. (The angles in the problem statement are not directed. It is assumed that $E$ and $F$ are chosen in such a way that the halflines $DE$ and $DF$ indeed intersect the line through $B$ and $C$.)

Two circles in a plane intersect. $A$ is one of the points of intersection. Starting simultaneously from $A$ two points move with constant speed, each travelling along its own circle in the same sense. The two points return to $A$ simultaneously after one revolution. Prove that there is a fixed point $P$ in the plane such that the two points are always equidistant from $P.$

The circle $\omega$ lies inside the circle $\Omega$ and touches it internally at $T.$ Let $XY{}$ be a variable chord of the circle $\Omega$ touching $\omega.$ Denote by $X'$ and $Y'$ the midpoints of the arcs $TY{}$ and $TX{}$ which do not contain $X{}$ and $Y{}$ respectively. Prove that all possible lines $X'Y'$ pass through a fixed point. [i]Proposed by F. Petrov[/i]

Two different points $A$ and $B$ have been marked on the circle $\omega$. We consider all points $X$ of the circle $\omega$, different from $A$ and $B$. Let $Y$ be the middpoint of the chord $AX$ and $Z$ be the projection of point $A$ on the line $BX$. Prove that all straight lines $YZ$ pass through a certain fixed point that does not depend on the choice of point $X$.

Two not necessarily equal non-intersecting wooden disks, one gray and one black, are glued to a plane. An infinite angle with one gray side and one black side can be moved along the plane so that the disks remain outside the angle, while the colored sides of the angle are tangent to the disks of the same color (the tangency points are not the vertices). Prove that it is possible to draw a ray in the angle, starting from the vertex of the angle and such that no matter how the angle is positioned, the ray passes through some fixed point of the plane. (Egor Bakaev, Ilya Bogdanov, Pavel Kozhevnikov, Vladimir Rastorguev) (Junior version [url=https://artofproblemsolving.com/community/c6h2094701p15140671]here[/url]) [hide=note]There was a mistake in the text of the problem 3, we publish here the correct version. The solutions were estimated according to the text published originally.[/hide]

Let $\omega$ be a circle with the center $O$ and $A$ and $C$ be two different points on $\omega$. For any third point $P$ of the circle let $X$ and $Y$ be the midpoints of the segments $AP$ and $CP$. Finally, let $H$ be the orthocenter (the point of intersection of the altitudes) of the triangle $OXY$ . Prove that the position of the point H does not depend on the choice of $P$. (Artemiy Sokolov)

A chord $AB$ is drawn in a circle. The line $\ell$ is parallel to $AB$ and does not intersect the circle. Let $C$ be a certain point on the circle (points $C$ located on one side of $AB$ are considered). Lines $CA$ and $CB$ intersect $\ell$ at points $D$ and $E$. Prove that there exists a fixed point $F$ of the plane, not lying on line $\ell$ , such that $\angle DFE$ is constant.

$G$ is a set of non-constant functions $f$. Each $f$ is defined on the real line and has the form $f(x)=ax+b$ for some real $a,b$. If $f$ and $g$ are in $G$, then so is $fg$, where $fg$ is defined by $fg(x)=f(g(x))$. If $f$ is in $G$, then so is the inverse $f^{-1}$. If $f(x)=ax+b$, then $f^{-1}(x)= \frac{x-b}{a}$. Every $f$ in $G$ has a fixed point (in other words we can find $x_f$ such that $f(x_f)=x_f$. Prove that all the functions in $G$ have a common fixed point.

A circle $\Gamma$ and a line $\ell$ which does not intersect $\Gamma$ are given. Suppose $P, Q,R, S$ are variable points on circle $\Gamma$ such that the points $A = PQ\cap RS$ and $B = PS \cap QR$ lie on $\ell$. Prove that the circle on $AB$ as a diameter passes through two fixed points.

Let $ABC$ be an acute triangle. Let $P$ be a point such that $PB$ and $PC$ are tangent to circumcircle of $ABC$. Let $X$ and $Y$ be variable points on $AB$ and $AC$, respectively, such that $\angle XPY = 2\angle BAC$ and $P$ lies in the interior of triangle $AXY$. Let $Z$ be the reflection of $A$ across $XY$. Prove that the circumcircle of $XYZ$ passes through a fixed point. (Dominik Burek, Poland)

A subset $S$ of size $n$ of a plane consisting of points with both coordinates integer is given, where $n$ is an odd number. The injective function $f\colon S\rightarrow S$ satisfies the following: for each pair of points $A, B\in S$, the distance between points $f(A)$ and $f(B)$ is not smaller than the distance between points $A$ and $B$. Prove there exists a point $X$ such that $f(X)=X$.

Let $ABC$ be triangle with circumcircle $(O)$ of fixed $BC$, $AB \ne AC$ and $BC$ not a diameter. Let $I$ be the incenter of the triangle $ABC$ and $D = AI \cap BC, E = BI \cap CA, F = CI \cap AB$. The circle passing through $D$ and tangent to $OA$ cuts for second time $(O)$ at $G$ ($G \ne A$). $GE, GF$ cut $(O)$ also at $M, N$ respectively. a) Let $H = BM \cap CN$. Prove that $AH$ goes through a fixed point. b) Suppose $BE, CF$ cut $(O)$ also at $L, K$ respectively and $AH \cap KL = P$. On $EF$ take $Q$ for $QP = QI$. Let $J$ be a point of the circimcircle of triangle $IBC$ so that $IJ \perp IQ$. Prove that the midpoint of $IJ$ belongs to a fixed circle.

Let $BC$ be a chord of a circle $(O)$ such that $BC$ is not a diameter. Let $AE$ be the diameter perpendicular to $BC$ such that $A$ belongs to the larger arc $BC$ of $(O)$. Let $D$ be a point on the larger arc $BC$ of $(O)$ which is different from $A$. Suppose that $AD$ intersects $BC$ at $S$ and $DE$ intersects $BC$ at $T$. Let $F$ be the midpoint of $ST$ and $I$ be the second intersection point of the circle $(ODF)$ with the line $BC$. 1. Let the line passing through $I$ and parallel to $OD$ intersect $AD$ and $DE$ at $M$ and $N$, respectively. Find the maximum value of the area of the triangle $MDN$ when $D$ moves on the larger arc $BC$ of $(O)$ (such that $D \ne A$). 2. Prove that the perpendicular from $D$ to $ST$ passes through the midpoint of $MN$

Given is a triangle $ABC$ with its circumscribed circle and $| AC | <| AB |$. On the short arc $AC$, there is a variable point $D\ne A$. Let $E$ be the reflection of $A$ wrt the inner bisector of $\angle BDC$. Prove that the line $DE$ passes through a fixed point, regardless of point $D$.

Let $ABC$ be a triangle with two isogonal points $ P$ and $Q$ . Let $D, E$ be the projection of $P$ on $AB$, $AC$. $G$ is the projection of $Q$ on $BC$. $U$ is the projection of $G$ on $DE$, $ L$ is the projection of $P$ on $AQ$, $K$ is the symmetric of $L$ wrt $UG$. Prove that $UK$ passes through a fixed point when $P$ and $Q$ vary.

The points $A \ne B$ are given on the plane. The point $C$ moves along the plane in such a way that $\angle ACB = \alpha$ , where $\alpha$ is the fixed angle from the interval ($0^o, 180^o$). The circle inscribed in triangle $ABC$ has center the point $I$ and touches the sides $AB, BC, CA$ at points $D, E, F$ accordingly. Rays $AI$ and $BI$ intersect the line $EF$ at points $M$ and $N$, respectively. Show that: a) the segment $MN$ has a constant length, b) all circles circumscribed around triangle $DMN$ have a common point

The chord $MN$ on the circle is fixed. For every diameter $AB$ of the circle consider the intersection point $C$ of the lines $AM$ and $BN$ and construct the line $\ell$ passing through $C$ perpendicularly to $AB$. Prove that all the lines $\ell$ pass through a fixed point. (E. Kulanin, Moscow)

Three distinct points $A$, $B$, and $C$ are fixed on a line in this order. Let $\Gamma$ be a circle passing through $A$ and $C$ whose center does not lie on the line $AC$. Denote by $P$ the intersection of the tangents to $\Gamma$ at $A$ and $C$. Suppose $\Gamma$ meets the segment $PB$ at $Q$. Prove that the intersection of the bisector of $\angle AQC$ and the line $AC$ does not depend on the choice of $\Gamma$.

Given two fixed points $A$ and $B$ .The point $M$ runs along the circumference containing $A$ and $B$. $K$ is the midpoint of the segment $[MB]$. $[KP]$ is a perpendicular to the line $(MA)$. a) Prove that all the possible lines $(KP)$ pass through one point. b) Find the set of all the possible points $P$.

On the side $AB$ of a triangle $ABC$, two points $X, Y$ are chosen so that $AX = BY$. Lines $CX$ and $CY$ meet the circumcircle of the triangle, for the second time, at points $U$ and $V$. Prove that all lines $UV$ (for all $X, Y$, given $A, B, C$) have a common point.