Given a point $P$ lying on a line $g,$ and given a circle $K.$ Construct a circle passing through the point $P$ and touching the circle $K$ and the line $g.$

For a given quadrilateral $A_1A_2B_1B_2,$ a point $P$ is called [i]phenomenal[/i], if line segments $A_1A_2$ and $B_1B_2$ subtend the same angle at point $P$ (i.e. triangles $PA_1A_2$ and $PB_1B_2$ which can be also also degenerate have equal inner angles at point $P$ disregarding orientation). Three non-collinear points, $A_1,A_2,$ and $B_1$ are given in the plane. Prove that it is possible to find a disc in the plane such that for every point $B_2$ on the disc, the quadrilateral $A_1A_2B_1B_2$ is convex and it is possible to construct seven distinct phenomenal points (with respect to $A_1A_2B_1B_2$) only using a right ruler. With a right ruler the following two operations are allowed: [list=1] [*]Given two points it is possible to draw the straight line connecting them; [*]Given a point and a straight line, it is possible to draw the straight line passing through the given point which is perpendicular to the given line. [/list] [i]Proposed by Á. Bán-Szabó, Budapest[/i]

In the plane we have a line $r$ and two points $A$ and $B$ outside the line and in the same half plane. Determine a point $M$ on the line such that the angle of $r$ with $AM$ is double that of $r$ with $BM$. (Consider the smaller angle of two lines of the angles they form).

A circle is drawn on the plane. How to use only a ruler to draw a perpendicular from a given point outside the circle to a given line passing through the center of this circle?

If possible, construct an equilateral triangle whose three vertices are on three given circles.

Let $I$ be the incenter of triangle $ABC$. $K_1$ and $K_2$ are the points on $BC$ and $AC$ respectively, at which the inscribed circle is tangent. Using a ruler and a compass, find the center of the inscribed circle for triangle $CK_1K_2$ in the minimal possible number of steps (each step is to draw a circle or a line). (Hryhorii Filippovskyi)

Let $\mathbb{N}_0$ denote the set of non-negative integers. Determine all non-negative integers $k$ for which there exists a function $f: \mathbb{N}_0 \to \mathbb{N}_0$ such that $f(2024) = k$ and $f(f(n)) \leq f(n+1) - f(n)$ for all non-negative integers $n$.

Construct a square from four points, one on each side.

Canmoo is trying to do constructions, but doesn't have a ruler or compass. Instead, Canmoo has a device that, given four distinct points $A$, $B$, $C$, $P$ in the plane, will mark the isogonal conjugate of $P$ with respect to triangle $ABC$, if it exists. Show that if two points are marked on the plane, then Canmoo can construct their midpoint using this device, a pencil for marking additional points, and no other tools. (Recall that the [i]isogonal conjugate[/i] of $P$ with respect to triangle $ABC$ is the point $Q$ such that lines $AP$ and $AQ$ are reflections around the bisector of $\angle BAC$, lines $BP$ and $BQ$ are reflections around the bisector of $\angle CBA$, lines $CP$ and $CQ$ are reflections around the bisector of $\angle ACB$. Additional points marked by the pencil can be assumed to be in general position, meaning they don't lie on any line through two existing points or any circle through three existing points.) [i]Maxim Li[/i]

Reconstruct a triangle, given the incenter, the midpoint of some side and the foot of the altitude drawn on this side.

With an unmarked ruler only, reconstruct the trapezoid $ABCD$ ($AD \parallel BC$) given the vertices $A$ and $B$, the intersection point $O$ of the diagonals of the trapezoid and the midpoint $M$ of the base $AD$. (Yukhim Rabinovych)

The diagonals of a convex quadrilateral $ABCD$ meet at point $L$. The orthocenter $H$ of the triangle $LAB$ and the circumcenters $O_1, O_2$, and $O_3$ of the triangles $LBC, LCD$, and $LDA$ were marked. Then the whole configuration except for points $H, O_1, O_2$, and $O_3$ was erased. Restore it using a compass and a ruler.

A set $A$ contains exactly $n$ integers, each of which is greater than $1$ and every of their prime factors is less than $10$. Determine the smallest $n$ such that $A$ must contain at least two distinct elements $a$ and $b$ such that $ab$ is the square of an integer.

In triangle $ABC$, points $D, E$, and $F$ are marked on sides $AC, BC$, and $AB$ respectively, so that $AD = AB$, $EC = DC$, $BF = BE$. After that, they erased everything except points $E, F$ and $D$. Reconstruct the triangle $ABC$ (no study required).

In an isosceles trapezoid $ABCD$ with bases $AD$ and $BC$, diagonals intersect at point $P$, and lines $AB$ and $CD$ intersect at point $Q$. $O_1$ and $O_2$ are the centers of the circles circumscribed around the triangles $ABP$ and $CDP$, $r$ is the radius of these circles. Construct the trapezoid ABCD given the segments $O_1O_2$, $PQ$ and radius $r$.

i) $ABCD$ is a rhombus. A tangent to the inscribed circle meets $AB, DA, BC, CD$ at $M, N, P, Q$ respectively. Find a relationship between $BM$ and $DN$. ii) $ABCD$ is a rhombus and $P$ a point inside. The circles through $P$ with centers $A, B, C, D$ meet the four sides $AB, BC, CD, DA$ in eight points. Find a property of the resulting octagon. Use it to construct a regular octagon. iii) Rotate the figure about the line $AC$ to form a solid. State a similar result.

Construct the $ \triangle ABC $, with $ AC <BC $, if the circumcircle is known, and the points $ D, E, F $ in it, where they intersect, respectively, the altitude, the median and the angle bisector that they start from the vertex $ C $.

Let $n$ be a positive integer and $N=n^{2021}$. There are $2021$ concentric circles centered at $O$, and $N$ equally-spaced rays are emitted from point $O$. Among the $2021N$ intersections of the circles and the rays, some are painted red while the others remain unpainted. It is known that, no matter how one intersection point from each circle is chosen, there is an angle $\theta$ such that after a rotation of $\theta$ with respect to $O$, all chosen points are moved to red points. Prove that the minimum number of red points is $2021n^{2020}$. [I]Proposed by usjl.[/i]

Let $ABC$ be a scalene triangle. Given the center $I$ of the inscribe circle and the points $K_1$, $K_2$ and $K_3$ where the inscribed circle is tangent to the sides $BC$, $AC$ and $AB$. Using only a ruler, construct the center of the circumscribed circle of triangle $ABC$. (Hryhorii Filippovskyi)

Bolek draw a trapezoid $ABCD$ trapezoid ($AB // CD$) on the board, with its midsegment line $EF$ in it. Point intersection of his diagonal $AC, BD$ denote by $P,$ and his rectangular projection on line $AB$ denote by $Q$. Lolek, wanting to tease Bolek, blotted from the board everything except segments $EF$ and $PQ$. When Bolek saw it, wanted to complete the drawing and draw the original trapezoid, but did not know how to do it. Can you help Bolek?

At a position $O$ of an airport in a plateau there is a gun which can rotate arbitrarily. Two tanks moving along two given segments $AB$ and $CD$ attack the airport. Determine, by a ruler and a compass, the reach of the gun, knowing that the total length of the parts of the trajectories of the two tanks reachable by the gun is equal to a given length $\ell$.

A given line passes through the center $O$ of a circle. The line intersects the circle at points $A$ and $B$. Point $P$ lies in the exterior of the circle and does not lie on the line $AB$. Using only an unmarked straightedge, construct a line through $P$, perpendicular to the line $AB$. Give complete instructions for the construction and prove that it works.