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 $.

Given two parallel lines $r$ and $r'$ and a point $P$ on the plane that contains them and that is not on them, determine an equilateral triangle whose vertex is point $P$ , and the other two, one on each of the two lines. [img]https://cdn.artofproblemsolving.com/attachments/9/3/1d475eb3e9a8a48f4a85a2a311e1bda978e740.png[/img]

In a given plane $\varrho$ consider a convex quadrilateral $ABCD$ and denote $E=AC\cap BD.$ Moreover, consider a point $V\notin\varrho$. On rays $VA,VB,VC,VD$ find points $A',B',C',D'$ respectively such that $E,A',B',C',D'$ are coplanar and $A'B'C'D'$ is a parallelogram. Discuss conditions of solvability.

Let $r>1$ be a rational number. Alice plays a solitaire game on a number line. Initially there is a red bead at $0$ and a blue bead at $1$. In a move, Alice chooses one of the beads and an integer $k \in \mathbb{Z}$. If the chosen bead is at $x$, and the other bead is at $y$, then the bead at $x$ is moved to the point $x'$ satisfying $x'-y=r^k(x-y)$. Find all $r$ for which Alice can move the red bead to $1$ in at most $2021$ moves.

Given two intersecting circles with centers $O_1, O_2$. Construct the circle touching one of them externally and the second one internally such that the distance from its center to $O_1O_2$ is maximal. (M.Volchkevich)

Given that a $60^{\circ}$ angle cannot be trisected with ruler and compass, prove that a $\frac{120^{\circ}}{n}$ angle cannot be trisected with ruler and compass for $n=1,2,\ldots$

Let $CH$ be the altitude of the triangle $ABC$ drawn on the board, in which $\angle C = 90^o$, $CA \ne CB$. The mathematics teacher drew the perpendicular bisectors of segments$ CA$ and $CB$, which cut the line CH at points $K$ and $M$, respectively, and then erased the drawing, leaving only the points $C, K$ and $M$ on the board. Restore triangle $ABC$, using only a compass and a ruler.

In making Euclidean constructions in geometry it is permitted to use a ruler and a pair of compasses. In the constructions considered in this question no compasses are permitted, but the ruler is assumed to have two parallel edges, which can be used for constructing two parallel lines through two given points whose distance is at least equal to the breadth of the rule. Then the distance between the parallel lines is equal to the breadth of the ruler. Carry through the following constructions with such a ruler. Construct: [b]a)[/b] The bisector of a given angle. [b]b)[/b] The midpoint of a given rectilinear line segment. [b]c)[/b] The center of a circle through three given non-collinear points. [b]d)[/b] A line through a given point parallel to a given line.

Construct a triangle given an angle, the side opposite the angle and the median to the other side (researching the number of solutions is not required).

Given a circle $k$ and two points $A$ and $B$ outside the circle. Specify how to can construct a circle with a compass and ruler, so that $A$ and $B$ lie on that circle and that circle is tangent to $k$.

Prove that among every ten consecutive natural numbers there is one that is coprime to each of the other nine.

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.

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 $r$.

Let $ABCD$ be a rectangle and let $M, N$ and $P, Q$ be the points of intersections of some line $e$ with the sides $AB, CD$ and $AD, BC$, respectively (or their extensions). Given the points $M, N, P, Q$ and the length $p$ of side $AB$, construct the rectangle. Under what conditions can this problem be solved, and how many solutions does it have?

We are given two concentric circles, Construct a square whose two vertices lie on one circle and the other two on the other circle.

Show how to construct a line segment length $(a^4 + b^4)^{1/4}$ given segments lengths $a$ and $b$.

The student drew a triangle $ABC$ on the board, in which $AB>BC$. On the side $AB$ is taken point $D$ such that $BD = AC$. Let points $E$ and $F$ be the midpoints of the segments $AD$ and $BC$ respectively. Then the whole picture was erased, leaving only dots $E$ and $F$. Restore triangle $ABC$.

A segment of length $\sqrt2 + \sqrt3 + \sqrt5$ is drawn. Is it possible to draw a segment of unit length using a compass and a straightedge? [i](3 points)[/i]

$9$ horizontal and $9$ vertical lines are drawn through a square. Prove that it is possible to select $20$ rectangles so that the sides of each rectangle is a segment of one of the given lines (including the sides of the square), and for any two of the $20$ rectangles, it is possible to cover one of them with the other (rotations are allowed).

In a right triangle $ABC$ with a hypotenuse $AB$, the angle $A$ is greater than the angle $B$. Point $N$ lies on the hypotenuse $AB$ , such that $BN = AC$. Construct this triangle $ABC$ given the point $N$, point $F$ on the side $AC$ and a straight line $\ell$ containing the bisector of the angle $A$ of the triangle $ABC$. (Grigory Filippovsky)

A $2\times 2$ square was cut from a squared sheet of paper. Using only a ruler without divisions and without going beyond the square, divide the diagonal of the square into $6$ equal parts.

A circle $k$ is drawn using a given disc (e.g. a coin). A point $A$ is chosen on $k$. Using just the given disc, determine the point $B$ on $k$ so that $AB$ is a diameter of $k$. (You are allowed to choose an arbitrary point in one of the drawn circles, and using the given disc it is possible to construct either of the two circles that passes through the points at a distance that is smaller than the radius of the circle.)