Peter made a paper rectangle, put it on an identical rectangle and pasted both rectangles along their perimeters. Then he cut the upper rectangle along one of its diagonals and along the perpendiculars to this diagonal from two remaining vertices. After this he turned back the obtained triangles in such a way that they, along with the lower rectangle form a new rectangle. Let this new rectangle be given. Restore the original rectangle using compass and ruler.

Using a ruler with a length of $20$ cm and a compass with a maximum deviation of $10$ cm to connect the segment given two points lying at a distance of $1$ m.

Three rays with a common origin are drawn on the plane, dividing the plane into three angles. One point is marked inside each corner. Using one ruler, construct a triangle whose vertices lie on the given rays and whose sides contain the given points.

On the plane, there is drawn a parallelogram $P$ and a point $X$ outside of $P$. Using only an ungraded rule, determine the point $W$ that is symmetric to $X$ with respect to the center $O$ of $P$.

Construct a triangle given the radii of the excircles.

A trapezoid $ABCD$ with bases $BC = a$ and $AD = 2a$ is drawn on the plane. Using only with a ruler, construct a triangle whose area is equal to the area of the trapezoid. With the help of a ruler you can draw straight lines through two known points. (Rozhkova Maria)

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

Let $n$ be a positive integer. The Georgian folk dance team consists of $2n$ dancers, with $n$ males and $n$ females. Each dancer, both male and female, is assigned a number from 1 to $n$. During one of their dances, all the dancers line up in a single line. Their wish is that, for every integer $k$ from 1 to $n$, there are exactly $k$ dancers positioned between the $k$th numbered male and the $k$th numbered female. Prove the following statements: a) If $n \equiv 1 \text{ or } 2 \mod{4}$, then the dancers cannot fulfill their wish. b) If $n \equiv 0 \text{ or } 3 \mod{4}$, then the dancers can fulfill their wish. [i]Proposed by Giorgi Arabidze, Georgia[/i]

Given a triangle $ABC$ ($AB> AC$) and a circle circumscribed around it. Construct with a compass and a ruler the midpoint of the arc $BC$ (not containing vertex $A$), with no more than two lines (straight or circles).

With a compass and a ruler, split a triangle into two smaller triangles with the same sum of squares of sides.

Let $ABC$ be an acute triangle, and its altitudes $AX,BY,CZ$ concurrent at $H$. Construct circles $(K_a), (K_b), (K_c)$ circumscribing the triangles $AY Z, BZX, CXY$ . Construct a circle $(K)$ that is internally tangent to all the three circles $(Ka), (K_b), (K_c)$. Prove that $(K)$ is tangent to the circumcircle $(O)$ of the triangle $ABC$. Đỗ Thanh Sơn

Albatross and Frankinfueter both own a circle. Frankinfueter also has stolen from Prof. Trugweg a ruler. Before that, Trugweg had two points with a distance of $1$ drawn his (infinitely large) board. For a natural number $n$, let A $(n)$ be the number of the construction steps that Albatross needs at least to create two points with a distance of $n$ to construct. Similarly, Frankinfueter needs at least $F(n)$ steps for this. How big can $\frac{A (n)}{F (n)}$ become? There are only the following three construction steps: a) Mark an intersection of two straight lines, two circles or a straight line with one circle. b) Pierce at a marked point $P$ and draw a circle around $P$ through one marked point . c) Draw a straight line through two marked points (this implies possession of a ruler ahead!).

Construct a point $N$ inside a given right triangle $ABC$ such that the angles $\angle NBC$, $\angle NCA$ and $\angle NAB$ are equal.

Given: point $ A $, line $ p $, and circle $ k $. Construct a triangle $ ABC $ with angles $ A = 60^\circ $, $ B = 90^\circ $, whose vertex $ B $ lies on line $ p $, and vertex $ C $ - on circle $ k $.

If an acute-angled triangle $ABC$ is given, construct an equilateral triangle $A'B'C'$ in space such that lines $AA',BB', CC'$ pass through a given point.

On the plane are given four distinct points $A,B,C,D$ on a line $g$ in this order, at the mutual distances $AB = a, BC = b, CD = c$. (a) Construct (if possible) a point $P$ outside line $g$ such that $\angle APB =\angle BPC =\angle CPD$. (b) Prove that such a point $P$ exists if and only if $ (a+b)(b+c) < 4ac$

Two circles meeting at points $A, B$ and a point $O$ lying outside them are given. Using a compass and a ruler construct a ray with origin $O$ meeting the first circle at point $C$ and the second one at point $D$ in such a way that the ratio $OC : OD$ be maximal.

Inside the triangle $ ABC $ a point $ P $ is given; find a point $ Q $ on the perimeter of this triangle such that the broken line $ APQ $ divides the triangle into two parts with equal areas.

Does there exist a positive integer $N$ which is divisible by at least $2024$ distinct primes and whose positive divisors $1 = d_1 < d_2 < \ldots < d_k = N$ are such that the number \[ \frac{d_2}{d_1}+\frac{d_3}{d_2}+\ldots+\frac{d_k}{d_{k-1}} \] is an integer?

The figure shows a triangle, a circle circumscribed around it and the center of its inscribed circle. Using only one ruler (one-sided, without divisions), construct the center of the circumscribed circle.

$(NET 5)$ The bisectors of the exterior angles of a pentagon $B_1B_2B_3B_4B_5$ form another pentagon $A_1A_2A_3A_4A_5.$ Construct $B_1B_2B_3B_4B_5$ from the given pentagon $A_1A_2A_3A_4A_5.$

Let \( \omega \) be the circumcircle of triangle \( ABC \), where \( AB > AC \). Let \( N \) be the midpoint of arc \( \smile\!BAC \), and \( D \) a point on the circle \( \omega \) such that \( ND \perp AB \). Let \( I \) be the incenter of triangle \( ABC \). Reconstruct triangle \( ABC \), given the marked points \( A, D, \) and \( I \). Proposed by Oleksii Karlyuchenko and Hryhorii Filippovskyi

The teacher drew a coordinate plane on the board and marked some points on this plane. Unfortunately, Vasya's second-grader, who was on duty, erased almost the entire drawing, except for two points $A (1, 2)$ and $B (3,1)$. Will the excellent Andriyko be able to follow these two points to construct the beginning of the coordinate system point $O (0, 0)$? Point A on the board located above and to the left of point $B$.

Construct a quadrangle along the given sides $a, b, c$, and $d$ and the distance $I$ between the midpoints of its diagonals.

For which positive integers $n$, one can find real numbers $x_1,x_2,\cdots ,x_n$ such that $$\dfrac{x_1^2+x_2^2+\cdots+x_n^2}{\left(x_1+2x_2+\cdots+nx_n\right)^2}=\dfrac{27}{4n(n+1)(2n+1)}$$ and $i\leq x_i\leq 2i$ for all $i=1,2,\cdots ,n$ ?