Construct triangle $ABC$, given $h_a$, $h_b$ (the altitudes from $A$ and $B$), and $m_a$, the median from vertex $A$.

Using only angle with angle $\frac{\pi}{7}$ and a ruler, constuct angle $\frac{\pi}{14}$

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!).

An infinite sequence $ \,x_{0},x_{1},x_{2},\ldots \,$ of real numbers is said to be [b]bounded[/b] if there is a constant $ \,C\,$ such that $ \, \vert x_{i} \vert \leq C\,$ for every $ \,i\geq 0$. Given any real number $ \,a > 1,\,$ construct a bounded infinite sequence $ x_{0},x_{1},x_{2},\ldots \,$ such that \[ \vert x_{i} \minus{} x_{j} \vert \vert i \minus{} j \vert^{a}\geq 1 \] for every pair of distinct nonnegative integers $ i, j$.

Let $n$ be a fixed positive integer and fix a point $O$ in the plane. There are $n$ lines drawn passing through the point $O$. Determine the largest $k$ (depending on $n$) such that we can always color $k$ of the $n$ lines red in such a way that no two red lines are perpendicular to each other. [i]Proposed by Nikola Velov[/i]

On the board was drawn a circle with a marked center, a quadrangle inscribed in it, and a circle inscribed in it, also with a marked center. Then they erased the quadrilateral (keeping one vertex) and the inscribed circle (keeping its center). Restore any of the erased vertices of the quadrilateral using only a ruler and no more than six lines.

Using only a compass and a ruler, reconstruct triangle $ABC$ given the following three points: point $M$ the intersection of its medians, point $I$ is the center of its inscribed circle and the point $Q_a$ is touch point of the inscribed circle to side $BC$.

Helen has a wooden rectangle of unknown dimensions, a straightedge, and a pencil (no compass). Is it possible for her to construct a line segment on the rectangle connecting the midpoints of two opposite sides, where she cannot draw any lines or points outside the rectangle? Note: Helen is allowed to draw lines between two points she has already marked, and mark the intersection of any two lines she has already drawn, if the intersection lies on the rectangle. Further, Helen is allowed to mark arbitrary points either on the rectangle or on a segment she has previously drawn. Assume that only the four vertices of the rectangle have been marked prior to the beginning of this process.

Determine the smallest natural number $n > 2$, or show that no such natural numbers $n$ exists, that satisfy the following condition: There exists natural numbers $a_1, a_2, \dots, a_n$ such that \[ \gcd(a_1, a_2, \dots, a_n) = \sum_{k = 1}^{n - 1} \underbrace{\left( \frac{1}{\gcd(a_k, a_{k + 1})} + \frac{1}{\gcd(a_k, a_{k + 2})} + \dots + \frac{1}{\gcd(a_k, a_n)} \right)}_{n - k \ \text{terms}} \]

Suppose there are several juice boxes, one of which is poisoned. You have $n$ guinea pigs to test the boxes. The testing happens in the following way: [list] [*] At each round, you can have the guinea pigs taste any number of juice boxes. [*] Conversely, a juice box can be tasted by any number of guinea pigs. [*] After the round ends, any guinea pigs who tasted the poisoned juice die. [/list] Suppose you have to find the poisoned juice box in at most $k$ rounds. What is the maximum number of juice boxes such that it is possible?

On a straight line, three distinct points $ M $, $ D $, $ H $ are given. Construct a right-angled triangle for which $ M $ is the midpoint of the hypotenuse, $ D $ is the point of intersection of the bisector of the right angle with the hypotenuse, and $ H $ is the foot of the altitude to the hypotenuse.

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

Let $ABC$ be a triangle right in $A$. Construct a point $P$ on the hypotenuse $BC$ such that if $Q$ is the foot of the perpendicular drawn from $P$ to side $AC$, then the area of the square of side $PQ$ is equal to the area of the rectangle of sides $PB$ and $PC$. Show construction steps.

Two chords are drawn in a circle of radius equal to unit, $AB$ and $AC$ of equal length. a) Describe how you can construct a third chord $DE$ that is divided into three equal parts by the intersections with $AB$ and $AC$. b) If $AB = AC =\sqrt2$, what are the lengths of the two segments that the chord $DE$ determines in $AB$?

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]

The points $S$ lie on side $AB$, $T$ on side $BC$, and $U$ on side $CA$ of a triangle so that the following applies: $\overline{AS} : \overline{SB} = 1 : 2$, $\overline{BT} : \overline{TC} = 2 : 3$ and $\overline{CU} : \overline{UA} = 3 : 1$. Construct the triangle $ABC$ if only the points $S, T$ and $U$ are given.

Prove that there exist infinitely many positive integers $k$ such that the equation $$\frac{n^2+m^2}{m^4+n}=k$$ don't have any positive integer solution.

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.

Exactly $n$ chords (i.e. diagonals and edges) of a regular $2n$-gon are coloured red, satisfying the following two conditions: (1) Each of the $2n$ vertices occurs exactly once as the endpoint of a red chord. (2) No two red chords have the same length. For which positive integers $n \ge 2$ is this possible?

Inside an acute angle is a circle. Investigate the possibility of constructing with only a compass and a ruler, a tangent to this circle that the point of contact will bisect the segment of the tangent that is cut off by the sides of the angle.

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

Given a circle and two points $P$ and $Q$, construct a right triangle inscribed in the circle such that its two legs pass through the points $P$ and $Q$ respectively. For what positions of $P$ and $Q$ is this construction impossible?

The triangle $ABC$ is drawn on the board such that $AB + AC = 2BC$. The bisectors $AL_1, BL_2, CL_3$ were drawn in this triangle, after which everything except the points $L_1, L_2, L_3$ was erased. Use a compass and a ruler to reconstruct triangle $ABC$.

In a trapezoid $ABCD$ with base $AB$ let $E$ be the midpoint of side $AD$. Suppose further that $2CD=EC=BC=b$. Let $\angle ECB=120^{\circ}$. Construct the trapezoid and determine its area based on $b$.

Consider a triangle $ABC$. For every point M $\in BC$ ,we define $N \in CA$ and $P \in AB$ such that $APMN$ is a parallelogram. Let $O$ be the intersection of $BN$ and $CP$. Find $M \in BC$ such that $\angle PMO=\angle OMN$