The convex set $F$ does not cover a semi-circle of radius $R$. Is it possible that two sets, congruent to $F$, cover the circle of radius $R$ ? What if $F$ is not convex? ( N . B . Vasiliev , A. G . Samosvat)

Show that no matter how $15$ points are plotted within a circle of radius $2$ (circle border included), there will be a circle with radius $1$ (circle border including) which contains at least three of the $15$ points.

The numbers from 1 to $n$ are arranged in some way on a circle. What’s the smallest value of $n$, for which no matter how the numbers are arranged there exist ten consecutively increasing numbers $A_1<A_2<A_3…<A_{10}$ such that $A_1 A_2…A_{10}$ is a convex decagon?

Assume there are $ n\ge3$ points in the plane, Prove that there exist three points $ A,B,C$ satisfying $ 1\le\frac{AB}{AC}\le\frac{n\plus{}1}{n\minus{}1}.$

Let $\mathcal S$ be a set of $16$ points in the plane, no three collinear. Let $\chi(S)$ denote the number of ways to draw $8$ lines with endpoints in $\mathcal S$, such that no two drawn segments intersect, even at endpoints. Find the smallest possible value of $\chi(\mathcal S)$ across all such $\mathcal S$. [i]Ankan Bhattacharya[/i]

(a) Let be given $2n$ distinct points on a circumference, $n$ of which are red and $n$ are blue. Prove that one can join these points pairwise by $n$ segments so that no two segments intersect and the endpoints of each segments have different colors. (b) Show that the statement from (a) remains valid if the points are in an arbitrary position in the plane so that no three of them are collinear.

Given $3n$ points in the plane, no three collinear, is it always possible to form $n$ triangles (with vertices at the points), so that no point in the plane lies in more than one triangle?

Let $A{}$ be a point in the Cartesian plane. At each step, Ann tells Bob a number $0< a\leqslant 1$ and he then moves $A{}$ in one of the four cardinal directions, at his choice, by a distance of $a{}$. This process cotinues as long as Ann wishes. Amongst every $100$ consecutive moves, each of the four possible moves should have been made at least once. Ann's goal is to force Bob to eventually choose a point at a distance greater than $100$ from the initial position of $A{}$. Can Ann achieve her goal?

An approximate construction of a regular pentagon goes as follows. Inscribe an arbitrary convex pentagon $P_1P_2P_3P_4P_5$ in a circle. Now choose an arror bound $\epsilon > 0$ and apply the following procedure. (a) Denote $P_0 = P_5$ and $P_6 = P_1$ and construct the midpoint $Q_i$ of the circular arc $P_{i-1}P_{i+1}$ containing $P_i$. (b) Rename the vertices $Q_1,...,Q_5$ as $P_1,...,P_5$. (c) Repeat this procedure until the difference between the lengths of the longest and the shortest among the arcs $P_iP_{i+1}$ is less than $\epsilon$. Prove this procedure must end in a finite time for any choice of $\epsilon$ and the points $P_i$.

There are $12$ points that are vertices of a regular polygon with $12$ sides. Rafael must draw segments that have their two ends at two of the points drawn. He is allowed to have each point be an endpoint of more than one segment and for the segments to intersect, but he is prohibited from drawing three segments that are the three sides of a triangle in which each vertex is one of the $12$ starting points. Find the maximum number of segments Rafael can draw and justify why he cannot draw a greater number of segments.

Consider the $8\times 8\times 8$ Rubik’s cube below. Each face is painted with a different color, and it is possible to turn any layer, as you can with smaller Rubik’s cubes. Let $X$ denote the move that turns the shaded layer shown (indicated by arrows going from the top to the right of the cube) clockwise by $90$ degrees, about the axis labeled $X$. When move $X$ is performed, the only layer that moves is the shaded layer. Likewise, define move $Y$ to be a clockwise $90$-degree turn about the axis labeled Y, of just the shaded layer shown (indicated by the arrows going from the front to the top, where the front is the side pierced by the $X$ rotation axis). Let $M$ denote the move “perform $X$, then perform $Y$.” [img]https://cdn.artofproblemsolving.com/attachments/e/f/951ea75a3dbbf0ca23c45cd8da372595c2de48.png[/img] Imagine that the cube starts out in “solved” form (so each face has just one color), and we start doing move $M$ repeatedly. What is the least number of repeats of $M$ in order for the cube to be restored to its original colors?

If $5$ points are placed in the plane at lattice points (i.e. points $(x,y)$ where $x $and $y$ are both integers) such that no three are collinear, then there are $10$ triangles whose vertices are among these points. What is the minimum possible number of these triangles that have area greater than $1/2$?

Cut an acute triangle, one of whose sides is equal to the altitude drawn, by two straight cuts, into four parts, from which you can fold a square.

Let $K$ be a convex planar set, symmetric about a point $O$, and let $X, Y , Z$ be three points in $K$. Show that $K$ contains the head of one of the vectors $\overrightarrow{OX} \pm \overrightarrow{OY} , \overrightarrow{OX} \pm \overrightarrow{OZ}, \overrightarrow{OY} \pm \overrightarrow{OZ}$.

The workers laid a floor of size $n \times n$ with tiles of two types: $2 \times 2$ and $3 \times 1$. It turned out that they were able to completely lay the floor in such a way that the same number of tiles of each type was used. Under what conditions could this happen? (You can’t cut tiles and also put them on top of each other.)

A lattice point in the plane is a point both of whose coordinates are integers. Each lattice point has four neighboring points: upper, lower, left, and right. Let $k$ be a circle with radius $r \geq 2$, that does not pass through any lattice point. An interior boundary point is a lattice point lying inside the circle $k$ that has a neighboring point lying outside $k$. Similarly, an exterior boundary point is a lattice point lying outside the circle $k$ that has a neighboring point lying inside $k$. Prove that there are four more exterior boundary points than interior boundary points.

The convex $1976$-gon is divided into $1975$ triangles. Prove that there is a straight line separating one of these triangles from the rest.

There are $1978$ points in the flat plane. Each point has a circular disk with that point as its center and the radius is the distance to a fixed point. Prove that there are five of these circular disks, which together cover all $1978$ points (circular disk means: the circle and its inner area).

Find all positive integers $n$ for which it is possible to partition a regular $n$-gon into triangles with diagonals not intersecting inside the $n$-gon such that at every vertex of the $n$-gon an odd number of triangles meet.

Does there exist a set $S$ of $100$ points in a plane such that the center of mass of any $10$ points in $S$ is also a point in $S$?

Let $C$ be a unit circle and $n \ge 1$ be a fixed integer. For any set $A$ of $n$ points $P_1,..., P_n$ on $C$ define $D(A) = \underset{d}{max}\, \underset{i}{min}\delta (P_i, d)$, where $d$ goes over all diameters of $C$ and $\delta (P, \ell)$ denotes the distance from point $P$ to line $\ell$. Let $F_n$ be the family of all such sets $A$. Determine $D_n = \underset{A\in F_n}{min} D(A)$ and describe all sets $A$ with $D(A) = D_n$.

Prove that given $n$ points in the plane, not all aligned, there exists a line that passes through exactly two of them. [hide=original wording]Demostrar que dados n puntos en el plano, no todos alineados, existe una recta que pasa por exactamente dos de ellos.[/hide]

Let $ k$ and $ n$ be integers with $ 0\le k\le n \minus{} 2$. Consider a set $ L$ of $ n$ lines in the plane such that no two of them are parallel and no three have a common point. Denote by $ I$ the set of intersections of lines in $ L$. Let $ O$ be a point in the plane not lying on any line of $ L$. A point $ X\in I$ is colored red if the open line segment $ OX$ intersects at most $ k$ lines in $ L$. Prove that $ I$ contains at least $ \dfrac{1}{2}(k \plus{} 1)(k \plus{} 2)$ red points. [i]Proposed by Gerhard Woeginger, Netherlands[/i]

Is it possible to cut a square of side $1$ into two parts and rearrange them so that one can cover a circle having diameter greater than $1$? (Note: any circle with diameter greater than $1$ suffices) [i](A. Shapovalov)[/i] (Translated from [url=http://sasja.shap.homedns.org/Turniry/TG/index.html]here.[/url])

Let $S = \{P_1,P_2,...,P_{2000}\}$ be a set of $2000$ points in the interior of a circle of radius $1$, one of which at its center. For $i = 1,2,...,2000$ denote by $x_i$ the distance from $P_i$ to the closest point $P_j \ne P_i$. Prove that $x_1^2 +x_2^2 +...+x_{2000}^2<9$ .