Let $ \mathbb{K} $ be a field and let be a matrix $ M\in\mathcal{M}_3(\mathbb{K} ) $ having the property that $ \text{tr} (A) =\text{tr} (A^2) =0 . $ Show that there is a $ \mu\in \mathbb{K} $ such that $ A^3=\mu A $ or $ A^3=\mu I. $ [i]Cristinel Mortici[/i]

Let $ABC$ be a triangle with incentre $I$. The circle through $B$ tangent to $AI$ at $I$ meets side $AB$ again at $P$. The circle through $C$ tangent to $AI$ at $I$ meets side $AC$ again at $Q$. Prove that $PQ$ is tangent to the incircle of $ABC.$

Three congruent equilateral triangles $T_1$, $T_2$, and $T_3$ are stacked from left to right inside rectangle $JHMT$ such that the bottom left vertex of $T_1$ is $T$, the bottom side of $T_1$ lies on $\overline{MT}$, the bottom left vertex of $T_2$ is the midpoint of a side of $T_1$, the bottom left vertex of $T_3$ is the midpoint of a side of $T_2$, and the other two vertices of $T_3$ lie on $\overline{JH}$ and $\overline{HM}$, as shown below. Given that rectangle $JHMT$ has area $2022$, find the area of any one of the triangles $T_1$, $T_2$, or $T_3$. [asy] unitsize(0.111111111111111111cm); real s = sqrt(4044/sqrt(75)); real l = 5s/2; real w = s * sqrt(3); pair J,H,M,T,V1,V2,V3,V4,V5,V6,V7,V8,C1,C2,C3; J = (0,w); H = (l,w); M = (l,0); T = (0,0); V1 = (s,0); V2 = (s/2,s * sqrt(3)/2); V3 = (V1+V2)/2; V4 = (3 * s/4+s,s * sqrt(3)/4); V5 = (3 * s/4+s/2,s * sqrt(3)/4+s * sqrt(3)/2); V6 = (V4+V5)/2; V7 = (l,s * sqrt(3)/4+s * sqrt(3)/4); V8 = (l-s/2,w); C1 = (T+V1+V2)/3; C2 = (V3+V4+V5)/3; C3 = (V6+V7+V8)/3; draw(J--H--M--T--cycle); draw(V1--V2--T); draw(V3--V4--V5--cycle); draw(V6--V7--V8--cycle); label("$J$", J, NW); label("$H$", H, NE); label("$M$", M, SE); label("$T$", T, SW); label("$T_1$", C1); label("$T_2$", C2); label("$T_3$", C3); [/asy]

Let $n\geq 3$ be an integer and $S_{n}$ the permutation group. $G$ is a subgroup of $S_{n}$, generated by $n-2$ transpositions. For all $k\in\{1,2,\ldots,n\}$, denote by $S(k)$ the set $\{\sigma(k) \ : \ \sigma\in G\}$. Show that for any $k$, $|S(k)|\leq n-1$.

Recall that a plane divides $\mathbb{R}^3$ into two regions, two parallel planes divide it into three regions, and two intersecting planes divide space into four regions. Consider the six planes which the faces of the cube $ABCD-A_1B_1C_1D_1$ lie on, and the four planes that the tetrahedron $ACB_1D_1$ lie on. How many regions do these ten planes split the space into?

Find the integer which is closest to the value of the following expression: $$\left((3 + \sqrt{1})^{2023} - \left(\frac{1}{3 - \sqrt{1}}\right)^{2023} \right) \cdot \left((3 + \sqrt{2})^{2023} - \left(\frac{1}{3 - \sqrt{2}}\right)^{2023} \right) \cdot \ldots \cdot \left((3 + \sqrt{8})^{2023} - \left(\frac{1}{3 - \sqrt{8}}\right)^{2023} \right)$$

In a fish shop with 28 kinds of fish, there are 28 fish sellers. In every seller, there exists only one type of each fish kind, depending on where it comes, Mediterranean or Black Sea. Each of the $k$ people gets exactly one fish from each seller and exactly one fish of each kind. For any two people, there exists a fish kind which they have different types of it (one Mediterranean, one Black Sea). What is the maximum possible number of $k$?

Let $\omega_1,\omega_2$ be two non-intersecting circles, with circumcenters $O_1,O_2$ respectively, and radii $r_1,r_2$ respectively where $r_1 < r_2$. Let $AB,XY$ be the two internal common tangents of $\omega_1,\omega_2$, where $A,X$ lie on $\omega_1$, $B,Y$ lie on $\omega_2$. The circle with diameter $AB$ meets $\omega_1,\omega_2$ at $P$ and $Q$ respectively. If $$\angle AO_1P+\angle BO_2Q=180^{\circ},$$ find the value of $\frac{PX}{QY}$ (in terms of $r_1,r_2$).

Let $p$ be a prime. For which $k$ can the set $\{1, 2, \dots , k\}$ be partitioned into $p$ subsets with equal sums of elements ?

Given four points $A,$ $B,$ $C,$ $D$ on a circle such that $AB$ is a diameter and $CD$ is not a diameter. Show that the line joining the point of intersection of the tangents to the circle at the points $C$ and $D$ with the point of intersection of the lines $AC$ and $BD$ is perpendicular to the line $AB.$

Pablo wrote $5$ numbers on one sheet and then wrote the numbers $6,7,8,8,9,9,10,10,11$ and $ 12$ on another sheet that he gave Sofia, indicating that those numbers are the possible sums of two of the numbers that he had hidden. Decide if with this information Sofia can determine the five numbers Pablo wrote .

Suppose that there is a point $P$ inside a convex quadrilateral $ABCD$ such that the triangles $PAB$, $PBC$, $PCD$, $PDA$ have equal areas. Prove that one of the diagonals bisects the area of $ABCD$.

Find all functions $f,g: \mathbb{Q}\to \mathbb{Q}$ such that the following two conditions hold: $$f(g(x)-g(y))=f(g(x))-y \ \ (1)$$ $$g(f(x)-f(y))=g(f(x))-y\ \ (2)$$ for all $x,y \in \mathbb{Q}$.

Find all integer solutions $(a,b)$ of the equation \[ (a+b+3)^2 + 2ab = 3ab(a+2)(b+2)\]

Consider $n$ lines in the plane in general position, that is, there are not three of the $n$ lines that pass through the same point. Determine if it is possible to label the $k$ points where these lines are inserted with the numbers $1$ through $k$ (using each number exactly once), so that on each line, the labels of the $n-1$ points of that line are arranged in increasing order (in one of the two directions in which they can be traversed).

Let $P$ be an arbitrary point on the diagonal $AC$ of cyclic quadrilateral $ABCD$, and $PK, PL, PM, PN, PO$ be the perpendiculars from $P$ to $AB, BC, CD, DA, BD$ respectively. Prove that the distance from $P$ to $KN$ is equal to the distance from $O$ to $ML$.

A cube of edge length $3$ consists of $27$ unit cubes. Find the number of lines passing through exactly three centers of these $27$ cubes, as well as the number of those passing through exactly two such centers.

A trapezoid has side lengths as indicated in the figure (the sides with length $11$ and $36$ are parallel). Calculate the area of the trapezoid.[img]https://1.bp.blogspot.com/-5PKrqDG37X4/XzcJtCyUv8I/AAAAAAAAMY0/tB0FObJUJdcTlAJc4n6YNEaVIDfQ91-eQCLcBGAsYHQ/s0/1995%2BMohr%2Bp1.png[/img]

Angles are equal in a hexagon, three main diagonals are equal and the other six diagonals are also equal. Is it true that it has equal sides?

Suppose that $2n$ points of an $n\times n$ grid are marked. Show that for some $k > 1$ one can select $2k$ distinct marked points, say $a_1,...,a_{2k}$, such that $a_{2i-1}$ and $a_{2i}$ are in the same row, $a_{2i}$ and $a_{2i+1}$ are in the same column, $\forall i$, indices taken mod 2n.

Consider $n$ segments in the plane which no two intersect and between their $2n$ endpoints no three are collinear. Is the following statement true? Statement: There exists a simple $2n$-gon such that it's vertices are the $2n$ endpoints of the segments and each segment is either completely inside the polygon or an edge of the polygon.

Prove that $\sqrt 2$ is irrational.

If five squares of a $3 \times 3$ board initially colored white are chosen at random and blackened, what is the expected number of edges between two squares of the same color?

In a certain language there are only two letters, $A$ and $B$. We know that (i) There are no words of length $1$, and the only words of length $2$ are $AB$ and $BB$. (ii) A segment of length $n > 2$ is a word if and only if it can be obtained from a word of length less than $n$ by replacing each letter $B$ by some (not necessarily the same) word. Prove that the number of words of length $n$ is equal to $\frac{2^n +2\cdot (-1)^n}{3}$

The cells of a square $2011 \times 2011$ array are labelled with the integers $1,2,\ldots, 2011^2$, in such a way that every label is used exactly once. We then identify the left-hand and right-hand edges, and then the top and bottom, in the normal way to form a torus (the surface of a doughnut). Determine the largest positive integer $M$ such that, no matter which labelling we choose, there exist two neighbouring cells with the difference of their labels at least $M$. (Cells with coordinates $(x,y)$ and $(x',y')$ are considered to be neighbours if $x=x'$ and $y-y'\equiv\pm1\pmod{2011}$, or if $y=y'$ and $x-x'\equiv\pm1\pmod{2011}$.) [i](Romania) Dan Schwarz[/i]