A sequence of real numbers $ a_{0},\ a_{1},\ a_{2},\dots$ is defined by the formula \[ a_{i + 1} = \left\lfloor a_{i}\right\rfloor\cdot \left\langle a_{i}\right\rangle\qquad\text{for}\quad i\geq 0; \]here $a_0$ is an arbitrary real number, $\lfloor a_i\rfloor$ denotes the greatest integer not exceeding $a_i$, and $\left\langle a_i\right\rangle=a_i-\lfloor a_i\rfloor$. Prove that $a_i=a_{i+2}$ for $i$ sufficiently large. Proposed by Harmel Nestra, Estionia

The sequence of real numbers $a_0,a_1,a_2,\ldots$ is defined recursively by \[a_0=-1,\qquad\sum_{k=0}^n\dfrac{a_{n-k}}{k+1}=0\quad\text{for}\quad n\geq 1.\]Show that $ a_{n} > 0$ for all $ n\geq 1$. Proposed by Mariusz Skalba, Poland

The sequence $c_{0}, c_{1}, . . . , c_{n}, . . .$ is defined by $c_{0}= 1, c_{1}= 0$, and $c_{n+2}= c_{n+1}+c_{n}$ for $n \geq 0$. Consider the set $S$ of ordered pairs $(x, y)$ for which there is a finite set $J$ of positive integers such that $x=\textstyle\sum_{j \in J}{c_{j}}$, $y=\textstyle\sum_{j \in J}{c_{j-1}}$. Prove that there exist real numbers $\alpha$, $\beta$, and $M$ with the following property: An ordered pair of nonnegative integers $(x, y)$ satisfies the inequality \[m < \alpha x+\beta y < M\] if and only if $(x, y) \in S$. Remark: A sum over the elements of the empty set is assumed to be $0$.

Prove the inequality: \[\sum_{i < j}{\frac {a_{i}a_{j}}{a_{i} + a_{j}}}\leq \frac {n}{2(a_{1} + a_{2} +\cdots + a_{n})}\cdot \sum_{i < j}{a_{i}a_{j}}\] for positive reals $ a_{1},a_{2},\ldots,a_{n}$. Proposed by Dusan Dukic, Serbia

If $a,b,c$ are the sides of a triangle, prove that \[\frac{\sqrt{b+c-a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}+\frac{\sqrt{c+a-b}}{\sqrt{c}+\sqrt{a}-\sqrt{b}}+\frac{\sqrt{a+b-c}}{\sqrt{a}+\sqrt{b}-\sqrt{c}}\leq 3 \] Proposed by Hojoo Lee, Korea

Determine the least real number $M$ such that the inequality \[|ab(a^{2}-b^{2})+bc(b^{2}-c^{2})+ca(c^{2}-a^{2})| \leq M(a^{2}+b^{2}+c^{2})^{2}\] holds for all real numbers $a$, $b$ and $c$.

We have $ n \geq 2$ lamps $ L_{1}, . . . ,L_{n}$ in a row, each of them being either on or off. Every second we simultaneously modify the state of each lamp as follows: if the lamp $ L_{i}$ and its neighbours (only one neighbour for $ i = 1$ or $ i = n$, two neighbours for other $ i$) are in the same state, then $ L_{i}$ is switched off; – otherwise, $ L_{i}$ is switched on. Initially all the lamps are off except the leftmost one which is on. $ (a)$ Prove that there are infinitely many integers $ n$ for which all the lamps will eventually be off. $ (b)$ Prove that there are infinitely many integers $ n$ for which the lamps will never be all off.

Let $P$ be a regular $2006$-gon. A diagonal is called good if its endpoints divide the boundary of $P$ into two parts, each composed of an odd number of sides of $P$. The sides of $P$ are also called good. Suppose $P$ has been dissected into triangles by $2003$ diagonals, no two of which have a common point in the interior of $P$. Find the maximum number of isosceles triangles having two good sides that could appear in such a configuration.

Let $ S$ be a finite set of points in the plane such that no three of them are on a line. For each convex polygon $ P$ whose vertices are in $ S$, let $ a(P)$ be the number of vertices of $ P$, and let $ b(P)$ be the number of points of $ S$ which are outside $ P$. A line segment, a point, and the empty set are considered as convex polygons of $ 2$, $ 1$, and $ 0$ vertices respectively. Prove that for every real number $ x$ \[\sum_{P}{x^{a(P)}(1 - x)^{b(P)}} = 1,\] where the sum is taken over all convex polygons with vertices in $ S$. Alternative formulation: Let $ M$ be a finite point set in the plane and no three points are collinear. A subset $ A$ of $ M$ will be called round if its elements is the set of vertices of a convex $ A -$gon $ V(A).$ For each round subset let $ r(A)$ be the number of points from $ M$ which are exterior from the convex $ A -$gon $ V(A).$ Subsets with $ 0,1$ and 2 elements are always round, its corresponding polygons are the empty set, a point or a segment, respectively (for which all other points that are not vertices of the polygon are exterior). For each round subset $ A$ of $ M$ construct the polynomial \[ P_A(x) = x^{|A|}(1 - x)^{r(A)}. \] Show that the sum of polynomials for all round subsets is exactly the polynomial $ P(x) = 1.$ Proposed by Federico Ardila, Colombia

A cake has the form of an $ n$ x $ n$ square composed of $ n^{2}$ unit squares. Strawberries lie on some of the unit squares so that each row or column contains exactly one strawberry; call this arrangement $\mathcal{A}$. Let $\mathcal{B}$ be another such arrangement. Suppose that every grid rectangle with one vertex at the top left corner of the cake contains no fewer strawberries of arrangement $\mathcal{B}$ than of arrangement $\mathcal{A}$. Prove that arrangement $\mathcal{B}$ can be obtained from $ \mathcal{A}$ by performing a number of switches, defined as follows: A switch consists in selecting a grid rectangle with only two strawberries, situated at its top right corner and bottom left corner, and moving these two strawberries to the other two corners of that rectangle.

An $ (n, k) -$ tournament is a contest with $ n$ players held in $ k$ rounds such that: $ (i)$ Each player plays in each round, and every two players meet at most once. $ (ii)$ If player $ A$ meets player $ B$ in round $ i$, player $ C$ meets player $ D$ in round $ i$, and player $ A$ meets player $ C$ in round $ j$, then player $ B$ meets player $ D$ in round $ j$. Determine all pairs $ (n, k)$ for which there exists an $ (n, k) -$ tournament. Proposed by Carlos di Fiore, Argentina

A holey triangle is an upward equilateral triangle of side length $n$ with $n$ upward unit triangular holes cut out. A diamond is a $60^\circ-120^\circ$ unit rhombus. Prove that a holey triangle $T$ can be tiled with diamonds if and only if the following condition holds: Every upward equilateral triangle of side length $k$ in $T$ contains at most $k$ holes, for $1\leq k\leq n$. Proposed by Federico Ardila, Colombia

Click for solution Every diamond contains an upward and a downward triangle with a common edge. Now in every upward triangle $ T$ with sidelength $ k$ we have $ \frac{k^{2}+k}2$ upward triangles and $ \frac{k^{2}-k}2$ downward triangles. The downward triangles are adjacent only with upward triangles from $ T$, so if more than $ k$ upward triangles are removed, there are not enough of them to match with the downward ones. This proves the easier part of the problem. Now let's prove the converse by induction on $ n$. Call a sequence $ 2k+1$ of consecutive upward and downward triangles (starting and ending with upward triangles) from the same row a "block". The block above it of $ 2k-1$ triangles be called the "above block". We wish to cover all the triangles in the last row of the big triangle $ T$ with diamonds. There can be horizontal diamond, but there also can be vertical ones, and for them we will need to borrow upward triangles from the above row. When we finish, it is easy to conclude there will be $ n-1$ holes in the remaining triangle $ T'$ of side $ n-1$. We just need to pick up the diamonds in such a way that the induction hypothesis is respected. Call a triangle which contains a part of the base a "base" triangle. Call a triangle which contains as many holes as its sidelength "full", and one which contains more than its sidelength "deadly". We must not create "deadly" triangles after cutting the last row. It is clear a deadly triangle may appear only as a "base" triangle for $ T'$. We claim that if a block from the base contained $ r$ holes, at most $ r$ holes are added to the block above it after removing the last row. We do the following procedure: the space between two consecutive holes in the last row can be filed in with diamonds except one downward triangle which has to be connected with a triangle above (clearly at least one of the downward triangles can be matched otherwise there would be too many holes breaking the condition). It is is easy to verify the claim after such a conclusion. Moreover it clear that any arrangement is of the above type otherwise we would create too many holes in $ T'$ (between any two consecutive holes we must crate a hole in $ T'$, and if we somewhere create more it would be over). Now we claim a "deadly" triangle arises from a full triangle. Indeed, if we have a base deadly triangle with sidelength $ j$ an $ j+l$ holes we can extend it one unit downward. This triangle has sidelength $ j+1$ and had originally also at least $ j+l$ holes according to our claim. So $ l=1$ and this triangle was full at the beginning. It only remains to manage the full base triangles, thus. If two full triangles of sidelength $ k,l$ intersect in a triangle of sidelength $ m$, we deduce this small triangle is also full and the big traingle of length $ k+l-m$ is too. So we may do such unions unless $ k+l-m$ is not $ n$. When we finish, we have two possible cases: a) we have some non-intersecting full triangles. Then every full base triangle is contained in one of them otherwise we could still unite it with one. Applying the induction hypothesis for them we can tile the bottom rows of them with diamonds without not making them full. Completing the tiling we achieve the goal. b) We have intersecting triangles, but uniting them would form the big triangle $ T$ for which we can not apply the induction step. It is easy to see this is possible only when there are two such triangles of sidelength $ k,l$ which intersect in a traingle of sidelength $ m$ and $ k+l-m=n$. Then we have at most $ m$ holes in the common part. So at least $ k+l-m=n$ holes in the union of these two traingles. It means the remaining parallelogram at the top contains no hole and then it can be tiled with diamonds. The two triangles can also be tiled. Moreover any tiling tiles also the common part of sidelength $ m$ completely. So we may tile one triangle, and tile the remaining part of the second. This is the desired tiling of $ T$. QED

Consider a convex polyhedron without parallel edges and without an edge parallel to any face other than the two faces adjacent to it. Call a pair of points of the polyhedron antipodal if there exist two parallel planes passing through these points and such that the polyhedron is contained between these planes. Let $A$ be the number of antipodal pairs of vertices, and let $B$ be the number of antipodal pairs of midpoint edges. Determine the difference $A-B$ in terms of the numbers of vertices, edges, and faces. Proposed by Kei Irei, Japan

Let $ABC$ be triangle with incenter $I$. A point $P$ in the interior of the triangle satisfies \[\angle PBA+\angle PCA = \angle PBC+\angle PCB.\] Show that $AP \geq AI$, and that equality holds if and only if $P=I$.

Let $ ABCD$ be a trapezoid with parallel sides $ AB > CD$. Points $ K$ and $ L$ lie on the line segments $ AB$ and $ CD$, respectively, so that $AK/KB=DL/LC$. Suppose that there are points $ P$ and $ Q$ on the line segment $ KL$ satisfying \[\angle{APB} = \angle{BCD}\qquad\text{and}\qquad \angle{CQD} = \angle{ABC}.\]Prove that the points $ P$, $ Q$, $ B$ and $ C$ are concyclic. Proposed by Vyacheslev Yasinskiy, Ukraine

Let $ ABCDE$ be a convex pentagon such that \[ \angle BAC = \angle CAD = \angle DAE\qquad \text{and}\qquad \angle ABC = \angle ACD = \angle ADE. \]The diagonals $BD$ and $CE$ meet at $P$. Prove that the line $AP$ bisects the side $CD$. Proposed by Zuming Feng, USA

A point $D$ is chosen on the side $AC$ of a triangle $ABC$ with $\angle C < \angle A < 90^\circ$ in such a way that $BD=BA$. The incircle of $ABC$ is tangent to $AB$ and $AC$ at points $K$ and $L$, respectively. Let $J$ be the incenter of triangle $BCD$. Prove that the line $KL$ intersects the line segment $AJ$ at its midpoint.

In triangle $ABC$, let $J$ be the center of the excircle tangent to side $BC$ at $A_{1}$ and to the extensions of the sides $AC$ and $AB$ at $B_{1}$ and $C_{1}$ respectively. Suppose that the lines $A_{1}B_{1}$ and $AB$ are perpendicular and intersect at $D$. Let $E$ be the foot of the perpendicular from $C_{1}$ to line $DJ$. Determine the angles $\angle{BEA_{1}}$ and $\angle{AEB_{1}}$. Proposed by Dimitris Kontogiannis, Greece

Circles $ w_{1}$ and $ w_{2}$ with centres $ O_{1}$ and $ O_{2}$ are externally tangent at point $ D$ and internally tangent to a circle $ w$ at points $ E$ and $ F$ respectively. Line $ t$ is the common tangent of $ w_{1}$ and $ w_{2}$ at $ D$. Let $ AB$ be the diameter of $ w$ perpendicular to $ t$, so that $ A, E, O_{1}$ are on the same side of $ t$. Prove that lines $ AO_{1}$, $ BO_{2}$, $ EF$ and $ t$ are concurrent.

In a triangle $ ABC$, let $ M_{a}$, $ M_{b}$, $ M_{c}$ be the midpoints of the sides $ BC$, $ CA$, $ AB$, respectively, and $ T_{a}$, $ T_{b}$, $ T_{c}$ be the midpoints of the arcs $ BC$, $ CA$, $ AB$ of the circumcircle of $ ABC$, not containing the vertices $ A$, $ B$, $ C$, respectively. For $ i \in \left\{a, b, c\right\}$, let $ w_{i}$ be the circle with $ M_{i}T_{i}$ as diameter. Let $ p_{i}$ be the common external common tangent to the circles $ w_{j}$ and $ w_{k}$ (for all $ \left\{i, j, k\right\}= \left\{a, b, c\right\}$) such that $ w_{i}$ lies on the opposite side of $ p_{i}$ than $ w_{j}$ and $ w_{k}$ do. Prove that the lines $ p_{a}$, $ p_{b}$, $ p_{c}$ form a triangle similar to $ ABC$ and find the ratio of similitude. Proposed by Tomas Jurik, Slovakia

Let $ABCD$ be a convex quadrilateral. A circle passing through the points $A$ and $D$ and a circle passing through the points $B$ and $C$ are externally tangent at a point $P$ inside the quadrilateral. Suppose that \[\angle{PAB}+\angle{PDC}\leq 90^\circ\qquad\text{and}\qquad\angle{PBA}+\angle{PCD}\leq 90^\circ.\] Prove that $AB+CD \geq BC+AD$. Proposed by Waldemar Pompe, Poland

Points $ A_{1}$, $ B_{1}$, $ C_{1}$ are chosen on the sides $ BC$, $ CA$, $ AB$ of a triangle $ ABC$ respectively. The circumcircles of triangles $ AB_{1}C_{1}$, $ BC_{1}A_{1}$, $ CA_{1}B_{1}$ intersect the circumcircle of triangle $ ABC$ again at points $ A_{2}$, $ B_{2}$, $ C_{2}$ respectively ($ A_{2}\neq A, B_{2}\neq B, C_{2}\neq C$). Points $ A_{3}$, $ B_{3}$, $ C_{3}$ are symmetric to $ A_{1}$, $ B_{1}$, $ C_{1}$ with respect to the midpoints of the sides $ BC$, $ CA$, $ AB$ respectively. Prove that the triangles $ A_{2}B_{2}C_{2}$ and $ A_{3}B_{3}C_{3}$ are similar.

Assign to each side $b$ of a convex polygon $P$ the maximum area of a triangle that has $b$ as a side and is contained in $P$. Show that the sum of the areas assigned to the sides of $P$ is at least twice the area of $P$.

Determine all pairs $(x, y)$ of integers such that \[1+2^{x}+2^{2x+1}= y^{2}.\]

For $ x \in (0, 1)$ let $ y \in (0, 1)$ be the number whose $ n$-th digit after the decimal point is the $ 2^{n}$-th digit after the decimal point of $ x$. Show that if $ x$ is rational then so is $ y$. Proposed by J.P. Grossman, Canada

We define a sequence $ \left(a_{1},a_{2},a_{3},\ldots \right)$ by \[ a_{n} = \frac {1}{n}\left(\left\lfloor\frac {n}{1}\right\rfloor + \left\lfloor\frac {n}{2}\right\rfloor + \cdots + \left\lfloor\frac {n}{n}\right\rfloor\right), \] where $\lfloor x\rfloor$ denotes the integer part of $x$. a) Prove that $a_{n+1}>a_n$ infinitely often. b) Prove that $a_{n+1}<a_n$ infinitely often. Proposed by Johan Meyer, South Africa

Let $P(x)$ be a polynomial of degree $n > 1$ with integer coefficients and let $k$ be a positive integer. Consider the polynomial $Q(x) = P(P(\ldots P(P(x)) \ldots ))$, where $P$ occurs $k$ times. Prove that there are at most $n$ integers $t$ such that $Q(t) = t$.

Find all integer solutions of the equation \[\frac {x^{7} - 1}{x - 1} = y^{5} - 1.\]

Let $ a > b > 1$ be relatively prime positive integers. Define the weight of an integer $ c$, denoted by $ w(c)$ to be the minimal possible value of $ |x| + |y|$ taken over all pairs of integers $ x$ and $ y$ such that \[ax + by = c.\] An integer $ c$ is called a local champion if $ w(c) \geq w(c \pm a)$ and $ w(c) \geq w(c \pm b)$. Find all local champions and determine their number. Proposed by Zoran Sunic, USA

For all positive integers $n$, show that there exists a positive integer $m$ such that $n$ divides $2^{m} + m$. Proposed by Juhan Aru, Estonia