If $a_{1}$, $a_{2}$, $\ldots$, $a_{n}\geq 0$ are such that \[a_{1}^{2}+\cdots+a_{n}^{2}=1,\] then find the maximum value of the product $(1-a_{1})\cdots (1-a_{n})$.

Let $f: \mathbb{Q}\rightarrow \mathbb{R}$ be a function such that \[|f(x)-f(y)|\leq (x-y)^{2}\] for all $x,y \in\mathbb{Q}$. Prove that $f$ is constant.

Let $A_{1}A_{2}\ldots A_{2n}$ be a convex polygon and let $P$ be a point in its interior such that it doesn't lie on any of the diagonals of the polygon. Prove that there is a side of the polygon such that none of the lines $PA_{1}$, $\ldots$, $PA_{2n}$ intersects it in its interior.

Let $\mathcal O_{1}$ and $\mathcal O_{2}$ two exterior circles. Let $A$, $B$, $C$ be points on $\mathcal O_{1}$ and $D$, $E$, $F$ points on $\mathcal O_{1}$ such that $AD$ and $BE$ are the common exterior tangents to these two circles and $CF$ is one of the interior tangents to these two circles, and such that $C$, $F$ are in the interior of the quadrilateral $ABED$. If $CO_{1}\cap AB=\{M\}$ and $FO_{2}\cap DE=\{N\}$ then prove that $MN$ passes through the middle of $CF$.

Let \[f = X^{n}+a_{n-1}X^{n-1}+\ldots+a_{1}X+a_{0}\] be an integer polynomial of degree $n \geq 3$ such that $a_{k}+a_{n-k}$ is even for all $k \in \overline{1,n-1}$ and $a_{0}$ is even. Suppose that $f = gh$, where $g,h$ are integer polynomials and $\deg g \leq \deg h$ and all the coefficients of $h$ are odd. Prove that $f$ has an integer root.

Click for solution You can prove g is linear by considering it in $\mathbb{Z}_{2}[x]$. The trick is that if h(x) divides f(x), then h(x) also divides $x^{n}f \left( \frac{1}{x}\right)$ (in $\mathbb{Z}_{2}[x]$). Edit, ok... as perfect_radio requested, I will provide further details. For this solution we'll be working only in $\mathbb{Z}_{2}[x]$. We have $f(x) \equiv g(x)h(x)$ and since $h(x) \equiv h(1/x) \cdot x^{\deg h}$, we also have $h(x)g(1/x)x^{\deg g}\equiv f(1/x)x^{n}$. Notice that since $a_{k}+a_{n-k}\equiv 0$, we have $f(1/x)x^{n}+f(x) \equiv x^{n}+1$. Now let $\deg h = d$, and we know $d \ge \frac{n}{2}$. By solving for the coefficients of g, we find that the first few are $g(x) = 1+x+x^{d+1}+x^{d+2}+\ldots$. However, if $\deg g \ge d+1$, then $\deg g+\deg h = 2d+1 > n$, so we must have $g(x) = 1+x$. So that pretty much shows that g is linear in $\mathbb{Z}[x]$, although we should probably also note that the leading coefficient on g must be odd (so that it doesn't vanish when reducing to $\mathbb{Z}_{2}[x]$).

Let $ABC$ be a triangle, $E$ and $F$ the points where the incircle and $A$-excircle touch $AB$, and $D$ the point on $BC$ such that the triangles $ABD$ and $ACD$ have equal in-radii. The lines $DB$ and $DE$ intersect the circumcircle of triangle $ADF$ again in the points $X$ and $Y$. Prove that $XY\parallel AB$ if and only if $AB=AC$.

Find all subsets $A$ of $\left\{ 1, 2, 3, 4, \ldots \right\}$, with $|A| \geq 2$, such that for all $x,y \in A, \, x \neq y$, we have that $\frac{x+y}{\gcd (x,y)}\in A$. Dan Schwarz

Let $S$ be the set of $n$-uples $\left( x_{1}, x_{2}, \ldots, x_{n}\right)$ such that $x_{i}\in \{ 0, 1 \}$ for all $i \in \overline{1,n}$, where $n \geq 3$. Let $M(n)$ be the smallest integer with the property that any subset of $S$ with at least $M(n)$ elements contains at least three $n$-uples \[\left( x_{1}, \ldots, x_{n}\right), \, \left( y_{1}, \ldots, y_{n}\right), \, \left( z_{1}, \ldots, z_{n}\right) \] such that \[\sum_{i=1}^{n}\left( x_{i}-y_{i}\right)^{2}= \sum_{i=1}^{n}\left( y_{i}-z_{i}\right)^{2}= \sum_{i=1}^{n}\left( z_{i}-x_{i}\right)^{2}. \] (a) Prove that $M(n) \leq \left\lfloor \frac{2^{n+1}}{n}\right\rfloor+1$. (b) Compute $M(3)$ and $M(4)$.

Let $\mathcal{F}$ be the set of all the functions $f : \mathcal{P}(S) \longrightarrow \mathbb{R}$ such that for all $X, Y \subseteq S$, we have $f(X \cap Y) = \min (f(X), f(Y))$, where $S$ is a finite set (and $\mathcal{P}(S)$ is the set of its subsets). Find \[\max_{f \in \mathcal{F}}| \textrm{Im}(f) |. \]

Prove that for $n, p$ integers, $n \geq 4$ and $p \geq 4$, the proposition $\mathcal{P}(n, p)$ \[\sum_{i=1}^{n}\frac{1}{{x_{i}}^{p}}\geq \sum_{i=1}^{n}{x_{i}}^{p}\quad \textrm{for}\quad x_{i}\in \mathbb{R}, \quad x_{i}> 0 , \quad i=1,\ldots,n \ ,\quad \sum_{i=1}^{n}x_{i}= n,\] is false. Dan Schwarz RemarkIn the competition, the students were informed (fact that doesn't actually relate to the problem's solution) that the propositions $\mathcal{P}(4, 3)$ are $\mathcal{P}(3, 4)$ true.

Click for solution Just take for $(4,4)$, $x_{1}=2$, $x_{2}=x_{3}=x_{4}=\frac{2}{3}$, the general case can be obtained by putting $x_{j}=1$, $j > 4$ and for $p>4$, take $x_{1}= 2^{\frac{4}{p}}$, etc.

Let $a_{i}$, $i = 1,2, \dots ,n$, $n \geq 3$, be positive integers, having the greatest common divisor 1, such that \[a_{j}\textrm{ divide }\sum_{i = 1}^{n}a_{i}\] for all $j = 1,2, \dots ,n$. Prove that \[\prod_{i = 1}^{n}a_{i}\textrm{ divides }\Big{(}\sum_{i = 1}^{n}a_{i}\Big{)}^{n-2}.\]

The points $M, N, P$ are chosen on the sides $BC, CA, AB$ of a triangle $\Delta ABC$, such that the triangle $\Delta MNP$ is acute-angled. We denote with $x$ the length of the shortest altitude of the triangle $\Delta ABC$, and with $X$ the length of the longest altitudes of the triangle $\Delta MNP$. Prove that $x \leq 2X$.

Click for solution For simplifying the notations, denote $y$ the longest altitude of $\triangle{MNP}$. Also denote $H$ the orthocenter of $\triangle{MNP}$ and $\triangle{A'B'C'}$ the pedal triangle of $H$. Thus $\frac{\sum{HA'}}{x}\geq \sum{\frac{HA'}{h_{a}}}= \sum{\frac{[HBC]}{[ABC]}=1}$ Therefore $x\leq \sum{HA'}$ $(1)$. Now let's see that $\sum{HA'}\leq \sum{HM}$, and try to prove that $\sum{HM}\leq 2y$ $(2)$. Suppose that $MM_{1}$ is the longest altitude of $\triangle{MNP}$, therefore $(NP)$ is the shortest side. Now call $D$ the symmetrical point of $H$, w.r.t to $M_{1}$. Thus $D$ is on the circumcircle of $\triangle{MNP}$. Observe that $\sum{HM}= HM+DN+DP$, and rewrite $(2)$ as: $DN+DP \leq MD$, which is obvious by the Ptolemy theorem in the cyclic quadrilateral $MNDP$ ($MD \cdot NP = DN \cdot MP+DP \cdot MN$, but $NP = \min\left\{NP,PM,MN\right\}$), so $(2)$ is proved. Thus by $(1)$ and $(2)$ we have that $x\leq 2y$. Remark: Another nice solution involves an homothety and a translation, I will post it after a month.

Prove that the function $f : \mathbb{N}\longrightarrow \mathbb{Z}$ defined by $f(n) = n^{2007}-n!$, is injective.

Let $ A_{1}A_{2}A_{3}A_{4}A_{5}$ be a convex pentagon, such that \[ [A_{1}A_{2}A_{3}] = [A_{2}A_{3}A_{4}] = [A_{3}A_{4}A_{5}] = [A_{4}A_{5}A_{1}] = [A_{5}A_{1}A_{2}].\] Prove that there exists a point $ M$ in the plane of the pentagon such that \[ [A_{1}MA_{2}] = [A_{2}MA_{3}] = [A_{3}MA_{4}] = [A_{4}MA_{5}] = [A_{5}MA_{1}].\] Here $ [XYZ]$ stands for the area of the triangle $ \Delta XYZ$.

Click for solution Denote $ B_{i}\in A_{i + 1}A_{i + 3}\cap A_{i + 2}A_{i + 4}$, where the indices are taken modulo $ 5$. The basic thing is that the equality of the areas imply the following parallelism: $ A_{i}A_{i + 1}\| A_{i + 2}A_{i + 4}$. It is straightforward to see that $ M \equiv X \in A_{1}B_{1}\cap A_{2}B_{2}\cap A_{3}B_{3}\cap A_{4}B_{4}\cap A_{5}B_{5}$, based on the above parallelism. Instead, another solution, more nicer I think, is by observing that there exists an affine transformations that maps the initial pentagon in a regular one, and the problem is now more like trivial. Remark: A very similar problem, related to this one, is the following I posted here: http://www.mathlinks.ro/Forum/viewtopic.php?t=149421

Consider the set $E = \{1,2,\ldots,2n\}$. Prove that an element $c \in E$ can belong to a subset $A \subset E$, having $n$ elements, and such that any two distinct elements in $A$ do not divide one each other, if and only if \[c > n \left( \frac{2}{3}\right )^{k+1},\] where $k$ is the exponent of $2$ in the factoring of $c$.

Click for solution Ok, I did't misunderstand the problem, I just swapped n with 2n (and e.lopes' mind is identical to mine ) Here is the proof. A comment: the problem is so strict that is not hard. But it's still very nice A definition: $\mbox{god}(n)$ is the greatest odd divisor of n Another definition: $v_{2}(n)$ is the greatest power of 2 that divides n. Then we have the decomposition $n = \mbox{god}(n) \cdot 2^{v_{2}(n)}$ Some facts: a) For each odd integer $1 \le k \le 2n$, there exist a unique power of 2 such that $n < 2^{e}k \le 2n$. Proof: take the greatest power of 2 such that $2^{i}k \le 2n$. Then $2^{i+1}> 2n$ because of maximality, and then $2^{i}> n$. b) If two distinct integers are such that $n <a,b \le 2n$, then $a \nmid b$ Proof: if $a\mid b$ and $a \neq b$ then $b \ge 2a$. c) If $a \mid b$ then $\mbox{god}(a) \mid \mbox{god}(b)$ d) If $\mbox{god}(a)=\mbox{god}(b)$ then $a \mid b$ or $b \mid a$. e) For each odd integer $1 \le c \le 2n$ and for each n-antichain A, there is an unique power of 2 such that $2^{e}c \in A$. Proof: A is an antichain, then using fact (d) we prove that the god's of its elements are distinct. There are n elements and n odd integers between 1 and 2n. Then they are all. And now to the proof, by induction on $k = v_{2}(c)$. $k=0$ means that c is odd. Suppose that $c \le \frac{2}{3}n$. Then $3c\le 2n$. By fact (e), a number of the form $2^{e}\cdot 3c \in A$. But $c \in A$ and $c \mid 2^{e}\cdot 3c$. Suppose that $c > \frac{2}{3}n$. Then, put c in A, and for each odd integer d, put the unique $n < 2^{e}d \le 2n$ in A (using fact a). A is an antichain, since if $a,b \in A, a\mid b$ then $\mbox{god}(a)=\mbox{god}(b)$, and either a,b are between n and 2n, or a = c. But then the greatest odd divisor of b is an odd multiple of c, and we obtain $b \ge 3c > 2n$. And now, suppose the problem is always true for all c such that $v_{2}(c) < i$ and let's prove it for $v_{2}(c) = k = i$, that is, $c=2^{i}c'$ with odd c'. Suppose $c \le \left(\frac{2}{3}\right)^{i+1}n$. Then $a = \frac{3}{2}c \le \left(\frac{2}{3}\right)^{i}n$ is odd, integer, such that $v_{2}(a) =i-1$. a, multiplied by a power of 2, must belong to A, and using the induction hypotesis, prove that this is impossible. Now suppose $c > \left(\frac{2}{3}\right)^{i+1}n$. We do this construction: - put c in A - If $3^{e}c' \le a < 3^{e+1}c'$, a is an odd integer and $c' \mid a$, then put $2^{i-e}a$ in A - if a is an odd integer and $c' \nmid a$, then put $n<2^{e}a \le 2n$ in A, as in fact (a) The proof that this works: - if a is an odd integer, then either $c' \mid a$ or $c' \nmid a$, therefor a number of the form $2^{e}a$ is in A. A has n elements. - If $a,b \in A, a \mid b$ then $\mbox{god}(a) \mid \mbox{god}(b)$. Then: * if $c' \nmid \mbox{god}(a)$, then by the construction $n < a \le 2n$ and b is too big. Impossible. * if $c' \mid \mbox{god}(a)$ then also $c' \mid \mbox{god}(b)$. But $\mbox{god}(b) \ge 3\mbox{god}(a)$, then, $\mbox{god}(a),\mbox{god}(b)$ are in two classes of the construction, then $v_{2}(b) < v_{2}(a)$ and it's impossible that $a \mid b$.

i) Find all infinite arithmetic progressions formed with positive integers such that there exists a number $N \in \mathbb{N}$, such that for any prime $p$, $p > N$, the $p$-th term of the progression is also prime. ii) Find all polynomials $f(X) \in \mathbb{Z}[X]$, such that there exist $N \in \mathbb{N}$, such that for any prime $p$, $p > N$, $| f(p) |$ is also prime. Dan Schwarz

In a circle with center $O$ is inscribed a polygon, which is triangulated. Show that the sum of the squares of the distances from $O$ to the incenters of the formed triangles is independent of the triangulation.

Let $ABC$ be a triangle, and $\omega_{a}$, $\omega_{b}$, $\omega_{c}$ be circles inside $ABC$, that are tangent (externally) one to each other, such that $\omega_{a}$ is tangent to $AB$ and $AC$, $\omega_{b}$ is tangent to $BA$ and $BC$, and $\omega_{c}$ is tangent to $CA$ and $CB$. Let $D$ be the common point of $\omega_{b}$ and $\omega_{c}$, $E$ the common point of $\omega_{c}$ and $\omega_{a}$, and $F$ the common point of $\omega_{a}$ and $\omega_{b}$. Show that the lines $AD$, $BE$ and $CF$ have a common point.

Click for solution Let $ U, V$ and $ W$ be the centers of the circles $ \Gamma_{A}, \Gamma_{B}$ and $ \Gamma_{C}$ respectively. The pairs of lines $ EF$ and $ VW$, $ FD$ and $ WU$, and $ DE$ and $ UV$ meet at $ X, Y$ and $ Z$ respectively (one or all three points may be at infinity, but the argument below works in those cases too). By applying Menelaus to triangle $ UVW$ and transversals $ \over arrow{E - F - X}, \over arrow{F - D - Y}$ and $ \over arrow{D - E - Z}$ yields that $ X, Y$ and $ Z$ are the intersection points of the common external tangents to $ \Gamma_{B}$ and $ \Gamma_{C}$, $ \Gamma_{C}$ and $ \Gamma_{A}$, and $ \Gamma_{A}$ and $ \Gamma_{B}$ respectively. Now by Monge theorem we have that the points $ X, Y$ and $ Z$ are collinear (see here http://mathworld.wolfram.com/MongesCircleTheorem.html ). Consequently, the triangles $ ABC$ and $ DEF$ are perspective, therefore the lines $ AD, BE$ and $ CF$ are concurrent (by Desargues' Theorem).

Let $ABCDE$ be a convex pentagon, such that $AB=BC$, $CD=DE$, $\angle B+\angle D=180^{\circ}$, and it's area is $\sqrt2$. a) If $\angle B=135^{\circ}$, find the length of $[BD]$. b) Find the minimum of the length of $[BD]$.

Let $ ABCD$ be a parallelogram with no angle equal to $ 60^{\textrm{o}}$. Find all pairs of points $ E, F$, in the plane of $ ABCD$, such that triangles $ AEB$ and $ BFC$ are isosceles, of basis $ AB$, respectively $ BC$, and triangle $ DEF$ is equilateral. Valentin Vornicu

The world-renowned Marxist theorist Joric is obsessed with both mathematics and social egalitarianism. Therefore, for any decimal representation of a positive integer $n$, he tries to partition its digits into two groups, such that the difference between the sums of the digits in each group be as small as possible. Joric calls this difference the defect of the number $n$. Determine the average value of the defect (over all positive integers), that is, if we denote by $\delta(n)$ the defect of $n$, compute \[\lim_{n \rightarrow \infty}\frac{\sum_{k = 1}^{n}\delta(k)}{n}.\] Iurie Boreico

Three travel companies provide transportation between $n$ cities, such that each connection between a pair of cities is covered by one company only. Prove that, for $n \geq 11$, there must exist a round-trip through some four cities, using the services of a same company, while for $n < 11$ this is not anymore necessarily true. Dan Schwarz

For $n\in\mathbb{N}$, $n\geq 2$, $a_{i}, b_{i}\in\mathbb{R}$, $1\leq i\leq n$, such that \[\sum_{i=1}^{n}a_{i}^{2}=\sum_{i=1}^{n}b_{i}^{2}=1, \sum_{i=1}^{n}a_{i}b_{i}=0. \] Prove that \[\left(\sum_{i=1}^{n}a_{i}\right)^{2}+\left(\sum_{i=1}^{n}b_{i}\right)^{2}\leq n. \] Cezar Lupu & Tudorel Lupu

Let $ ABC$ be a triangle, let $ E, F$ be the tangency points of the incircle $ \Gamma(I)$ to the sides $ AC$, respectively $ AB$, and let $ M$ be the midpoint of the side $ BC$. Let $ N = AM \cap EF$, let $ \gamma(M)$ be the circle of diameter $ BC$, and let $ X, Y$ be the other (than $ B, C$) intersection points of $ BI$, respectively $ CI$, with $ \gamma$. Prove that \[ \frac {NX} {NY} = \frac {AC} {AB}. \] Cosmin Pohoata

The problem is about real polynomial functions, denoted by $f$, of degree $\deg f$. a) Prove that a polynomial function $f$ can`t be wrriten as sum of at most $\deg f$ periodic functions. b) Show that if a polynomial function of degree $1$ is written as sum of two periodic functions, then they are unbounded on every interval (thus, they are "wild"). c) Show that every polynomial function of degree $1$ can be written as sum of two periodic functions. d) Show that every polynomial function $f$ can be written as sum of $\deg f+1$ periodic functions. e) Give an example of a function that can`t be written as a finite sum of periodic functions. Dan Schwarz