Let $ n$ be an integer, $ n\geq 2$. Find all sets $ A$ with $ n$ integer elements such that the sum of any nonempty subset of $ A$ is not divisible by $ n+1$.

Let $ a_i, b_i$ be positive real numbers, $ i=1,2,\ldots,n$, $ n\geq 2$, such that $ a_i<b_i$, for all $ i$, and also \[ b_1+b_2+\cdots + b_n < 1 + a_1+\cdots + a_n.\] Prove that there exists a $ c\in\mathbb R$ such that for all $ i=1,2,\ldots,n$, and $ k\in\mathbb Z$ we have \[ (a_i+c+k)(b_i+c+k) > 0.\]

Click for solution The author of this problem is prof. Vasile Pop, from Cluj. I'll post here the official solution. The condition is equivalent to no integer $ k$ to be found in any interval $ [a_i+c;b_i+c]$ , i.e. in the translates by $ c$ of $ [a_i;b_i]$. Moreover, the sum of the lengths of these intervals is known to be less than 1, since $ \sum_{i=1}^n(b_i-a_i)<1.$ We are thus asked to show that the union $ S$ of a (finite) set of non-degenerate intervals, of total length less than 1, may be translated such that it becomes disjunct of $ \mathbb{Z}$. For all $ k \in \mathbb{Z}$, consider the sets $ S_k=S \cap [k;k+1)$ and their translates $ S_k-k \subset [0;1)$. Due to possible overlaps, the total length of the union $ \cap \{S_k; k \in \mathbb{Z} \}=S$, hence less than $ 1$. Therefore, there exists $ -c \in [0;1) \cup \{ S_k-k; k \in \mathbb{Z} \}$, hence $ (S+c) \cap \mathbb{Z}= \emptyset$.

Let $ ABCDEF$ be a convex hexagon with all the sides of length 1. Prove that one of the radii of the circumcircles of triangles $ ACE$ or $ BDF$ is at least 1.

Prove that there exists a set $ S$ of $ n - 2$ points inside a convex polygon $ P$ with $ n$ sides, such that any triangle determined by $3$ vertices of $ P$ contains exactly one point from $ S$ inside or on the boundaries.

Find the greatest common divisor of the numbers \[ 2^{561}-2, 3^{561}-3, \ldots, 561^{561}-561.\]

Let $ n \geq 3$ be an odd integer. Determine the maximum value of \[ \sqrt{|x_{1}-x_{2}|}+\sqrt{|x_{2}-x_{3}|}+\ldots+\sqrt{|x_{n-1}-x_{n}|}+\sqrt{|x_{n}-x_{1}|},\] where $ x_{i}$ are positive real numbers from the interval $ [0,1]$.

Are there any sequences of positive integers $ 1 \leq a_{1} < a_{2} < a_{3} < \ldots$ such that for each integer $ n$, the set $ \left\{a_{k} + n\ |\ k = 1, 2, 3, \ldots\right\}$ contains finitely many prime numbers?

Show that each convex pentagon has a vertex from which the distance to the opposite side of the pentagon is strictly less than the sum of the distances from the two adjacent vertices to the same side. Note. If the pentagon is labeled $ ABCDE$, the adjacent vertices of $ A$ are $ B$ and $ E$, the ones of $ B$ are $ A$ and $ C$ etc.

Let $ G$ be a connected graph with $ n$ vertices and $ m$ edges such that each edge is contained in at least one triangle. Find the minimum value of $ m$.

Let $ ABC$ be a triangle with $ \measuredangle{BAC} < \measuredangle{ACB}$. Let $ D$, $ E$ be points on the sides $ AC$ and $ AB$, such that the angles $ ACB$ and $ BED$ are congruent. If $ F$ lies in the interior of the quadrilateral $ BCDE$ such that the circumcircle of triangle $ BCF$ is tangent to the circumcircle of $ DEF$ and the circumcircle of $ BEF$ is tangent to the circumcircle of $ CDF$, prove that the points $ A$, $ C$, $ E$, $ F$ are concyclic. Author: Cosmin Pohoata

Click for solution pohoatza wrote: Let $ ABC$ be a triangle with $ \measuredangle{BAC} < \measuredangle{ACB}$. Let $ D$, $ E$ be points on the sides $ AC$ and $ AB$, such that the angles $ ACB$ and $ BED$ are congruent. If $ F$ lies in the interior of the quadrilateral $ BCDE$ such that the circumcircle of triangle $ BCF$ is tangent to the circumcircle of $ DEF$ and the circumcircle of $ BEF$ is tangent to the circumcircle of $ CDF$, prove that the points $ A$, $ C$, $ E$, $ F$ are concyclic. Invert around $ F$, which maps point $ P$ to $ P'$. Now, since the circumcircles of $ \triangle FBE$ and $ \triangle DFC$ are tangent, we see that $ B'E'\parallel D'C'$. Similarly, $ E'D'\parallel B'C'$, so $ E'D'C'B'$ is a parallelogram. This means that $ \angle B'E'D' = \angle D'C'B'$. Also, the inversion maps $ \angle BED = \angle FEB + \angle FED = \angle FD'E' + \angle FB'E'$. Similarly,\[ \angle ACB = \angle DCB = \angle FCD + \angle FCB = \angle FB'C' + \angle FD'C'\]so\[ \angle FD'E' + \angle FB'E' = \angle BED = \angle ACB = \angle FB'C' + \angle FD'C'\]Also,\[ \angle B'FD' = 360 - (\angle B'E'D' + \angle FB'E' + \angle FD'E') = 360 - (\angle B'C'D' + \angle FB'C' + \angle FD'C') = 360 - \angle B'FD'\]Which implies that $ \angle B'FD' = 180$, so $ B'$, $ F$, and $ D'$ are collinear. Now, notice that $ A$ is the intersection between $ BE$ and $ CD$, so $ A'$ is the intersection between the circumcircles of $ \triangle FE'B'$ and $ \triangle FD'C'$. Now,\[ \angle FA'E' = 180 - \angle FB'E' = 180 - \angle FD'C' = 180 - \angle FA'C'\]so $ \angle FA'C' + \angle FA'E' = 180$, which means that $ E'$, $ A'$, and $ C'$ are collinear, so $ FCAE$ is cyclic.

Let $ ABC$ be an acute triangle with orthocenter $ H$ and let $ X$ be an arbitrary point in its plane. The circle with diameter $ HX$ intersects the lines $ AH$ and $ AX$ at $ A_{1}$ and $ A_{2}$, respectively. Similarly, define $ B_{1}$, $ B_{2}$, $ C_{1}$, $ C_{2}$. Prove that the lines $ A_{1}A_{2}$, $ B_{1}B_{2}$, $ C_{1}C_{2}$ are concurrent. Click to reveal hidden textRemark. The triangle obviously doesn't need to be acute.

Let $ m,\ n \geq 3$ be positive odd integers. Prove that $ 2^{m}-1$ doesn't divide $ 3^{n}-1$.

Click for solution @Rust: divides means "is a divisor of", not "is divisible by". Nice problem . Assume, to the contrary, that there are such odd numbers $ m,n\ge3$ for which $ 2^m-1|3^n-1$. Let $ p$ be a prime divisor of $ 2^m-1$ of the form $ 4k+3$. Because $ p|3^n-1$, we have that $ d=\text{ord}_3(p)$ is odd. Since $ d|p-1=4k+2$, we have $ d|2k+1$, hence $ 3^{\frac{p-1}2}=\binom{\underline3}p=1$. Then, by the Quadratic Reciprocity Law, we have $ \binom{\underline3}p\cdot\binom{\underline p}3=(-1)^{\frac{3-1}2\cdot\frac{p-1}2}=-1$ hence $ \binom{\underline p}3=-1$, so $ p=3t+2$. Let now $ p$ be a prime divisor of $ 2^m-1$ of the form $ 4k+1$. A reasoning just as above and $ \binom{\underline3}p\cdot\binom{\underline p}3=(-1)^{\frac{3-1}2\cdot\frac{p-1}2}=1$ leads to $ \binom{\underline p}3=1$, hence $ p=3t+1$. So let $ M$ be the multiset of prime divisors $ p$ of $ 2^m-1$ of the form $ 4k+3$, containing each prime with multiplicity equal to its exponent in the prime factorization of $ 2^m-1$. Because $ 2^m-1\equiv 3\pmod 4$, $ |M|$ is odd. But $ M$ contains precisely all prime divisors $ p$ of the form $ 3t+2$ of $ 2^m-1$. Then considering $ \text{mod}$ $ 3$, we have $ 2^m-1\equiv 2^{|M|}=2\pmod 3$, Contradiction.

Let $ n$ be a nonzero positive integer. A set of persons is called a $ n$-balanced set if in any subset of $ 3$ persons there exists at least two which know each other and in each subset of $ n$ persons there are two which don't know each other. Prove that a $ n$-balanced set has at most $ (n - 1)(n + 2)/2$ persons.

Let $ ABCD$ be a convex quadrilateral and let $ O \in AC \cap BD$, $ P \in AB \cap CD$, $ Q \in BC \cap DA$. If $ R$ is the orthogonal projection of $ O$ on the line $ PQ$ prove that the orthogonal projections of $ R$ on the sidelines of $ ABCD$ are concyclic.

Let $ m, n \geq 1$ be two coprime integers and let also $ s$ an arbitrary integer. Determine the number of subsets $ A$ of $ \{1, 2, ..., m + n - 1\}$ such that $ |A| = m$ and $ \sum_{x \in A} x \equiv s \pmod{n}$.

Let $ n \geq 3$ be a positive integer and let $ m \geq 2^{n-1}+1$. Prove that for each family of nonzero distinct subsets $ (A_j)_{j \in \overline{1, m}}$ of $ \{1, 2, ..., n\}$ there exist $ i$, $ j$, $ k$ such that $ A_i \cup A_j = A_k$.

Let $ n$ be a nonzero positive integer. Find $ n$ such that there exists a permutation $ \sigma \in S_{n}$ such that \[ \left| \{ |\sigma(k) - k| \ : \ k \in \overline{1, n} \}\right | = n.\]

Let $ ABC$ be a triangle and let $ \mathcal{M}_{a}$, $ \mathcal{M}_{b}$, $ \mathcal{M}_{c}$ be the circles having as diameters the medians $ m_{a}$, $ m_{b}$, $ m_{c}$ of triangle $ ABC$, respectively. If two of these three circles are tangent to the incircle of $ ABC$, prove that the third is tangent as well.

Let $ \mathcal{P}$ be a square and let $ n$ be a nonzero positive integer for which we denote by $ f(n)$ the maximum number of elements of a partition of $ \mathcal{P}$ into rectangles such that each line which is parallel to some side of $ \mathcal{P}$ intersects at most $ n$ interiors (of rectangles). Prove that \[ 3 \cdot 2^{n-1} - 2 \le f(n) \le 3^n - 2.\]