If $n$ is a positive integer, let $A = \{n,n+1,...,n+17 \}$. Does there exist some values of $n$ for which we can divide $A$ into two disjoints subsets $B$ and $C$ such that the product of the elements of $B$ is equal to the product of the elements of $C$?

Let $ABCD$ be a parallelogram. Let $M$ be a point on the side $AB$ and $N$ be a point on the side $BC$ such that the segments $AM$ and $CN$ have equal lengths and are non-zero. The lines $AN$ and $CM$ meet at $Q$. Prove that the line $DQ$ is the bisector of the angle $\measuredangle ADC$. Alternative formulation. Let $ABCD$ be a parallelogram. Let $M$ and $N$ be points on the sides $AB$ and $BC$, respectively, such that $AM=CN\neq 0$. The lines $AN$ and $CM$ intersect at a point $Q$. Prove that the point $Q$ lies on the bisector of the angle $\measuredangle ADC$.

Click for solution Let $P$ be the intersection point between $AN$ and $CM$, and let $AN$ intersect $CD$ in $Q$ (I supposed that $AB<BC$, otherwise we switch letters, but it's the same proof). Then $DP$ is the interior angle bisector if and only if (the bisector theorem) \[ \frac{ AP}{PQ} = \frac{AD}{DQ} \quad (1) \] But the triangles $AMP$ and $QPC$ are similar, thus \[ \frac{AP}{PQ} = \frac{AM}{CQ} = \frac{CN} {CQ } = \frac{AD}{DQ} \] where the last equality is derived from Thales for the triangle $QAD$ and the parallel $CN$.

Each point of the plane with two integer coordinates is the center of a disk with radius $ \frac {1} {1000}$. Prove that there exists an equilateral triangle whose vertices belong to distinct disks. Prove that such a triangle has side-length greater than 96.

Let $n$ be a positive integer, and $a_1,...,a_n, b_1,..., b_n$ be $2n$ positive real numbers such that $a_1 + ... + a_n = b_1 + ... + b_n = 1$. Find the minimal value of $ \frac {a_1^2} {a_1 + b_1} + \frac {a_2^2} {a_2 + b_2} + ...+ \frac {a_n^2} {a_n + b_n}$.

Let $P$, $Q$, and $R$ be the points where the incircle of a triangle $ABC$ touches the sides $AB$, $BC$, and $CA$, respectively. Prove the inequality $\frac{BC} {PQ} + \frac{CA} {QR} + \frac{AB} {RP} \geq 6$.

Let $P$ be the set of prime numbers. Consider a subset $M$ of $P$ with at least three elements. We assume that, for each non empty and finite subset $A$ of $M$, with $A \neq M$, the prime divisors of the integer $( \prod_{p \in A} ) - 1$ belong to $M$. Prove that $M = P$.