Let $A$ be a $101$-element subset of the set $S=\{1,2,\ldots,1000000\}$. Prove that there exist numbers $t_1$, $t_2, \ldots, t_{100}$ in $S$ such that the sets \[ A_j=\{x+t_j\mid x\in A\},\qquad j=1,2,\ldots,100 \] are pairwise disjoint.

Click for solution Assume that we have already found t_1, t_2, ..., t_k (k<=99) and are searching for t_{k+1}. We can not take t_{k+1} only of form t_i+a_j-a_l, where 1<=i<=k and a_j and a_l are elements of A. So, we have k*101*100 forbidden values of t_{k+1}, which correspond to a_j<>a_l, and k forbidden values which correspond to a_j=a_l. So, at most 99*101*100+99=1000000-1 forbidden values, and at least 1 admissible.

Determine all pairs of positive integers $(a,b)$ such that \[ \dfrac{a^2}{2ab^2-b^3+1} \] is a positive integer.

Click for solution First note that we must have 1=< 2ab^2-b^3+1=b^2(2a-b)+1 and so, 1<=b<=2a. Now, we'll prove that the pairs that solve this problem are (i) (a,b)=(n,1), (ii) (a,b)=(n,2n), and (iii) (a,b)=(8n^4-n,2n) where n is a positive Indeed, if b=1, we get the pairs in (i), while if b=2a we get the pairs in (ii). Now assume that 1<b<2a and set r=a^2/( 2ab^2-b^3+1) Then the quadratic equation (*) x^2-2xrb^2+r(b^3-1)=0 has two solution that are positive integers, since one of them is a and their product is r(b^3-1)>0. Evaluating the discriminant D we have D=4(r^2b^4-rb^3+r). Thus r^2b^4-rb^3+r=m^2 for some natural number and the above equation has solution rb^2+m and rb^2-m. Now note that m^2 = r^2b^4-rb^3+r > r^2b^4-2rb^3+b^2=(rb^2-b)^2, with rb^2-b>0. Hence, b>rb^2-m. However, since a^2> b^2(2a-b>=b^2, we have a>b. Therefore, a=rb^2+m > b and the other (also positve) root, say c, is c=rb^2-m < b. Furthermore, if 2c>b, then we have r=c^2/(b^2(2c-b)+1 <= c^2/b^2 <1, a contradiction. Thus b=2c and r=c^2. Since ac=r(b^3-1), we get a=c(b^3-1)=8c^4-c. It is easy to verify that if c>=1, then the integers b=2c and a=c(b^3-1) solve the problem, since a^2/(2ab^2-b^3+1) = c^2(b^3-1)^2 / [(b^3-1)(2cb^2-1)]=c^2. The proof is complete.

Each pair of opposite sides of a convex hexagon has the following property: the distance between their midpoints is equal to $\dfrac{\sqrt{3}}{2}$ times the sum of their lengths. Prove that all the angles of the hexagon are equal.

Click for solution Let a1,..,a6 by radius-vectors of hexagon vertices. We have abs((a1+a2)-(a4+a5))/2=sqrt(3/4)*(abs(a1-a2)+abs(a4-a5))>= sqrt(3/4)*abs(a1+a5-a2-a4). Take the square, ans sum up three such inequalities. Then simplify. We get 0>=(a1+a3+a5-a2-a4-a6)^2. So, a1+a3+a5=a2+a4+a6 and above inequalities were equalities, so opposite sides are parallel. From a1+a3+a5=a2+a4+a6 we have (projecting to the perpendicular to a1a2) that a6 and a3 are equidistant from the line a1a2. Then consider the triangle formed by continuations of sides a1a2, a3a4, a5a6. The hexagon is gotten from the triangle by cutting small similar to big and equal to each other triangles near vertives. We get for this triangle, that each of its medians equals sqrt(3/4)*respective side. The rest is easy (one of the ways is the formula for the length of the median).

Let $ABCD$ be a cyclic quadrilateral. Let $P$, $Q$, $R$ be the feet of the perpendiculars from $D$ to the lines $BC$, $CA$, $AB$, respectively. Show that $PQ=QR$ if and only if the bisectors of $\angle ABC$ and $\angle ADC$ are concurrent with $AC$.

Click for solution in two cyclic quadrilatrals APRD & DRCQ ,AD & CD are diameters respectively and we have: RQ/sin(<RCQ)=CD/2 or RQ/sin(<BCA)=CD/2 PR/sin(<PAR)=AD/2 or PR/sin(<BAC)=AD/2 By dividing: CD/AD= sin(<BAC)/sin(<BCA)=BC/AB Since CD/AD=BC/AB thus bisectors of <CBA and <ADC intersect AC in the same poin. S.Ebadollahi

Let $n$ be a positive integer and let $x_1\le x_2\le\cdots\le x_n$ be real numbers. Prove that \[ \left(\sum_{i,j=1}^{n}|x_i-x_j|\right)^2\le\frac{2(n^2-1)}{3}\sum_{i,j=1}^{n}(x_i-x_j)^2. \] Show that the equality holds if and only if $x_1, \ldots, x_n$ is an arithmetic sequence.

Click for solution We can rewrite the given expression as \( \sum_i (2i-n-1)x_i \)^2 \leq (n^2 - 1)/3 \( n\sum x_i^2 - (\sum x_i )^2\). As the expression is translation invariant (obvious from the original form), we can assume that \sum x_i =0. The rest follows by Cauchy-Schwartz.

Let $p$ be a prime number. Prove that there exists a prime number $q$ such that for every integer $n$, the number $n^p-p$ is not divisible by $q$.

Click for solution For p=2 take n=5. Let now p be odd. We have p^p-1=(p-1)(1+p+..+p^{p-1}}. The second bracket is odd and is not equal 1 modulo p^2. Then there exists its prime divisor q, which is odd and not of form k*p^2+1. Lets prove that q satisfies condition. If n^p=p (mod q), then (n^p)^p=p^p=1 (mod q), so n^{p^2}=1 (mod q). Since n^{q-1}=1 (mod q), we have n^gcd(q-1,p^2)=1 (mod q). But q-1 is not divisible by q^2, then gcd(q-1,p^2) is p or 1. so, n^p=1 (mod q) -- a contradiction, unless p-1 is divisible by q. But the 1+p+..+p^{p-1}=p (mod p-1), so coprime with p-1.