Let $\mathcal{C}_1(O_1)$ and $\mathcal{C}_2(O_2)$ be two circles which intersect in the points $A$ and $B$. The tangent in $A$ at $\mathcal{C}_2$ intersects the circle $\mathcal{C}_1$ in $C$, and the tangent in $A$ at $\mathcal{C}_1$ intersects $\mathcal{C}_2$ in $D$. A ray starting from $A$ and lying inside the $\angle CAD$ intersects the circles $\mathcal{C}_1$, $\mathcal{C}_2$ in the points $M$ and $N$ respectively, and the circumcircle of $\triangle ACD$ in $P$. Prove that $AM=NP$.

Find the largest positive integer $n>10$ such that the residue of $n$ when divided by each perfect square between $2$ and $\dfrac n2$ is an odd number.

In a country 6 cities are connected two by two with round-trip air routes operated by exactly one of the two air companies in that country. Prove that there exist 4 cities $A$, $B$, $C$ and $D$ such that each of the routes $A\leftrightarrow B$, $B\leftrightarrow C$, $C\leftrightarrow D$ and $D\leftrightarrow A$ are operated by the same company. Dan Schwartz

Let $a,b,c$ be positive numbers such that $a+b+c \geq \dfrac 1a + \dfrac 1b + \dfrac 1c$. Prove that \[ a+b+c \geq \frac 3{abc}. \]

On the sides $AD$ and $BC$ of a rhombus $ABCD$ we consider the points $M$ and $N$ respectively. The line $MC$ intersects the segment $BD$ in the point $T$, and the line $MN$ intersects the segment $BD$ in the point $U$. We denote by $Q$ the intersection between the line $CU$ and the side $AB$ and with $P$ the intersection point between the line $QT$ and the side $CD$. Prove that the triangles $QCP$ and $MCN$ have the same area.

Let $ABC$ be an equilateral triangle and $M$ be a point inside the triangle. We denote by $A'$, $B'$, $C'$ the projections of the point $M$ on the sides $BC$, $CA$ and $AB$ respectively. Prove that the lines $AA'$, $BB'$ and $CC'$ are concurrent if and only if $M$ belongs to an altitude of the triangle.

A phone company starts a new type of service. A new customer can choose $k$ phone numbers in this network which are call-free, whether that number is calling or is being called. A group of $n$ students want to use the service. (a) If $n\geq 2k+2$, show that there exist 2 students who will be charged when speaking. (b) It $n=2k+1$, show that there is a way to arrange the free calls so that everybody can speak free to anybody else in the group. Valentin Vornicu

Let $a$, $b$, $c$ be three positive reals such that $(a+b)(b+c)(c+a)=1$. Prove that the following inequality holds: \[ ab+bc+ca \leq \frac 34 . \] Cezar Lupu

Let $ABC$ be a triangle with $BC>CA>AB$ and let $G$ be the centroid of the triangle. Prove that \[ \angle GCA+\angle GBC<\angle BAC<\angle GAC+\angle GBA . \] Dinu Serbanescu

Let $k,r \in \mathbb N$ and let $x\in (0,1)$ be a rational number given in decimal representation \[ x = 0.a_1a_2a_3a_4 \ldots . \] Show that if the decimals $a_k, a_{k+r}, a_{k+2r}, \ldots$ are canceled, the new number obtained is still rational. Dan Schwarz

Three circles $\mathcal C_1(O_1)$, $\mathcal C_2(O_2)$ and $\mathcal C_3(O_3)$ share a common point and meet again pairwise at the points $A$, $B$ and $C$. Show that if the points $A$, $B$, $C$ are collinear then the points $Q$, $O_1$, $O_2$ and $O_3$ lie on the same circle.

Find all positive integers $n$ and $p$ if $p$ is prime and \[ n^8 - p^5 = n^2+p^2 . \] Adrian Stoica

Click for solution There are no solutions for $p = 2$ so assume $p$ is odd. Note that $n < p$ or else $p^2(p^3+1) = n^2(n^6-1) \geq p^2(n^3+1)(n^3-1) > p^2(p^3+1)$. Thus, $p^2 | n^6 - 1 = n^2(n+1)(n-1)(n^2+n+1)(n^2-n+1)$. If $n = p-1$, we can easily show $n^2(n^6-1) > p^2(p^3+1)$ unless $p =3$ in which case we have the solution $(n,p) = (2,3)$. Now, $p$ divides exactly one of $n^2 +n+1$ and $n^2-n+1$ so in fact $p^2$ is a divisor. We have $p^2 < n^2 + n$ Now, $n^2(n+1)(n-1) | p^3+1$ so $n^4 - n^2 < p^3$. Thus, we have $(n^4 - n^2)^2 < (n^2+n)^3$ or $n^4(n^2-1)^2 < n^3(n+1)^3$ $\iff n(n^4-2n^2+1) < n^3+3n^2+3n+1$ $\iff n^5 - 3n^3 - 3n^2 - 2n - 1 < 0$ which cannot hold for large enough $n$, say $n \geq 3$ Thus the only solution is $(n,p) = (2,3)$.

The positive integers from 1 to $n^2$ are placed arbitrarily on the $n^2$ squares of a $n\times n$ chessboard. Two squares are called adjacent if they have a common side. Show that two opposite corner squares can be joined by a path of $2n-1$ adjacent squares so that the sum of the numbers placed on them is at least $\left\lfloor \frac{n^3} 2 \right\rfloor + n^2 - n + 1$. Radu Gologan

Let $a,b,c$ be three positive real numbers with $a+b+c=3$. Prove that \[ (3-2a)(3-2b)(3-2c) \leq a^2b^2c^2 . \] Robert Szasz

Click for solution This is equivalent to $(a+b+c)^3(-a+b+c)(a-b+c)(a+b-c) \leq 27a^2b^2c^2$ Let $x = -a + b + c, y = a-b+c, z = a+b-c$ and note that at most one of $x,y,z$ can be negative (since the sum of any two is positive). If $x < 0$ then $(a+b+c)^3xyz \leq 0 < 27a^2b^2c^2$. If $x,y,z > 0$, note $x+y+z = a+b+c, x+y = 2c$ etc so our inequality becomes $64xyz(x+y+z)^3 \leq 27(x+y)^2(y+z)^2(z+x)^2$ Note that $9(x+y)(y+z)(z+x) \geq 8(x+y+z)(xy+yz+zx)$ and $(xy+yz+zx)^2 \geq 3xyz(x+y+z)$. Combining these completes our proof!

Let $n>3$ be a positive integer. Consider $n$ sets, each having two elements, such that the intersection of any two of them is a set with one element. Prove that the intersection of all sets is non-empty. Sever Moldoveanu

Click for solution Let the sets be $A_1, A_2, \dots , A_n$. Let $A_1 = \{a,b\}$ and suppose $A_2 = \{a,c\}$ where $a,b,c$ are distinct elements. Now, any other $A_i$ must contain either $a$ or $b$ (and not both) or $a$ or $c$ (and not both). Thus, $A_i = \{b,c\}$ or $\{a,x\}$ for some element $x$. Because $n > 3$, no $A_i$ can be $\{b,c\}$ or else some other $A_j = \{a,x\}$ with $x \neq b,c$ and has no common element with $A_i = \{b,c\}$. Thus, all the sets contain $a$... qed

Let $AB$ and $BC$ be two consecutive sides of a regular polygon with 9 vertices inscribed in a circle of center $O$. Let $M$ be the midpoint of $AB$ and $N$ be the midpoint of the radius perpendicular to $BC$. Find the measure of the angle $\angle OMN$.

A piece of cardboard has the shape of a pentagon $ABCDE$ in which $BCDE$ is a square and $ABE$ is an isosceles triangle with a right angle at $A$. Prove that the pentagon can be divided in two different ways in three parts that can be rearranged in order to recompose a right isosceles triangle.

Consider two distinct positive integers $a$ and $b$ having integer arithmetic, geometric and harmonic means. Find the minimal value of $|a-b|$. Mircea Fianu

Click for solution Clearly $a$ and $b$ must have the same parity so let $a = k+m$ and $b = k - m$ for integers $k > m > 0$. We have $|a-b| = 2m$. Note that $ab = k^2 - m^2 = n^2$ is a perfect square. Thus, $(n,m,k)$ is a Pythagorean triple. Now, $\left( \frac{1}{2a} + \frac{1}{2b} \right)^{-1} = \left( \frac{k}{k^2-m^2} \right)^{-1} = \frac{k^2-m^2}{k}$. Thus, $k | m^2$ and $k | n^2$. Starting with a primitive Pythagorean triple $(r,s,t)$ we want $t | r^2$ and $t | s^2$ so we must scale everything by $t$ since $(r,s,t) = 1$. From this idea, we use $(20, 15, 25) = (n,m,k)$. Finally, $a = 40, b = 10$ and $|a-b| = 30$. We can check $\frac{a+b}{2} = 25$, $\sqrt{ab} = 20$, and $\frac{2ab}{a+b} = 16$!