Let $ a$, $ b$, $ c$ be positive real numbers such that $ abc = 1$. Prove that \[ \frac {1}{a^{3}\left(b + c\right)} + \frac {1}{b^{3}\left(c + a\right)} + \frac {1}{c^{3}\left(a + b\right)}\geq \frac {3}{2}. \]

Click for solution From Cauchy we have \[ \left( \frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(c+a)}+\frac{1}{c^{3}(a+b)}\right) \Big( a(b+c)+b(c+a)+c(a+b) \Big) \geq \left( \frac 1a +\frac 1b+ \frac 1c \right)^2 \] and as $\frac 1a + \frac 1b + \frac 1c = ab+bc+ca$ we obtain that \[ \left( \frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(c+a)}+\frac{1}{c^{3}(a+b)}\right) \geq \frac{ ab+bc+ca} 2 \geq \frac{\sqrt [3]{a^2b^2c^2}}{2} = \frac 32 \] which is what we wanted to prove.

Let $ a$ and $ b$ be non-negative integers such that $ ab \geq c^2,$ where $ c$ is an integer. Prove that there is a number $ n$ and integers $ x_1, x_2, \ldots, x_n, y_1, y_2, \ldots, y_n$ such that \[ \sum^n_{i=1} x^2_i = a, \sum^n_{i=1} y^2_i = b, \text{ and } \sum^n_{i=1} x_iy_i = c.\]

Let $ n$ be an integer, $ n \geq 3.$ Let $ a_1, a_2, \ldots, a_n$ be real numbers such that $ 2 \leq a_i \leq 3$ for $ i = 1, 2, \ldots, n.$ If $ s = a_1 + a_2 + \ldots + a_n,$ prove that \[ \frac{a^2_1 + a^2_2 - a^2_3}{a_1 + a_2 - a_3} + \frac{a^2_2 + a^2_3 - a^2_4}{a_2 + a_3 - a_4} + \ldots + \frac{a^2_n + a^2_1 - a^2_2}{a_n + a_1 - a_2} \leq 2s - 2n.\]

Click for solution first of all define $ A_i$ as: $ A_i = \frac {a_i^2 + a_{i + 1}^2 - a_{i + 2}^2}{a_i + a_{i + 1} - a_{i + 2}}$ (with $ a_{n + 1} = a_1,a_{n + 2} = a_2$) now we have: $ \begin{eqnarray*}A_i & = & a_i + a_{i + 1} + a_{i + 2} - 2\cdot\frac {a_{i}a_{i + 1}}{a_i + a_{i + 1} - a_{i + 2}} \\ & = & a_i + a_{i + 1} + a_{i + 2} - 2\left(2 + \frac {(a_i - 2)(a_{i + 1} - 2) + 2(a_i + a_{i + 1}) - 4}{a_i + a_{i + 1} - a_{i + 2}}\right)$ now note that $ 1\leq a_i + a_{i + 1} - a_{i + 2}\leq 4$ and $ 0\leq (a_i - 2)(a_{i + 1} - 2)$,thus: $ A_i\leq a_i + a_{i + 1} + a_{i + 2} - 2\left(2 + \frac {2a_{i + 2} - 4}{4}\right) = a_i + a_{i + 1} - 2$ which yields us: $ \sum_{i = 1}^n A_i\leq 2s - 2n$ QED

Find all of the positive real numbers like $ x,y,z,$ such that : 1.) $ x + y + z = a + b + c$ 2.) $ 4xyz = a^2x + b^2y + c^2z + abc$ Proposed to Gazeta Matematica in the 80s by VASILE CÎRTOAJE and then by Titu Andreescu to IMO 1995.

Let $ \mathbb{R}$ be the set of real numbers. Does there exist a function $ f: \mathbb{R} \mapsto \mathbb{R}$ which simultaneously satisfies the following three conditions? (a) There is a positive number $ M$ such that $ \forall x:$ $ - M \leq f(x) \leq M.$ (b) The value of $f(1)$ is $1$. (c) If $ x \neq 0,$ then \[ f \left(x + \frac {1}{x^2} \right) = f(x) + \left[ f \left(\frac {1}{x} \right) \right]^2 \]

Let $ n$ be an integer,$ n \geq 3.$ Let $ x_1, x_2, \ldots, x_n$ be real numbers such that $ x_i < x_{i+1}$ for $ 1 \leq i \leq n - 1$. Prove that \[ \frac{n(n-1)}{2} \sum_{i < j} x_ix_j > \left(\sum^{n-1}_{i=1} (n-i)\cdot x_i \right) \cdot \left(\sum^{n}_{j=2} (j-1) \cdot x_j \right)\]

Let $ A,B,C,D$ be four distinct points on a line, in that order. The circles with diameters $ AC$ and $ BD$ intersect at $ X$ and $ Y$. The line $ XY$ meets $ BC$ at $ Z$. Let $ P$ be a point on the line $ XY$ other than $ Z$. The line $ CP$ intersects the circle with diameter $ AC$ at $ C$ and $ M$, and the line $ BP$ intersects the circle with diameter $ BD$ at $ B$ and $ N$. Prove that the lines $ AM,DN,XY$ are concurrent.

Click for solution we know that ${XY}\perp{AD}$ at $Z$. let ${AM}\cap{XY} = R$. we have $\angle{RMC} = 90$ (AC is diameter) so, $RMCZ$ is cyclic, that implies $PR.PZ = PM.PC$. let ${ND}\cap{XY} = S$. analogous, we have $PS.PZ = PN.PB$. $XY$ is the radical axes of the two circles, so, $PM.PC = PN.PB \Longrightarrow PR = PS \Longrightarrow R = S$.

Let $ A, B$ and $ C$ be non-collinear points. Prove that there is a unique point $ X$ in the plane of $ ABC$ such that \[ XA^2 + XB^2 + AB^2 = XB^2 + XC^2 + BC^2 = XC^2 + XA^2 + CA^2.\]

The incircle of triangle $ \triangle ABC$ touches the sides $ BC$, $ CA$, $ AB$ at $ D, E, F$ respectively. $ X$ is a point inside triangle of $ \triangle ABC$ such that the incircle of triangle $ \triangle XBC$ touches $ BC$ at $ D$, and touches $ CX$ and $ XB$ at $ Y$ and $ Z$ respectively. Show that $ E, F, Z, Y$ are concyclic.

An acute triangle $ ABC$ is given. Points $ A_1$ and $ A_2$ are taken on the side $ BC$ (with $ A_2$ between $ A_1$ and $ C$), $ B_1$ and $ B_2$ on the side $ AC$ (with $ B_2$ between $ B_1$ and $ A$), and $ C_1$ and $ C_2$ on the side $ AB$ (with $ C_2$ between $ C_1$ and $ B$) so that \[ \angle AA_1A_2 = \angle AA_2A_1 = \angle BB_1B_2 = \angle BB_2B_1 = \angle CC_1C_2 = \angle CC_2C_1.\] The lines $ AA_1,BB_1,$ and $ CC_1$ bound a triangle, and the lines $ AA_2,BB_2,$ and $ CC_2$ bound a second triangle. Prove that all six vertices of these two triangles lie on a single circle.

Let $ ABCDEF$ be a convex hexagon with $ AB = BC = CD$ and $ DE = EF = FA$, such that $ \angle BCD = \angle EFA = \frac {\pi}{3}$. Suppose $ G$ and $ H$ are points in the interior of the hexagon such that $ \angle AGB = \angle DHE = \frac {2\pi}{3}$. Prove that $ AG + GB + GH + DH + HE \geq CF$.

Click for solution Let $C',F'$ be the reflections of $C,F$ respectively in $BE$. We have $GB+GA=GC',\ HD+HE=HF'$, so $GB+GA+HD+HE+GH=C'G+GH+HF'\ge C'F'=CF$, Q.E.D.

Let $ A_1A_2A_3A_4$ be a tetrahedron, $ G$ its centroid, and $ A'_1, A'_2, A'_3,$ and $ A'_4$ the points where the circumsphere of $ A_1A_2A_3A_4$ intersects $ GA_1,GA_2,GA_3,$ and $ GA_4,$ respectively. Prove that \[ GA_1 \cdot GA_2 \cdot GA_3 \cdot GA_ \cdot4 \leq GA'_1 \cdot GA'_2 \cdot GA'_3 \cdot GA'_4\] and \[ \frac{1}{GA'_1} + \frac{1}{GA'_2} + \frac{1}{GA'_3} + \frac{1}{GA'_4} \leq \frac{1}{GA_1} + \frac{1}{GA_2} + \frac{1}{GA_3} + \frac{1}{GA_4}.\]

Let ABCD be a convex quadrilateral and O a point inside it. Let the parallels to the lines BC, AB, DA, CD through the point O meet the sides AB, BC, CD, DA of the quadrilateral ABCD at the points E, F, G, H, respectively. Then, prove that $ \sqrt {\left|AHOE\right|} + \sqrt {\left|CFOG\right|}\leq\sqrt {\left|ABCD\right|}$, where $ \left|P_1P_2...P_n\right|$ is an abbreviation for the non-directed area of an arbitrary polygon $ P_1P_2...P_n$.

Suppose that $ ABCD$ is a cyclic quadrilateral. Let $ E = AC\cap BD$ and $ F = AB\cap CD$. Denote by $ H_{1}$ and $ H_{2}$ the orthocenters of triangles $ EAD$ and $ EBC$, respectively. Prove that the points $ F$, $ H_{1}$, $ H_{2}$ are collinear. Original formulation: Let $ ABC$ be a triangle. A circle passing through $ B$ and $ C$ intersects the sides $ AB$ and $ AC$ again at $ C'$ and $ B',$ respectively. Prove that $ BB'$, $CC'$ and $ HH'$ are concurrent, where $ H$ and $ H'$ are the orthocentres of triangles $ ABC$ and $ AB'C'$ respectively.

Let $ k$ be a positive integer. Show that there are infinitely many perfect squares of the form $ n \cdot 2^k - 7$ where $ n$ is a positive integer.

Let $ \mathbb{Z}$ denote the set of all integers. Prove that for any integers $ A$ and $ B,$ one can find an integer $ C$ for which $ M_1 = \{x^2 + Ax + B : x \in \mathbb{Z}\}$ and $ M_2 = {2x^2 + 2x + C : x \in \mathbb{Z}}$ do not intersect.

Determine all integers $ n > 3$ for which there exist $ n$ points $ A_{1},\cdots ,A_{n}$ in the plane, no three collinear, and real numbers $ r_{1},\cdots ,r_{n}$ such that for $ 1\leq i < j < k\leq n$, the area of $ \triangle A_{i}A_{j}A_{k}$ is $ r_{i} + r_{j} + r_{k}$.

Click for solution Let's prove that only $n=4$ works. Obviously, $n=4$ is Ok, since we can take $A_1A_2A_3A_4$ to be a square and all the $r_i$'s equal to one third of the area of a triangle formed by three of its vertices. Since if we have $n\ge 5$ such points we also have $5$, all we need to do is prove that $n=5$ is impossible. First notice that if $A_1A_2A_3A_4$ is a convex quadrilateral, then $r_1+r_3=r_2+r_4$ and if $A_4$ lies inside $A_1A_2A_3$, then $r_4=-\frac{r_1+r_2+r_3}3$. I'll use these without mentioning it from now on. Case 1 $A_1A_2A_3A_4A_5$ is a convex pentagon In this case $r_i+r_{i+2}$ is the same for all $i$ (the indices are modulo $5$), which, in turn, implies that all the $r_i$ are equal. This is impossible, since $[A_1A_2A_3]=[A_1A_2A_5]$ clearly implies $[A_1A_2A_4]>[A_1A_2A_3]$. Case 2 $A_5$ lies inside the convex quadrilateral $A_1A_2A_3A_4$ Suppose furthermore that $A_5$ lies inside the triangles $A_1A_2A_3,A_1A_2A_4$, and outside the other two formed by the vertices $A_1,A_2,A_3,A_4$. We then get $r_5=-\frac{r_1+r_2+r_3}3=-\frac{r_1+r_2+r_4}3\Rightarrow r_3=r_4$, which means that $A_3A_4\|A_1A_2\Rightarrow r_1=r_2$. We also have $r_5+r_3=r_2+r_4=r_2+r_3\Rightarrow r_1=r_2=r_5$. This is impossible, since it would mean that $A_5$ lies on the parallel $A_1A_2$ to $A_3A_4$. Case 3 $A_4,A_5$ lie inside $A_1A_2A_3$ We get $r_4=r_5=-\frac{r_1+r_2+r_3}3$, so the areas of the following pairs of triangles are equal: $(A_1A_2A_4,A_1A_2A_5),(A_2A_3A_4,A_2A_3A_5)$. This means that $A_4A_5$ is parallel to both $A_1A_2$ and $A_2A_3$, which is, of course, absurd.

Find all $ x,y$ and $ z$ in positive integer: $ z + y^{2} + x^{3} = xyz$ and $ x = \gcd(y,z)$.

At a meeting of $ 12k$ people, each person exchanges greetings with exactly $ 3k+6$ others. For any two people, the number who exchange greetings with both is the same. How many people are at the meeting?

Let $ p$ be an odd prime number. How many $ p$-element subsets $ A$ of $ \{1,2,\dots,2p\}$ are there, the sum of whose elements is divisible by $ p$?

Click for solution Consider the polynomial $f(x,y)=(1+xy)(1+x^2y)\ldots(1+x^{2p-1}y)$, the $k$-th factor refering to whether $k$ is present in the set or not. Then the monomial $x^ky^l$ will refer to the set having $l$ elements and sum of elements $k$, so we need to compute the sum of coefficients of $x^{kp}y^p$ of $f$ to find the answer. To do this, we need to compute the sum of coefficients $x^{pk}y^{pl}$ and substract 2, since there are two terms for $l\neq 1$:$1, x^{p(2p-1)}y^{2p}$. To compute the sum of coefficients of $x^{pk}y^{pl}$ we use Multisection formula that says us that it equals $\frac{1}{p^2}\sum_{i,j}f(w^i,w^j)$($w$ is $p$th root of unity). Now if $w^i$ is not $1$, $f(w^i,w^j)=\prod(1+w^{ik}w^j)=\prod (1+w^k)^2$ (every $w^m$ will be written twice as $w^{ki}w^j$). This equals $\prod (-1-w^k)^2=g(-1)^2=((-1)^p-1)^2=4$ ($g(x)=x^p-1$ is the polynomial with roots $w^i$). Since there are $p-1$ choices for $w^i$, $p$ choices for $w^j$ we get $4p(p-1)$. Finally, if $w^i=1$, $f(1,w^j)=(1+w^j)^{2p}$. To evaluate $\sum_{j=0}^{p-1} (1+w^j)^{2p}$, note that it equals $p$ times the coefficients of $x^p$ in the polynomial $(1+x)^{2p}$ by the same multisection formula, so it equals $p\binom{2p}{p}$. So our total sum is $\frac 1{p^2} (4p(p-1)+p\binom{2p}{p}) =\frac{\binom{2p}{p}-2}p+4$. Substracting 2 we get the desired answer.

Does there exist an integer $ n > 1$ which satisfies the following condition? The set of positive integers can be partitioned into $ n$ nonempty subsets, such that an arbitrary sum of $ n - 1$ integers, one taken from each of any $ n - 1$ of the subsets, lies in the remaining subset.

Let $ p$ be an odd prime. Determine positive integers $ x$ and $ y$ for which $ x \leq y$ and $ \sqrt{2p} - \sqrt{x} - \sqrt{y}$ is non-negative and as small as possible.

Does there exist a sequence $ F(1), F(2), F(3), \ldots$ of non-negative integers that simultaneously satisfies the following three conditions? (a) Each of the integers $ 0, 1, 2, \ldots$ occurs in the sequence. (b) Each positive integer occurs in the sequence infinitely often. (c) For any $ n \geq 2,$ \[ F(F(n^{163})) = F(F(n)) + F(F(361)). \]

Find the maximum value of $ x_{0}$ for which there exists a sequence $ x_{0},x_{1}\cdots ,x_{1995}$ of positive reals with $ x_{0} = x_{1995}$, such that \[ x_{i - 1} + \frac {2}{x_{i - 1}} = 2x_{i} + \frac {1}{x_{i}}, \] for all $ i = 1,\cdots ,1995$.

Click for solution Taking sum both side from 1 to 1995 and some simplifying job we would obtain $ \sum_{i = 0}^{1994}x_i - \frac {1}{x_i} = 0$ ..........(#) Also , simplfying the expression also gives us $ (x_i\cdot x_{i - 1} - 1)(2x_i - x_{i - 1}) = 0$ If $ x_i\cdot x_{i - 1} = 1$ then (#) would gives us $ x_{1994} = \frac {1}{x_{1994}}$ or $ x_{1994} = 1$ . So this means $ x_0 = x_{1995} = \frac {1}{x_{1994}} = 1$ If $ 2x_i = x_{i - 1}$ $ \implies x_i = x_0\left(\frac {1}{2}\right)^i$ . Hence from (#) we get $ \sum_{i = 0}^{1994}x_0\left(\frac {1}{2}\right)^i - \frac {2^i}{x_0} = 0$ simplify to get $ x_0 = 2^{997}$ . So the maximum is $ x_0 = 2^{997}$

For an integer $x \geq 1$, let $p(x)$ be the least prime that does not divide $x$, and define $q(x)$ to be the product of all primes less than $p(x)$. In particular, $p(1) = 2.$ For $x$ having $p(x) = 2$, define $q(x) = 1$. Consider the sequence $x_0, x_1, x_2, \ldots$ defined by $x_0 = 1$ and \[ x_{n+1} = \frac{x_n p(x_n)}{q(x_n)} \] for $n \geq 0$. Find all $n$ such that $x_n = 1995$.

Suppose that $ x_1, x_2, x_3, \ldots$ are positive real numbers for which \[ x^n_n = \sum^{n-1}_{j=0} x^j_n\] for $ n = 1, 2, 3, \ldots$ Prove that $ \forall n,$ \[ 2 - \frac{1}{2^{n-1}} \leq x_n < 2 - \frac{1}{2^n}.\]

For positive integers $ n,$ the numbers $ f(n)$ are defined inductively as follows: $ f(1) = 1,$ and for every positive integer $ n,$ $ f(n+1)$ is the greatest integer $ m$ such that there is an arithmetic progression of positive integers $ a_1 < a_2 < \ldots < a_m = n$ for which \[ f(a_1) = f(a_2) = \ldots = f(a_m).\] Prove that there are positive integers $ a$ and $ b$ such that $ f(an+b) = n+2$ for every positive integer $ n.$

Let $ \mathbb{N}$ denote the set of all positive integers. Prove that there exists a unique function $ f: \mathbb{N} \mapsto \mathbb{N}$ satisfying \[ f(m + f(n)) = n + f(m + 95) \] for all $ m$ and $ n$ in $ \mathbb{N}.$ What is the value of $ \sum^{19}_{k = 1} f(k)?$