Solve the system of equation $$x+y+z=2;$$$$(x+y)(y+z)+(y+z)(z+x)+(z+x)(x+y)=1;$$$$x^2(y+z)+y^2(z+x)+z^2(x+y)=-6.$$

Find all triples $(x,y,z)$ of positive real numbers such that $$ \begin{cases} 2x^2+y^3=3 \\ 3y^2+z^3=4 \\ 4z^2+x^3=5 \\ \end{cases} $$ [i]M. Zorka[/i]

If the price of an article is increased by percent $ p$, then the decrease in percent of sales must not exceed $ d$ in order to yield the same income. The value of $ d$ is: $ \textbf{(A)}\ \frac{1}{1\plus{}p} \qquad\textbf{(B)}\ \frac{1}{1\minus{}p} \qquad\textbf{(C)}\ \frac{p}{1\plus{}p} \qquad\textbf{(D)}\ \frac{p}{p\minus{}1} \qquad\textbf{(E)}\ \frac{1\minus{}p}{1\plus{}p}$

Find all triples $(x, y, z)$ of real numbers satisfying: $x + y - z = -1$ , $x^2 - y^2 + z^2 = 1$ and $- x^3 + y^3 + z^3 = -1$

Given natural numbers $n > k > 1$, find all real solutions $x_1,..., x_n$ of the system $$x_i^3(x_i^2 + x_{i+1}^2+... +x_{i+k-1}^2) = x_{i-1}^2$$ for 1 $\le i \le n$. Here $x_{n+i} = x_i$ for all$ i$.

Find all sets of four real numbers $x_1, x_2, x_3, x_4$ such that the sum of any one and the product of the other three is equal to 2.

The cost of $1000$ grams of chocolate is $x$ dollars and the cost of $1000$ grams of potatoes is $y$ dollars, the numbers $x$ and $y$ are positive integers and have not more than $2$ digits. Mother said to Maria to buy $200$ grams of chocolate and $1000$ grams of potatoes that cost exactly $N$ dollars. Maria got confused and bought $1000$ grams of chocolate and $200$ grams of potatoes that cost exactly $M$ dollars ($M >N$). It turned out that the numbers $M$ and $N$ have no more than two digits and are formed of the same digits but in a different order. Find $x$ and $y$.

Determine all positive integers $n$ for which there exist positive integers $a_1,a_2, ..., a_n$ with $a_1 + 2a_2 + 3a_3 +... + na_n = 6n$ and $\frac{1}{a_1}+\frac{2}{a_2}+\frac{3}{a_3}+ ... +\frac{n}{a_n}= 2 + \frac1n$

Let $ X: \equal{} \{x_1,x_2,\ldots,x_{29}\}$ be a set of $ 29$ boys: they play with each other in a tournament of Pro Evolution Soccer 2009, in respect of the following rules: [list]i) every boy play one and only one time against each other boy (so we can assume that every match has the form $ (x_i \text{ Vs } x_j)$ for some $ i \neq j$); ii) if the match $ (x_i \text{ Vs } x_j)$, with $ i \neq j$, ends with the win of the boy $ x_i$, then $ x_i$ gains $ 1$ point, and $ x_j$ doesn’t gain any point; iii) if the match $ (x_i \text{ Vs } x_j)$, with $ i \neq j$, ends with the parity of the two boys, then $ \frac {1}{2}$ point is assigned to both boys. [/list] (We assume for simplicity that in the imaginary match $ (x_i \text{ Vs } x_i)$ the boy $ x_i$ doesn’t gain any point). Show that for some positive integer $ k \le 29$ there exist a set of boys $ \{x_{t_1},x_{t_2},\ldots,x_{t_k}\} \subseteq X$ such that, for all choice of the positive integer $ i \le 29$, the boy $ x_i$ gains always a integer number of points in the total of the matches $ \{(x_i \text{ Vs } x_{t_1}),(x_i \text{ Vs } x_{t_2}),\ldots, (x_i \text{ Vs } x_{t_k})\}$. [i](Paolo Leonetti)[/i]

Let $\alpha,\beta,\gamma$ be angles of a triangle. Determine all real triplets $x,y,z$ satisfying the system \begin{align*} x\cos\beta+\frac1z\cos\alpha &=1, \\ y\cos\gamma+\frac1x\cos\beta &=1, \\ z\cos\alpha+\frac1y\cos\gamma &=1. \end{align*}

The equations are given $$ \begin{array}{c} x^2 + p_1x + q_1 = 0\\ x^2 + p_2x + q_2 = 0\\ x^2 + p_3x + q_3 = 0 \end{array}$$ each two of which have a common root, but all three have no common root. Prove that: 1) $2 (p_1p_2 + p_2p_3 + p_3p_1) - (p_1^2 + p_2^2 + p_3^2) = 4 (q_1 + q_2+ q_3)$ 2) he roots of these equations are rational when the numbers $p_1$, $p_2$ and $p_3$ are rational}.

For some positive numbers $A, B, C$ and $D$, the system of equations $$\begin{cases} x^2 + y^2 = A \\ |x| + |y| = B \end{cases}$$ has $m$ solutions, while the system of equations $$\begin{cases} x^2 + y^2 +z^2= X\\ |x| + |y| +|z| = D \end{cases}$$ has $n$ solutions. If $m > n > 1$, find $m$ and $n$. ( G Galperin)

Prove that if the numbers $ x $, $ y $, $ z $ satisfy the equationw $$x + y + z = a,$$ $$ \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1}{a},$$ then at least one of them is equal to $ a $.

Find all quadruplets $(x_1, x_2, x_3, x_4)$ of real numbers such that the next six equalities apply: $$\begin{cases} x_1 + x_2 = x^2_3 + x^2_4 + 6x_3x_4\\ x_1 + x_3 = x^2_2 + x^2_4 + 6x_2x_4\\ x_1 + x_4 = x^2_2 + x^2_3 + 6x_2x_3\\ x_2 + x_3 = x^2_1 + x^2_4 + 6x_1x_4\\ x_2 + x_4 = x^2_1 + x^2_3 + 6x_1x_3 \\ x_3 + x_4 = x^2_1 + x^2_2 + 6x_1x_2 \end{cases}$$

Find all real pairs $(a,b)$ that solve the system of equation \begin{align*} a^2+b^2 &= 25, \\ 3(a+b)-ab &= 15. \end{align*} [i](German MO 2016 - Problem 1)[/i]

Solve the system of equations: $ \sqrt {3x}(1 \plus{} \frac {1}{x \plus{} y}) \equal{} 2$ $ \sqrt {7y}(1 \minus{} \frac {1}{x \plus{} y}) \equal{} 4\sqrt {2}$

Find all solutions $x_1, x_2, x_3, x_4, x_5$ of the system \[ x_5+x_2=yx_1 \] \[ x_1+x_3=yx_2 \] \[ x_2+x_4=yx_3 \] \[ x_3+x_5=yx_4 \] \[ x_4+x_1=yx_5 \] where $y$ is a parameter.

The number of solutions, in real numbers $a$, $b$, and $c$, to the system of equations $$a+bc=1,$$$$b+ac=1,$$$$c+ab=1,$$ is $\text{(A) }3\qquad\text{(B) }4\qquad\text{(C) }5\qquad\text{(D) more than }5\text{, but finitely many}\qquad\text{(E) infinitely many}$

Three players $A,B$ and $C$ play a game with three cards and on each of these $3$ cards it is written a positive integer, all $3$ numbers are different. A game consists of shuffling the cards, giving each player a card and each player is attributed a number of points equal to the number written on the card and then they give the cards back. After a number $(\geq 2)$ of games we find out that A has $20$ points, $B$ has $10$ points and $C$ has $9$ points. We also know that in the last game B had the card with the biggest number. Who had in the first game the card with the second value (this means the middle card concerning its value).

Determine all pairs of real numbers $(k, d)$ such that the system of equations $$\begin{cases} x^3 + y^3 = 2 \\ kx + d = y\end{cases}$$ has no solutions $(x, y)$ with $x$ and $y$ real numbers.

Let $a$ be a real number and $A = \{(x, y) \in R \times R | \, x + y = a\}$, $B = \{(x,y) \in R \times R | \, x^3 + y^3 < a\}$ . Find all values of $a$ such that $A \cap B = \emptyset$ .

We consider the following system with $q=2p$: \[\begin{matrix} a_{11}x_{1}+\ldots+a_{1q}x_{q}=0,\\ a_{21}x_{1}+\ldots+a_{2q}x_{q}=0,\\ \ldots ,\\ a_{p1}x_{1}+\ldots+a_{pq}x_{q}=0,\\ \end{matrix}\] in which every coefficient is an element from the set $\{-1,0,1\}$$.$ Prove that there exists a solution $x_{1}, \ldots,x_{q}$ for the system with the properties: [b]a.)[/b] all $x_{j}, j=1,\ldots,q$ are integers$;$ [b]b.)[/b] there exists at least one j for which $x_{j} \neq 0;$ [b]c.)[/b] $|x_{j}| \leq q$ for any $j=1, \ldots ,q.$

Solve the system of equations: $ y^2\equal{}(x\plus{}8)(x^2\plus{}2),$ $ y^2\minus{}(8\plus{}4x)y\plus{}(16\plus{}16x\minus{}5x^2)\equal{}0.$

Determine all $4$-tuples $(a,b, c, d)$ of positive real numbers satisfying $a + b +c + d = 1$ and $\max (\frac{a^2}{b},\frac{b^2}{a}) \cdot \max (\frac{c^2}{d},\frac{d^2}{c}) = (\min (a + b, c + d))^4$

For the system of equations $x^2 + x^2y^2 + x^2y^4 = 525$ and $x + xy + xy^2 = 35$, the sum of the real $y$ values that satisfy the equations is $ \mathrm{(A) \ } 2 \qquad \mathrm{(B) \ } \frac{5}{2} \qquad \mathrm {(C) \ } 5 \qquad \mathrm{(D) \ } 20 \qquad \mathrm{(E) \ } \frac{55}{2}$