This website contains problems from math contests. Problems and corresponding tags were obtained from the Art of Problem Solving website.

Tags were heavily modified to better represent problems.

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Found problems: 91

2016 Hanoi Open Mathematics Competitions, 13

Find all triples $(a,b,c)$ of real numbers such that $|2a + b| \ge 4$ and $|ax^2 + bx + c| \le 1$ $ \forall x \in [-1, 1]$.

2010 Saudi Arabia BMO TST, 4

Let $a > 0$. If the system $$\begin{cases} a^x + a^y + a^z = 14 - a \\ x + y + z = 1 \end{cases}$$ has a solution in real numbers, prove that $a \le 8$.

1974 Swedish Mathematical Competition, 5

Find the smallest positive real $t$ such that \[\left\{ \begin{array}{l} x_1 + x_3 = 2t x_2 \\ x_2 + x_4 = 2t x_3 \\ x_3 + x_5=2t x_4 \\ \end{array} \right. \] has a solution $x_1$, $x_2$, $x_3$, $x_4$, $x_5$ in non-negative reals, not all zero.

1971 Czech and Slovak Olympiad III A, 1

Let $a,b,c$ real numbers. Show that there are non-negative $x,y,z,xyz\neq0$ such that \begin{align*} cy-bz &\ge 0, \\ az-cx &\ge 0, \\ bx-ay &\ge 0. \end{align*}

1992 Swedish Mathematical Competition, 3

Solve: $$\begin{cases} 2x_1 - 5x_2 + 3x_3 \ge 0 \\ 2x_2 - 5x_3 + 3x4 \ge 0 \\ ...\\ 2x_{23} - 5x_{24} + 3x_{25} \ge 0\\ 2x_{24} - 5x_{25} + 3x_1 \ge 0\\ 2x_{25} - 5x_1 + 3x_2 \ge 0 \end{cases}$$

2021 Dutch IMO TST, 2

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

2014 NZMOC Camp Selection Problems, 7

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.

2021 Junior Balkan Team Selection Tests - Moldova, 6

Solve the system of equations $$\begin{cases} (x+y)(x^2-y^2)=32 \\ (x-y)(x^2+y^2)=20 \end{cases}$$

1982 Spain Mathematical Olympiad, 1

On the puzzle page of a newspaper this problem is proposed: “Two children, Antonio and José, have $160$ comics. Antonio counts his by $7$ by $7$ and there are $4$ left over. José counts his $ 8$ by $8$ and he also has $4$ left over. How many comics does he have each?" In the next issue of the newspaper this solution is given: “Antonio has $60$ comics and José has $100$.” Analyze this solution and indicate what a mathematician would do with this problem.

1999 Romania National Olympiad, 1

Tags: algebra , system
Solve the system $$\begin{cases} \displaystyle 4^{-x}+27^{-y}= \frac{5}{6} \\ \displaystyle 27^y-4^x \le 1 \\ \displaystyle \log_{27}y-\log_4 x \ge \frac{1}{6} \end{cases}.$$

1977 Swedish Mathematical Competition, 3

Show that the only integral solution to \[\left\{ \begin{array}{l} xy + yz + zx = 3n^2 - 1\\ x + y + z = 3n \\ \end{array} \right. \] with $x \geq y \geq z$ is $x=n+1$, $y=n$, $z=n-1$.

1963 Poland - Second Round, 1

Prove that if the numbers $ p $, $ q $, $ r $ satisfy the equality $$ p+q + r=1$$ $$ \frac{1}{p} + \frac{1}{q} + \frac{1}{r} = 0$$ then for any numbers $ a $, $ b $, $ c $ equality holds $$a^2 + b^2 + c^2 = (pa + qb + rc)^2 + (qa + rb + pc)^2 + (ra + pb + qc)^2.$$

1972 Swedish Mathematical Competition, 1

Find the largest real number $a$ such that \[\left\{ \begin{array}{l} x - 4y = 1 \\ ax + 3y = 1\\ \end{array} \right. \] has an integer solution.

2014 Hanoi Open Mathematics Competitions, 13

Let $a, b,c$ satis es the conditions $\begin{cases} 5 \ge a \ge b \ge c \ge 0 \\ a + b \le 8 \\ a + b + c = 10 \end{cases}$ Prove that $a^2 + b^2 + c^2 \le 38$

1973 Spain Mathematical Olympiad, 2

Determine all solutions of the system $$\begin{cases} 2x - 5y + 11z - 6 = 0 \\ -x + 3y - 16z + 8 = 0 \\ 4x - 5y - 83z + 38 = 0 \\ 3x + 11y - z + 9 > 0 \end{cases}$$ in which the first three are equations and the last one is a linear inequality.

2017 NZMOC Camp Selection Problems, 8

Find all possible real values for $a, b$ and $c$ such that (a) $a + b + c = 51$, (b) $abc = 4000$, (c) $0 < a \le 10$ and $c \ge 25$.

2000 Junior Balkan Team Selection Tests - Moldova, 4

Find the smallest natural number nonzero n so that it exists in real numbers $x_1, x_2,..., x_n$ which simultaneously check the conditions: 1) $x_i \in [1/2 , 2]$ , $i = 1, 2,... , n$ 2) $x_1+x_2+...+x_n \ge \frac{7n}{6}$ 3) $\frac{1}{x_1}+\frac{1}{x_2}+...+\frac{1}{x_n}\ge \frac{4n}{3}$

2010 Saudi Arabia BMO TST, 4

Find all triples $(x,y, z)$ of integers such that $$\begin{cases} x^2y + y^2z + z^2x= 2010^2 \\ xy^2 + yz^2 + zx^2= -2010 \end{cases}$$

1986 Swedish Mathematical Competition, 4

Prove that $x = y = z = 1$ is the only positive solution of the system \[\left\{ \begin{array}{l} x+y^2 +z^3 = 3\\ y+z^2 +x^3 = 3\\ z+x^2 +y^3 = 3\\ \end{array} \right. \]

2015 Swedish Mathematical Competition, 4

Solve the system of equations $$ \left\{\begin{array}{l} x \log x+y \log y+z \log x=0\\ \\ \dfrac{\log x}{x}+\dfrac{\log y}{y}+\dfrac{\log z}{z}=0 \end{array} \right. $$

2010 Saudi Arabia Pre-TST, 4.4

Find all pairs $(x, y)$ of real numbers that satisfy the system of equations $$\begin{cases} x^4 + 2z^3 - y =\sqrt3 - \dfrac14 \\ y^4 + 2y^3 - x = - \sqrt3 - \dfrac14 \end{cases}$$

2003 Junior Balkan Team Selection Tests - Moldova, 6

The real numbers x and у satisfy the equations $$\begin{cases} \sqrt{3x}\left(1+\dfrac{1}{x+y}\right)=2 \\ \\ \sqrt{7y}\left(1-\dfrac{1}{x+y}\right)=4\sqrt2 \end{cases}$$ Find the numerical value of the ratio $y/x$.

2003 Swedish Mathematical Competition, 1

If $x, y, z, w$ are nonnegative real numbers satisfying \[\left\{ \begin{array}{l}y = x - 2003 \\ z = 2y - 2003 \\ w = 3z - 2003 \\ \end{array} \right. \] find the smallest possible value of $x$ and the values of $y, z, w$ corresponding to it.

1979 Dutch Mathematical Olympiad, 2

Solve in $N$: $$\begin{cases} a^3=b^3+c^3+12a \\ a^2=5(b+c) \end{cases}$$

2009 Mathcenter Contest, 4

Find the values of the real numbers $x,y,z$ that correspond to the system of equations. $$8(x+\frac{1}{x}) =15(y+\frac{1}{y}) = 17(z+\frac{1}{z})$$ $$xy + yz + zx=1$$ [i](Heir of Ramanujan)[/i]