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: 9

2014 Czech-Polish-Slovak Match, 1

Prove that if the positive real numbers $a, b, c$ satisfy the equation \[a^4 + b^4 + c^4 + 4a^2b^2c^2 = 2 (a^2b^2 + a^2c^2 + b^2c^2),\] then there is a triangle $ABC$ with internal angles $\alpha, \beta, \gamma$ such that \[\sin \alpha = a, \qquad \sin \beta = b, \qquad \sin \gamma= c.\]

1975 Vietnam National Olympiad, 1

The roots of the equation $x^3 - x + 1 = 0$ are $a, b, c$. Find $a^8 + b^8 + c^8$.

1987 Spain Mathematical Olympiad, 6

For all natural numbers $n$, consider the polynomial $P_n(x) = x^{n+2}-2x+1$. (a) Show that the equation $P_n(x)=0$ has exactly one root $c_n$ in the open interval $(0,1)$. (b) Find $lim_{n \to \infty}c_n$.

2017 Bulgaria National Olympiad, 5

Let $n$ be a natural number and $f(x)$ be a polynomial with real coefficients having $n$ different positive real roots. Is it possible the polynomial: $$x(x+1)(x+2)(x+4)f(x)+a$$ to be presented as the $k$-th power of a polynomial with real coefficients, for some natural $k\geq 2$ and real $a$?

2012 India Regional Mathematical Olympiad, 6

Let $a$ and $b$ be real numbers such that $a \ne 0$. Prove that not all the roots of $ax^4 + bx^3 + x^2 + x + 1 = 0$ can be real.

2003 Poland - Second Round, 3

Let $W(x) = x^4 - 3x^3 + 5x^2 - 9x$ be a polynomial. Determine all pairs of different integers $a$, $b$ satisfying the equation $W(a) = W(b)$.

1985 Spain Mathematical Olympiad, 7

Find the values of $p$ for which the equation $x^5 - px-1 = 0$ has two roots $r$ and $s$ which are the roots of equation $x^2-ax+b= 0$ for some integers $a,b$.

2020 Greece National Olympiad, 1

Find all non constant polynomials $P(x),Q(x)$ with real coefficients such that: $P((Q(x))^3)=xP(x)(Q(x))^3$

2001 Nordic, 3

Determine the number of real roots of the equation ${x^8 -x^7 + 2x^6- 2x^5 + 3x^4 - 3x^3 + 4x^2 - 4x + \frac{5}{2}= 0}$