Olympiad problems

2006 IMC, 5

Tags: algebra , polynomial , Vieta , IMC , college contests

Show that there are an infinity of integer numbers $m,n$, with $gcd(m,n)=1$ such that the equation $(x+m)^{3}=nx$ has 3 different integer sollutions.

2000 Baltic Way, 17

Tags: Vieta , algebra , polynomial , quadratics , system of equations , algebra proposed

Find all real solutions to the following system of equations: \[\begin{cases} x+y+z+t=5\\xy+yz+zt+tx=4\\xyz+yzt+ztx+txy=3\\xyzt=-1\end{cases}\]

2010 Brazil National Olympiad, 1

Tags: function , Vieta , algebra unsolved , algebra , functional equation

Find all functions $f$ from the reals into the reals such that \[ f(ab) = f(a+b) \] for all irrational $a, b$.

2010 Contests, 2

Tags: geometry , function , rhombus , Vieta , geometry proposed

Each of two different lines parallel to the the axis $Ox$ have exactly two common points on the graph of the function $f(x)=x^3+ax^2+bx+c$. Let $\ell_1$ and $\ell_2$ be two lines parallel to $Ox$ axis which meet the graph of $f$ in points $K_1, K_2$ and $K_3, K_4$, respectively. Prove that the quadrilateral formed by $K_1, K_2, K_3$ and $ K_4$ is a rhombus if and only if its area is equal to $6$ units.

2010 Harvard-MIT Mathematics Tournament, 3

Tags: calculus , algebra , polynomial , Vieta

Let $p$ be a monic cubic polynomial such that $p(0)=1$ and such that all the zeroes of $p^\prime (x)$ are also zeroes of $p(x)$. Find $p$. Note: monic means that the leading coefficient is $1$.

1994 Turkey Team Selection Test, 3

Tags: algebra , polynomial , Vieta , quadratics , number theory proposed , number theory

Find all integer pairs $(a,b)$ such that $a\cdot b$ divides $a^2+b^2+3$.

2011 Mongolia Team Selection Test, 3

Tags: modular arithmetic , algebra , polynomial , Vieta , number theory unsolved , number theory

Let $m$ and $n$ be positive integers such that $m>n$ and $m \equiv n \pmod{2}$. If $(m^2-n^2+1) \mid n^2-1$, then prove that $m^2-n^2+1$ is a perfect square. (proposed by G. Batzaya, folklore)

2008 IMO, 2

Tags: inequalities , algebra , Vieta , calculus , IMO 2008 , IMO , IMO Shortlist

[b](a)[/b] Prove that \[\frac {x^{2}}{\left(x \minus{} 1\right)^{2}} \plus{} \frac {y^{2}}{\left(y \minus{} 1\right)^{2}} \plus{} \frac {z^{2}}{\left(z \minus{} 1\right)^{2}} \geq 1\] for all real numbers $x$, $y$, $z$, each different from $1$, and satisfying $xyz=1$. [b](b)[/b] Prove that equality holds above for infinitely many triples of rational numbers $x$, $y$, $z$, each different from $1$, and satisfying $xyz=1$. [i]Author: Walther Janous, Austria[/i]

2015 Purple Comet Problems, 10

Tags: algebra , polynomial , Vieta

Find the sum of all the real values of x satisfying $(x+\frac{1}{x}-17)^2$ $= x + \frac{1}{x} + 17.$

2002 AMC 10, 14

Tags: quadratics , Vieta , algebra , number theory , prime numbers , special factorizations

Both roots of the quadratic equation $ x^2 \minus{} 63x \plus{} k \equal{} 0$ are prime numbers. The number of possible values of $ k$ is $ \textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ \textbf{more than four}$

2003 Purple Comet Problems, 19

Tags: Vieta

Let $x_1$ and $x_2$ be the roots of the equation $x^2 + 3x + 1 = 0$. Compute \[\left(\frac{x_1}{x_2 + 1}\right)^2 + \left(\frac{x_2}{x_1 + 1}\right)^2\]

2003 All-Russian Olympiad, 1

Tags: geometry , algebra , polynomial , Vieta , area of a triangle , Heron's formula , algebra unsolved

The side lengths of a triangle are the roots of a cubic polynomial with rational coefficients. Prove that the altitudes of this triangle are roots of a polynomial of sixth degree with rational coefficients.

1959 AMC 12/AHSME, 44

Tags: LaTeX , algebra , polynomial , Vieta , AMC

The roots of $x^2+bx+c=0$ are both real and greater than $1$. Let $s=b+c+1$. Then $s:$ $ \textbf{(A)}\ \text{may be less than zero}\qquad\textbf{(B)}\ \text{may be equal to zero}\qquad$ $\textbf{(C)}\ \text{must be greater than zero}\qquad\textbf{(D)}\ \text{must be less than zero}\qquad $ $\textbf{(E)}\text{ must be between -1 and +1} $

2016 SDMO (High School), 1

Tags: quadratics , Vieta , algebra

Quadratic equation $ x^2\plus{}ax\plus{}b\plus{}1\equal{}0$ have 2 positive integer roots, for integers $ a,b$. Show that $ a^2\plus{}b^2$ is not a prime.

MathLinks Contest 7th, 4.3

Tags: inequalities , function , parameterization , algebra , polynomial , Vieta , Cauchy Inequality

Let $ a,b,c$ be positive real numbers such that $ ab\plus{}bc\plus{}ca\equal{}3$. Prove that \[ \frac 1{1\plus{}a^2(b\plus{}c)} \plus{} \frac 1{1\plus{}b^2(c\plus{}a)} \plus{} \frac 1 {1\plus{}c^2(a\plus{}b) } \leq \frac 3 {1\plus{}2abc} .\]

2011 All-Russian Olympiad, 2

Tags: quadratics , invariant , Vieta , induction , algebra , polynomial , arithmetic sequence

In the notebooks of Peter and Nick, two numbers are written. Initially, these two numbers are 1 and 2 for Peter and 3 and 4 for Nick. Once a minute, Peter writes a quadratic trinomial $f(x)$, the roots of which are the two numbers in his notebook, while Nick writes a quadratic trinomial $g(x)$ the roots of which are the numbers in [i]his[/i] notebook. If the equation $f(x)=g(x)$ has two distinct roots, one of the two boys replaces the numbers in his notebook by those two roots. Otherwise, nothing happens. If Peter once made one of his numbers 5, what did the other one of his numbers become?

2009 Harvard-MIT Mathematics Tournament, 8

Tags: algebra , polynomial , Vieta , sum of roots

Let $a$, $b$, and $c$ be the $3$ roots of $x^3-x+1=0$. Find $\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}.$

2013 Iran MO (3rd Round), 2

Tags: algebra , polynomial , Vieta , number theory proposed , number theory , Vieta Jumping

Suppose that $a,b$ are two odd positive integers such that $2ab+1 \mid a^2 + b^2 + 1$. Prove that $a=b$. (15 points)

2012 ELMO Shortlist, 6

Tags: floor function , algebra , polynomial , Vieta , quadratics , inequalities , number theory proposed

Prove that if $a$ and $b$ are positive integers and $ab>1$, then \[\left\lfloor\frac{(a-b)^2-1}{ab}\right\rfloor=\left\lfloor\frac{(a-b)^2-1}{ab-1}\right\rfloor.\]Here $\lfloor x\rfloor$ denotes the greatest integer not exceeding $x$. [i]Calvin Deng.[/i]

2013 Iran MO (3rd Round), 4

Tags: algebra , polynomial , Vieta , number theory proposed , number theory , Diophantine equation , Pell equations

Prime $p=n^2 +1$ is given. Find the sets of solutions to the below equation: \[x^2 - (n^2 +1)y^2 = n^2.\] (25 points)

1995 AIME Problems, 2

Tags: logarithms , quadratics , modular arithmetic , algebra , polynomial , Vieta , search

Find the last three digits of the product of the positive roots of \[ \sqrt{1995}x^{\log_{1995}x}=x^2. \]

2016 CMIMC, 8

Tags: 2016 , CMIMC , algebra , polynomial , Vieta

Let $r_1$, $r_2$, $\ldots$, $r_{20}$ be the roots of the polynomial $x^{20}-7x^3+1$. If \[\dfrac{1}{r_1^2+1}+\dfrac{1}{r_2^2+1}+\cdots+\dfrac{1}{r_{20}^2+1}\] can be written in the form $\tfrac mn$ where $m$ and $n$ are positive coprime integers, find $m+n$.

2008 Harvard-MIT Mathematics Tournament, 7

Tags: quadratics , HMMT , trigonometry , algebra , polynomial , FTW , Vieta

A [i]root of unity[/i] is a complex number that is a solution to $ z^n \equal{} 1$ for some positive integer $ n$. Determine the number of roots of unity that are also roots of $ z^2 \plus{} az \plus{} b \equal{} 0$ for some integers $ a$ and $ b$.

2017 Azerbaijan JBMO TST, 2

Tags: algebra , Vieta

Let $x,y,z$ be 3 different real numbers not equal to $0$ that satisfiying $x^2-xy=y^2-yz=z^2-zx$. Find all the values of $\frac{x}{z}+\frac{y}{x}+\frac{z}{y}$ and $(x+y+z)^3+9xyz$.

2008 IMO Shortlist, 2

Tags: inequalities , algebra , Vieta , calculus , IMO 2008 , IMO , IMO Shortlist

[b](a)[/b] Prove that \[\frac {x^{2}}{\left(x \minus{} 1\right)^{2}} \plus{} \frac {y^{2}}{\left(y \minus{} 1\right)^{2}} \plus{} \frac {z^{2}}{\left(z \minus{} 1\right)^{2}} \geq 1\] for all real numbers $x$, $y$, $z$, each different from $1$, and satisfying $xyz=1$. [b](b)[/b] Prove that equality holds above for infinitely many triples of rational numbers $x$, $y$, $z$, each different from $1$, and satisfying $xyz=1$. [i]Author: Walther Janous, Austria[/i]