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

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Found problems: 7

Three distinct real numbers form (in some order) a 3-term arithmetic sequence, and also form (in possibly a different order) a 3-term geometric sequence. Compute the greatest possible value of the common ratio of this geometric sequence.
Let $ p(x) \equal{} x^3 \plus{} ax^2 \plus{} bx \plus{} c$, where $ a$, $ b$, and $ c$ are complex numbers. Suppose that \[ p(2009 \plus{} 9002\pi i) \equal{} p(2009) \equal{} p(9002) \equal{} 0 \]What is the number of nonreal zeros of $ x^{12} \plus{} ax^8 \plus{} bx^4 \plus{} c$? $ \textbf{(A)}\ 4\qquad \textbf{(B)}\ 6\qquad \textbf{(C)}\ 8\qquad \textbf{(D)}\ 10\qquad \textbf{(E)}\ 12$
Let the roots of $x^2-3x+1=0$ be $r$ and $s$. Then the expression $r^2+s^2$ is: $ \textbf{(A)}\ \text{a positive integer} \qquad\textbf{(B)}\ \text{a positive fraction greater than 1}\qquad\textbf{(C)}\ \text{a positive fraction less than 1}$ $\textbf{(D)}\ \text{an irrational number}\qquad\textbf{(E)}\ \text{an imaginary number}$
Let be a natural number $ n\ge 2 $ and an imaginary number $ z $ having the property that $ |z-1|=|z+1|\cdot\sqrt[n]{2} . $ Denote with $ A,B,C $ the points in the Euclidean plane whose representation in the complex plane are the affixes of $ z,\frac{1-\sqrt[n]{2}}{1+\sqrt[n]{2}} ,\frac{1+\sqrt[n]{2}}{1-\sqrt[n]{2}} , $ respectively. Prove that $ AB $ is perpendicular to $ AC. $
Let $N$ be a positive integer. Define a sequence $a_0,a_1,\ldots$ by $a_0=0$, $a_1=1$, and $a_{n+1}+a_{n-1}=a_n(2-1/N)$ for $n\ge1$. Prove that $a_n<\sqrt{N+1}$ for all $n$. [i]Evan O'Dorney.[/i]
Equation $(1-\text{i})x^2+(\lambda+\text{i})x+(1+\text{i}\lambda)=0(\lambda\in\mathbb{R})$ has two imaginary roots, then the range value of $\lambda$ is________.
Let $N$ be a positive integer. Define a sequence $a_0,a_1,\ldots$ by $a_0=0$, $a_1=1$, and $a_{n+1}+a_{n-1}=a_n(2-1/N)$ for $n\ge1$. Prove that $a_n<\sqrt{N+1}$ for all $n$. [i]Evan O'Dorney.[/i]