Mock AIME I 2015 Problems/Problem 8

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Let $a,b,c$ be consecutive terms (in that order) in an arithmetic sequence with common difference $d$. Suppose $\cos b$ and $\cos d$ are roots of a monic quadratic $p(x)$ with $p(-\tfrac{1}{2})=\tfrac1{2014}$. Then \[|\cos a+\cos b+\cos c+\cos d\,|=\frac pq\] for positive relatively prime integers $p$ and $q$. Find the remainder when $p+q$ is divided by $1000$.


Let $a=b-d$ and $c=b+d$ and substitute, then use the trigonometric identities $\cos{b-d}=\cos{b}\cos{d}+\sin{b}\sin{d}$ and $\cos{b+d}=\cos{b}\cos{d}-\sin{b}\sin{d}$ to find that


Furthermore, we have that $p(x)=x^2+\alpha x+\beta$ for some numbers $\alpha$ and $\beta$ and that $p(-\frac{1}{2})=-\frac{1}{2}\alpha+\beta+\frac{1}{4}=\frac{1}{2014}$. We also know that the roots of $p(x)$ are $\cos{b}$ and $\cos{d}$, so it follows by Vieta's formulas that $\beta=\cos{b}\cos{d}$ and that $\alpha=-\cos{b}-\cos{d}$. Hence

\begin{align*}-\frac{1}{2}\alpha+\beta+\frac{1}{4}=\frac{1}{2014}&\implies-\frac{1}{2}(-\cos{b}-\cos{d})+\cos{b}\cos{d}=\frac{1}{2014}-\frac{1}{4}\\ &\implies\cos{b}+\cos{d}+2\cos{b}\cos{d}=\frac{1}{1007}-\frac{1}{2}=\frac{-1005}{2014}\end{align*}

and $|\cos{b}+\cos{d}+2\cos{b}\cos{d}|=\frac{1005}{2014}.$ We see that $1005$ and $2014$ are already relatively prime; hence, $p+q=1005+2014=3019\equiv\boxed{019}\pmod{1000}.$

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