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Difference between revisions of "Mock AIME I 2015 Problems/Problem 8"

(Created page with "==Problem== Let <math>a,b,c</math> be consecutive terms (in that order) in an arithmetic sequence with common difference <math>d</math>. Suppose <math>\cos b</math> and <math...")
 
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<cmath>\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*}</cmath>
 
<cmath>\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*}</cmath>
  
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and <math>|\cos{b}+\cos{d}+2\cos{b}\cos{d}|=\frac{1005}{2014}.</math> We see that <math>1005</math> and <math>2014</math> are already relatively prime; hence, <math>m+n=1005+2014=3019\equiv\boxed{019}\pmod{1000}.</math>
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and <math>|\cos{b}+\cos{d}+2\cos{b}\cos{d}|=\frac{1005}{2014}.</math> We see that <math>1005</math> and <math>2014</math> are already relatively prime; hence, <math>p+q=1005+2014=3019\equiv\boxed{019}\pmod{1000}.</math>

Latest revision as of 18:46, 27 March 2016

Problem

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 $1000$.

Solution

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

$|\cos{a}+\cos{b}+\cos{c}+\cos{d}|=|\cos{b}+\cos{d}+2\cos{b}\cos{d}|.$

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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