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Difference between revisions of "1983 AHSME Problems/Problem 20"

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== Solution ==
 
== Solution ==
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By Vieta's Formulas, we have <math>\tan(\alpha)\tan(\beta)=q</math> and <math>\cot(\alpha)\cot(\beta)=s</math>. Recalling that <math>\cot(\theta)=\frac{1}{\tan(\theta)}</math>, we have <math>\frac{1}{\tan(\alpha)\tan(\beta)}=\frac{1}{q}=s</math>.  
+
By Vieta's Formulas, we have <math>\tan(\alpha)\tan(\beta)=q</math> and <math>\cot(\alpha)\cot(\beta)=s</math>. Recalling that <math>\cot\theta=\frac{1}{\tan\theta}</math>, we have <math>\frac{1}{\tan(\alpha)\tan(\beta)}=\frac{1}{q}=s</math>.  
  
−
By Vieta's Formulas, we have <math>\tan(\alpha)+\tan(\beta)=p</math> and <math>\cot(\alpha)+\cot(\beta)=r</math>. Recalling that <math>\cot(\theta)=\frac{1}{\tan(\theta)}</math>, we have <math>\tan(\alpha)+\tan(\beta)=r(\tan(\alpha)\tan(\beta))</math>. Using <math>\tan(\alpha)\tan(\beta)=q</math> and <math>\tan(\alpha)+\tan(\beta)=p</math>, we get that <math>r=\frac{p}{q}</math>, which yields a product of <math>\frac{1}{q}\cdot\frac{p}{q}=\frac{p}{q^2}</math>.
+
By Vieta's Formulas, we have <math>\tan(\alpha)+\tan(\beta)=p</math> and <math>\cot(\alpha)+\cot(\beta)=r</math>. Recalling that <math>\cot\theta=\frac{1}{\tan\theta}</math>, we have <math>\tan(\alpha)+\tan(\beta)=r(\tan(\alpha)\tan(\beta))</math>. Using <math>\tan(\alpha)\tan(\beta)=q</math> and <math>\tan(\alpha)+\tan(\beta)=p</math>, we get that <math>r=\frac{p}{q}</math>, which yields a product of <math>\frac{1}{q}\cdot\frac{p}{q}=\frac{p}{q^2}</math>.
  
 
Thus, the answer is <math>(\text{C}) \ \frac{p}{q^2}</math>
 
Thus, the answer is <math>(\text{C}) \ \frac{p}{q^2}</math>

Revision as of 20:19, 14 April 2016

Problem 20

If $\tan{\alpha}$ and $\tan{\beta}$ are the roots of $x^2 - px + q = 0$, and $\cot{\alpha}$ and $\cot{\beta}$ are the roots of $x^2 - rx + s = 0$, then $rs$ is necessarily

$\text{(A)} \ pq \qquad \text{(B)} \ \frac{1}{pq} \qquad \text{(C)} \ \frac{p}{q^2} \qquad \text{(D)}\ \frac{q}{p^2}\qquad \text{(E)}\ \frac{p}{q}$

Solution

By Vieta's Formulas, we have $\tan(\alpha)\tan(\beta)=q$ and $\cot(\alpha)\cot(\beta)=s$. Recalling that $\cot\theta=\frac{1}{\tan\theta}$, we have $\frac{1}{\tan(\alpha)\tan(\beta)}=\frac{1}{q}=s$.

By Vieta's Formulas, we have $\tan(\alpha)+\tan(\beta)=p$ and $\cot(\alpha)+\cot(\beta)=r$. Recalling that $\cot\theta=\frac{1}{\tan\theta}$, we have $\tan(\alpha)+\tan(\beta)=r(\tan(\alpha)\tan(\beta))$. Using $\tan(\alpha)\tan(\beta)=q$ and $\tan(\alpha)+\tan(\beta)=p$, we get that $r=\frac{p}{q}$, which yields a product of $\frac{1}{q}\cdot\frac{p}{q}=\frac{p}{q^2}$.

Thus, the answer is $(\text{C}) \ \frac{p}{q^2}$

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