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

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are the roots of <math>x^2 - rx + s = 0</math>, then <math>rs</math> is necessarily  
 
are the roots of <math>x^2 - rx + s = 0</math>, then <math>rs</math> is necessarily  
  
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<math>\text{(A)} \ pq \qquad  
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<math>\textbf{(A)} \ pq \qquad  
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\text{(B)} \ \frac{1}{pq} \qquad  
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\textbf{(B)} \ \frac{1}{pq} \qquad  
−
\text{(C)} \ \frac{p}{q^2} \qquad  
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\textbf{(C)} \ \frac{p}{q^2} \qquad  
−
\text{(D)}\ \frac{q}{p^2}\qquad
+
\textbf{(D)}\ \frac{q}{p^2}\qquad
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\text{(E)}\ \frac{p}{q}</math>     
+
\textbf{(E)}\ \frac{p}{q}</math>     
  
 
== 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 Formulae, 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>.
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Also by Vieta's Formulae, we have <math>\tan(\alpha)+\tan(\beta)=p</math> and <math>\cot(\alpha)+\cot(\beta)=r</math>, and again using <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 therefore deduce that <math>r=\frac{p}{q}</math>, which yields <math>rs = \frac{1}{q}\cdot\frac{p}{q}=\frac{p}{q^2}</math>.
  
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Thus, the answer is <math>(\text{C}) \ \frac{p}{q^2}</math>
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Thus, the answer is <math>\boxed{\textbf{(C)} \ \frac{p}{q^2}}</math>.
 +
 
 +
==See Also==
 +
{{AHSME box|year=1983|num-b=19|num-a=21}}
 +
 
 +
{{MAA Notice}}

Latest revision as of 23:58, 19 February 2019

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

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

Solution

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

Also by Vieta's Formulae, we have $\tan(\alpha)+\tan(\beta)=p$ and $\cot(\alpha)+\cot(\beta)=r$, and again using $\cot\theta=\frac{1}{\tan\theta}$, we have $\tan(\alpha)+\tan(\beta)=r(\tan(\alpha)\tan(\beta))$. Using $\tan(\alpha)\tan(\beta)=q$ and $\tan(\alpha)+\tan(\beta)=p$, we therefore deduce that $r=\frac{p}{q}$, which yields $rs = \frac{1}{q}\cdot\frac{p}{q}=\frac{p}{q^2}$.

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

See Also

1983 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions


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