1983 AHSME Problems/Problem 20: Difference between revisions

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 $\boxed{\textbf{(C)} \ \frac{p}{q^2}}$ .

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

@@ Line 4: / Line 4: @@
 are the roots of <math>x^2 - rx + s = 0</math>, then <math>rs</math> is necessarily
-<math>\text{(A)} \ pq \qquad
+<math>\textbf{(A)} \ pq \qquad
-\text{(B)} \ \frac{1}{pq} \qquad
+\textbf{(B)} \ \frac{1}{pq} \qquad
-\text{(C)} \ \frac{p}{q^2} \qquad
+\textbf{(C)} \ \frac{p}{q^2} \qquad
-\text{(D)}\ \frac{q}{p^2}\qquad
+\textbf{(D)}\ \frac{q}{p^2}\qquad
-\text{(E)}\ \frac{p}{q}</math>
+\textbf{(E)}\ \frac{p}{q}</math>
 == Solution ==
-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>.
+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>.
-Thus, the answer is <math>(\text{C}) \ \frac{p}{q^2}</math>
+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}}

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1983 AHSME Problems/Problem 20: Difference between revisions

Latest revision as of 23:58, 19 February 2019

Problem 20

Solution

See Also