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

Revision as of 05:36, 18 May 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 $(\text{C}) \ \frac{p}{q^2}$

1983 AHSME (Problems • Answer Key • Resources)
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The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 16: / Line 16: @@
 Thus, the answer is <math>(\text{C}) \ \frac{p}{q^2}</math>
+==See Also==
+{{AHSME box|year=1983|num-b=19|num-a=21}}
+{{MAA Notice}}

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

Revision as of 05:36, 18 May 2016

Problem 20

Solution

See Also