Difference between revisions of "1983 AHSME Problems/Problem 20"
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== Solution == | == 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 | + | 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 | + | 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 21:19, 14 April 2016
Problem 20
If and
are the roots of
, and
and
are the roots of
, then
is necessarily
Solution
By Vieta's Formulas, we have and
. Recalling that
, we have
.
By Vieta's Formulas, we have and
. Recalling that
, we have
. Using
and
, we get that
, which yields a product of
.
Thus, the answer is