1983 AHSME Problems/Problem 20

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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1983 AHSME Problems/Problem 20

Problem 20

Solution