Difference between revisions of "1979 AHSME Problems/Problem 20"

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Solution by e_power_pi_times_i
 
Solution by e_power_pi_times_i
  
Since <math>a=\frac{1}{2}</math>, <math>b=\frac{1}{3}</math>. Now we evaluate <math>\arctan a</math> and <math>\arctan b</math>. Denote <math>x</math> and <math>\theta</math> such that <math>\arctan x = \theta</math>. Then <math>\tan(\arctan(x)) = \tan(\theta)</math>, and simplifying gives <math>x = \tan(\theta)</math>. So <math>a = \tan(\theta_a) = \frac{1}{2}</math> and <math>b = \tan(\theta_b) = \frac{1}{3}</math>. The question asks for <math>\theta_a + \theta_b</math>, so we try to find <math>\tan(\theta_a + \theta_b)</math> in terms of <math>\tan(\theta_a)</math> and <math>\tan(\theta_b)</math>. Using the angle addition formula for <math>\tan(\theta)</math>, we get that <math>\tan(\theta_a + \theta_b) = \frac{\tan(\theta_a)+\tan(\theta_b)}{1-\tan(\theta_a)\tan(\theta_b)}</math>. Plugging <math>\tan(\theta_a) = \frac{1}{2}</math> and <math>\tan(\theta_b) = \frac{1}{3}</math> in, we have <math>\tan(\theta_a + \theta_b) = \frac{\frac{1}{2}+\frac{1}{3}}{1-(\frac{1}{2})(\frac{1}{3})}</math>. Simplifying, <math>\tan(\theta_a + \theta_b) = 1</math>, so <math>\theta_a + \theta_b</math> in radians is <math>\boxed{\textbf{(C) } \frac{\pi}{4}}</math>.
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Since <math>a=\frac{1}{2}</math>, <math>b=\frac{1}{3}</math>. Now we evaluate <math>\arctan a</math> and <math>\arctan b</math>. Denote <math>x</math> and <math>\theta</math> such that <math>\arctan x = \theta</math>. Then <math>\tan(\arctan(x)) = \tan(\theta)</math>, and simplifying gives <math>x = \tan(\theta)</math>. So <math>a = \tan(\theta_a) = \frac{1}{2}</math> and <math>b = \tan(\theta_b) = \frac{1}{3}</math>. The question asks for <math>\theta_a + \theta_b</math>, so we try to find <math>\tan(\theta_a + \theta_b)</math> in terms of <math>\tan(\theta_a)</math> and <math>\tan(\theta_b)</math>. Using the angle addition formula for <math>\tan(\alpha+\beta)</math>, we get that <math>\tan(\theta_a + \theta_b) = \frac{\tan(\theta_a)+\tan(\theta_b)}{1-\tan(\theta_a)\tan(\theta_b)}</math>. Plugging <math>\tan(\theta_a) = \frac{1}{2}</math> and <math>\tan(\theta_b) = \frac{1}{3}</math> in, we have <math>\tan(\theta_a + \theta_b) = \frac{\frac{1}{2}+\frac{1}{3}}{1-(\frac{1}{2})(\frac{1}{3})}</math>. Simplifying, <math>\tan(\theta_a + \theta_b) = 1</math>, so <math>\theta_a + \theta_b</math> in radians is <math>\boxed{\textbf{(C) } \frac{\pi}{4}}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 11:51, 9 January 2017

Problem 20

If $a=\tfrac{1}{2}$ and $(a+1)(b+1)=2$ then the radian measure of $\arctan a + \arctan b$ equals

$\textbf{(A) }\frac{\pi}{2}\qquad \textbf{(B) }\frac{\pi}{3}\qquad \textbf{(C) }\frac{\pi}{4}\qquad \textbf{(D) }\frac{\pi}{5}\qquad \textbf{(E) }\frac{\pi}{6}$

Solution

Solution by e_power_pi_times_i

Since $a=\frac{1}{2}$, $b=\frac{1}{3}$. Now we evaluate $\arctan a$ and $\arctan b$. Denote $x$ and $\theta$ such that $\arctan x = \theta$. Then $\tan(\arctan(x)) = \tan(\theta)$, and simplifying gives $x = \tan(\theta)$. So $a = \tan(\theta_a) = \frac{1}{2}$ and $b = \tan(\theta_b) = \frac{1}{3}$. The question asks for $\theta_a + \theta_b$, so we try to find $\tan(\theta_a + \theta_b)$ in terms of $\tan(\theta_a)$ and $\tan(\theta_b)$. Using the angle addition formula for $\tan(\alpha+\beta)$, we get that $\tan(\theta_a + \theta_b) = \frac{\tan(\theta_a)+\tan(\theta_b)}{1-\tan(\theta_a)\tan(\theta_b)}$. Plugging $\tan(\theta_a) = \frac{1}{2}$ and $\tan(\theta_b) = \frac{1}{3}$ in, we have $\tan(\theta_a + \theta_b) = \frac{\frac{1}{2}+\frac{1}{3}}{1-(\frac{1}{2})(\frac{1}{3})}$. Simplifying, $\tan(\theta_a + \theta_b) = 1$, so $\theta_a + \theta_b$ in radians is $\boxed{\textbf{(C) } \frac{\pi}{4}}$.

See also

1979 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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