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

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== Problem #20 ==
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== Problem ==
 
   
 
   
 
If <math>a=\tfrac{1}{2}</math> and <math>(a+1)(b+1)=2</math> then the radian measure of <math>\arctan a + \arctan b</math> equals
 
If <math>a=\tfrac{1}{2}</math> and <math>(a+1)(b+1)=2</math> then the radian measure of <math>\arctan a + \arctan b</math> equals
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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>.
 
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>.
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== Solution 1.1 (Guiding through the thought process) ==
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Thinking through the problem, we can see <math>\arctan a</math> is the angle whose tangent is <math>0.5</math>. <math>\arctan b</math> is the angle whose tangent is <math>\frac{1}{3}</math>. Call the former angle <math>\theta_1</math>, the latter <math>\theta_2</math>. So we are trying to find <math>\theta_1+\theta_2</math>. So given two tangent measures, it is natural for us to think about the sum of tangent measures (what else can we try? Remember: we are not allowed to use calculators). Plug in as above and continue on.
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~hastapasta
  
 
== See also ==
 
== See also ==

Latest revision as of 12:33, 16 July 2024

Problem

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 $\boxed{\textbf{(C) } \frac{\pi}{4}}$.

Solution 1.1 (Guiding through the thought process)

Thinking through the problem, we can see $\arctan a$ is the angle whose tangent is $0.5$. $\arctan b$ is the angle whose tangent is $\frac{1}{3}$. Call the former angle $\theta_1$, the latter $\theta_2$. So we are trying to find $\theta_1+\theta_2$. So given two tangent measures, it is natural for us to think about the sum of tangent measures (what else can we try? Remember: we are not allowed to use calculators). Plug in as above and continue on.

~hastapasta

See also

1979 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

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