Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

1984 AIME Problems/Problem 13

Revision as of 15:52, 6 March 2007 by Chess64 (talk | contribs)

Problem

Find the value of $\displaystyle 10\cot(\cot^{-1}3+\cot^{-1}7+\cot^{-1}13+\cot^{-1}21).$

Solution

We know that $\tan(\arctan(x)) = x$ so we can repeatedly apply the addition formula, $\tan(x+y) = \frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)}$. Let $a = \arccot(3)$ (Error compiling LaTeX. Unknown error_msg), $b=\arccot(7)$ (Error compiling LaTeX. Unknown error_msg), $c=\arccot(13)$ (Error compiling LaTeX. Unknown error_msg), and $d=\arccot(21)$ (Error compiling LaTeX. Unknown error_msg). We have

$\tan(a)=\frac{1}{3},\quad\tan(b)=\frac{1}{7},\quad\tan(c)=\frac{1}{13},\quad\tan(d)=\frac{1}{21}$,

So

$\tan(a+b) = \frac{\frac{1}{3}+\frac{1}{7}}{1-\frac{1}{21}} = \frac{1}{2}$

and

$\tan(c+d) = \frac{\frac{1}{13}+\frac{1}{21}}{1-\frac{1}{273}} = \frac{1}{8}$,

so

$\tan((a+b)+(c+d)) = \frac{\frac{1}{2}+\frac{1}{8}}{1-\frac{1}{16}} = \frac{2}{3}$.

Thus our answer is $10\cdot\frac{3}{2}=15$.

See also

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1984_AIME_Problems/Problem_13&oldid=13718"