Difference between revisions of "1984 AIME Problems/Problem 13"

Revision as of 22:18, 19 November 2007

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 $10\cdot\frac{3}{2}=15$ .

Alternative method

Apply the formula $\cot^{-1}x + \cot^{-1} y = \cot^{-1}\left(\frac {xy-1}{x+y}\right)$ repeatedly.

@@ Line 21: / Line 21: @@
 Thus our answer is <math>10\cdot\frac{3}{2}=15</math>.
+----
+== Alternative method ==
+Apply the formula <math>\cot^{-1}x + \cot^{-1} y = \cot^{-1}\left(\frac {xy-1}{x+y}\right)</math> repeatedly.
 == See also ==

1984 AIME (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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Difference between revisions of "1984 AIME Problems/Problem 13"

Revision as of 22:18, 19 November 2007

Contents

Problem

Solution

Alternative method

See also