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

Revision as of 23:08, 17 August 2012

Problem

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

Solution

Solution 1

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 $a = \cot^{-1}(3)$ , $b=\cot^{-1}(7)$ , $c=\cot^{-1}(13)$ , and $d=\cot^{-1}(21)$ . 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 $10\cdot\frac{3}{2}=15$ .

Solution 2

Apply the formula $\cot^{-1}x + \cot^{-1} y = \cot^{-1}\left(\frac {xy-1}{x+y}\right)$ repeatedly. Using it twice on the inside, the desired sum becomes $\cot (\cot^{-1}2+\cot^{-1}8)$ . This sum can then be tackled by taking the cotangent of both sides of the inverse cotangent addition formula shown at the beginning.

@@ Line 25: / Line 25: @@
 === Solution 2 ===
-Apply the formula <math>\cot^{-1}x + \cot^{-1} y = \cot^{-1}\left(\frac {xy-1}{x+y}\right)</math> repeatedly. Using it twice on the inside, the desired sum becomes <math>\cot (\cot^{-1}2+\cot^{-1}8)</math>
+Apply the formula <math>\cot^{-1}x + \cot^{-1} y = \cot^{-1}\left(\frac {xy-1}{x+y}\right)</math> repeatedly. Using it twice on the inside, the desired sum becomes <math>\cot (\cot^{-1}2+\cot^{-1}8)</math>. This sum can then be tackled by taking the cotangent of both sides of the inverse cotangent addition formula shown at the beginning.
 == 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
All AIME Problems and Solutions

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