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1984 AIME Problems/Problem 13

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Problem

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

Contents

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 = \arccot(3), b=\arccot(7), c=\arccot(13), and d=\arccot(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.

See also

1984 AIME (Problems • 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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