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

Revision as of 14:26, 6 May 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 $10\cdot\frac{3}{2}=15$ .

@@ Line 23: / Line 23: @@
 == See also ==
-* [[1984 AIME Problems/Problem 12 | Previous problem]]
+{{AIME box|year=1984|num-b=12|num-a=14}}
-* [[1984 AIME Problems/Problem 14 | Next problem]]
+* [[AIME Problems and Solutions]]
-* [[1984 AIME Problems]]
+* [[American Invitational Mathematics Examination]]
+* [[Mathematics competition resources]]

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

Revision as of 14:26, 6 May 2007

Problem

Solution

See also