Difference between revisions of "1984 AIME Problems/Problem 13"
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== Solution == | == Solution == | ||
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| + | We know that <math>\tan(\arctan(x)) = x</math> so we can repeatedly apply the addition formula, <math>\tan(x+y) = \frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)}</math>. Let <math>a = \arccot(3)</math>, <math>b=\arccot(7)</math>, <math>c=\arccot(13)</math>, and <math>d=\arccot(21)</math>. We have <center><p><math>\tan(a)=\frac{1}{3},\quad\tan(b)=\frac{1}{7},\quad\tan(c)=\frac{1}{13},\quad\tan(d)=\frac{1}{21}</math>,</p></center> So <center><p><math>\tan(a+b) = \frac{\frac{1}{3}+\frac{1}{7}}{1-\frac{1}{21}} = \frac{1}{2}</math></p></center> and <center><p><math>\tan(c+d) = \frac{\frac{1}{13}+\frac{1}{21}}{1-\frac{1}{273}} = \frac{1}{8}</math>,</p></center> so <center><p><math>\tan((a+b)+(c+d)) = \frac{\frac{1}{2}+\frac{1}{8}}{1-\frac{1}{16}} = \frac{2}{3}</math>.</p></center> Thus our answer is <math>10\cdot\frac{3}{2}=15</math>. | ||
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== See also == | == See also == | ||
* [[1984 AIME Problems/Problem 12 | Previous problem]] | * [[1984 AIME Problems/Problem 12 | Previous problem]] | ||
* [[1984 AIME Problems/Problem 14 | Next problem]] | * [[1984 AIME Problems/Problem 14 | Next problem]] | ||
* [[1984 AIME Problems]] | * [[1984 AIME Problems]] | ||
Revision as of 16:52, 6 March 2007
Problem
Find the value of 
Solution
We know that 
so we can repeatedly apply the addition formula, 
. 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
,
So
and
,
so
.
Thus our answer is 
.
