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Difference between revisions of "1984 AIME Problems/Problem 13"
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== Solution ==
 
== Solution ==
  
−
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>.
+
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>.
  
 
== See also ==
 
== See also ==

Revision as of 15:54, 6 March 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 $10\cdot\frac{3}{2}=15$.

See also

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