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

Latest revision as of 11:06, 18 December 2018

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 $10\cdot\frac{3}{2}=\boxed{015}$ .

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.

Solution 3

On the coordinate plane, let $O=(0,0)$ , $A_1=(3,0)$ , $A_2=(3,1)$ , $B_1=(21,7)$ , $B_2=(20,10)$ , $C_1=(260,130)$ , $C_2=(250,150)$ , $D_1=(5250,3150)$ , $D_2=(5100,3400)$ , and $H=(5100,0)$ . We see that $\cot^{-1}(\angle A_2OA_1)=3$ , $\cot^{-1}(\angle B_2OB_1)=7$ , $\cot^{-1}(\angle C_2OC_1)=13$ , and $\cot^{-1}(\angle D_2OD_1)=21$ . The sum of these four angles forms the angle of triangle $OD_2H$ , which has a cotangent of $\frac{5100}{3400}=\frac{3}{2}$ , which must mean that $\cot( \cot^{-1}3+\cot^{-1}7+\cot^{-1}13+\cot^{-1}21)=\frac{3}{2}$ . So the answer is $10\cdot\left(\frac{3}{2}\right)=\boxed{015}.$

Solution 4

Recall that $\cot^{-1}\theta = \frac{\pi}{2} - \tan^{-1}\theta$ and that $\arg(a + bi) = \tan^{-1}\frac{b}{a}$ . Then letting $w = 1 + 3i, x = 1 + 7i, y = 1 + 13i,$ and $z = 1 + 21i$ , we are left with

$10\cot(\frac{\pi}{2} - \arg w + \frac{\pi}{2} - \arg x + \frac{\pi}{2} - \arg y + \frac{\pi}{2} - \arg z) = 10\cot(2\pi - \arg wxyz)$ $= -10\cot(\arg wxyz).$

Expanding $wxyz$ , we are left with $(3+i)(7+i)(13+i)(21+i) = (20+10i)(13+i)(21+i)$ $= (2+i)(13+i)(21+i)$ $= (25+15i)(21+i)$ $= (5+3i)(21+i)$ $= (102+68i)$ $= (3+2i)$ $= 10\cot \tan^{-1}\frac{2}{3}$ $= 10 \cdot \frac{3}{2} = \boxed{015}$

@@ Line 1: / Line 1: @@
 == Problem ==
-{{empty}}
+Find the value of <math>10\cot(\cot^{-1}3+\cot^{-1}7+\cot^{-1}13+\cot^{-1}21).</math>
+__TOC__
 == Solution ==
-{{solution}}
+=== Solution 1 ===
+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 = \cot^{-1}(3)</math>, <math>b=\cot^{-1}(7)</math>, <math>c=\cot^{-1}(13)</math>, and <math>d=\cot^{-1}(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}=\boxed{015}</math>.
+=== 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>. This sum can then be tackled by taking the cotangent of both sides of the inverse cotangent addition formula shown at the beginning.
+=== Solution 3 ===
+On the coordinate plane, let <math>O=(0,0)</math>, <math>A_1=(3,0)</math>, <math>A_2=(3,1)</math>, <math>B_1=(21,7)</math>, <math>B_2=(20,10)</math>, <math>C_1=(260,130)</math>, <math>C_2=(250,150)</math>, <math>D_1=(5250,3150)</math>, <math>D_2=(5100,3400)</math>, and <math>H=(5100,0)</math>. We see that <math>\cot^{-1}(\angle A_2OA_1)=3</math>, <math>\cot^{-1}(\angle B_2OB_1)=7</math>, <math>\cot^{-1}(\angle C_2OC_1)=13</math>, and <math>\cot^{-1}(\angle D_2OD_1)=21</math>. The sum of these four angles forms the angle of triangle <math>OD_2H</math>, which has a cotangent of <math>\frac{5100}{3400}=\frac{3}{2}</math>, which must mean that <math> \cot( \cot^{-1}3+\cot^{-1}7+\cot^{-1}13+\cot^{-1}21)=\frac{3}{2}</math>. So the answer is <math>10\cdot\left(\frac{3}{2}\right)=\boxed{015}.</math>
+===Solution 4===
+Recall that <math>\cot^{-1}\theta = \frac{\pi}{2} - \tan^{-1}\theta</math> and that <math>\arg(a + bi) = \tan^{-1}\frac{b}{a}</math>. Then letting <math>w = 1 + 3i, x = 1 + 7i, y = 1 + 13i,</math> and <math>z = 1 + 21i</math>, we are left with
+<cmath>10\cot(\frac{\pi}{2} - \arg w + \frac{\pi}{2} - \arg x + \frac{\pi}{2} - \arg y + \frac{\pi}{2} - \arg z) = 10\cot(2\pi - \arg wxyz)</cmath>
+<cmath>= -10\cot(\arg wxyz).</cmath>
+Expanding <math>wxyz</math>, we are left with
+<cmath>(3+i)(7+i)(13+i)(21+i) = (20+10i)(13+i)(21+i)</cmath>
+<cmath>= (2+i)(13+i)(21+i)</cmath>
+<cmath>= (25+15i)(21+i)</cmath>
+<cmath>= (5+3i)(21+i)</cmath>
+<cmath>= (102+68i)</cmath>
+<cmath>= (3+2i)</cmath>
+<cmath>= 10\cot \tan^{-1}\frac{2}{3}</cmath>
+<cmath> = 10 \cdot \frac{3}{2} = \boxed{015}</cmath>
 == 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]]
-* [[1984 AIME Problems]]
+[[Category:Intermediate Trigonometry Problems]]

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