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

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(→‎Solution 4)
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Expanding <math>wxyz</math>, we are left with
 
Expanding <math>wxyz</math>, we are left with
−
<cmath>(1 + 3i)(1 + 13i)(1 + 7i)(1 + 21i) = (1 + 16i - 39)(1 + 28i - 147)</cmath>
+
<cmath>(3+i)(7+i)(13+i)(21+i) = (20+10i)(13+i)(21+i)</cmath>
−
<cmath>= (16i - 38)(28i - 146)</cmath>
+
<cmath>= (2+i)(13+i)(21+i)</cmath>
−
<cmath>= 4(8i - 19)(14i - 73)</cmath>
+
<cmath>= (25+15i)(21+i)</cmath>
−
<cmath>= 4(1275 - 850i)</cmath>
+
<cmath>= (5+3i)(21+i)</cmath>
−
<cmath>-10\cot \tan^{-1}-\frac{850}{1275} = 10\cot \tan^{-1}\frac{850}{1275}</cmath>
+
<cmath>= (102+68i)</cmath>
−
<cmath> = 10\frac{1275}{850}</cmath>
+
<cmath>= (3+2i)</cmath>
−
<cmath> = 10\frac{3}{2} = 15.</cmath>
+
<cmath>= 10\cot \tan^{-1}\frac{2}{3}</cmath>
 +
<cmath> = 10\frac{3}{2} = \boxed{015}</cmath>
  
 
== See also ==
 
== See also ==

Revision as of 11:22, 9 May 2017

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 $10\cdot\frac{3}{2}=15$.

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 $10*\left(\frac{3}{2}\right)=\boxed{015}$.


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 \[(3+i)(7+i)(13+i)(21+i) = (20+10i)(13+i)(21+i)\] \[= (2+i)(13+i)(21+i)\] \[= (25+15i)(21+i)\] \[= (5+3i)(21+i)\] \[= (102+68i)\] \[= (3+2i)\] \[= 10\cot \tan^{-1}\frac{2}{3}\] \[= 10\frac{3}{2} = \boxed{015}\]

See also

1984 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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