Difference between revisions of "1984 AIME Problems/Problem 13"
m (cleanup) |
(→Solution 3) |
||
(17 intermediate revisions by 10 users not shown) | |||
Line 5: | Line 5: | ||
== Solution == | == Solution == | ||
=== Solution 1 === | === 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 = \ | + | 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> | <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> | <center><p><math>\tan(a+b) = \frac{\frac{1}{3}+\frac{1}{7}}{1-\frac{1}{21}} = \frac{1}{2}</math></p></center> | ||
Line 21: | Line 21: | ||
<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> | <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}= | + | Thus our answer is <math>10\cdot\frac{3}{2}=\boxed{015}</math>. |
=== Solution 2 === | === Solution 2 === | ||
− | Apply the formula <math>\cot^{-1}x + \cot^{-1} y = \cot^{-1}\left(\frac {xy-1}{x+y}\right)</math> repeatedly. | + | 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 == | == See also == |
Latest revision as of 11:06, 18 December 2018
Problem
Find the value of
Contents
Solution
Solution 1
We know that so we can repeatedly apply the addition formula, . Let , , , and . We have
,
so
and
,
so
.
Thus our answer is .
Solution 2
Apply the formula repeatedly. Using it twice on the inside, the desired sum becomes . 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 , , , , , , , , , and . We see that , , , and . The sum of these four angles forms the angle of triangle , which has a cotangent of , which must mean that . So the answer is
Solution 4
Recall that and that . Then letting and , we are left with
Expanding , we are left with
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 |