Difference between revisions of "1984 AIME Problems/Problem 13"
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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*\left(\frac{3}{2}\right)=\boxed{015}</math>. | 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*\left(\frac{3}{2}\right)=\boxed{015}</math>. | ||
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+ | ===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>(1 + 3i)(1 + 13i)(1 + 7i)(1 + 21i) = (1 + 16i - 39)(1 + 28i - 147)</cmath> | ||
+ | <cmath>= (16i - 38)(28i - 146)</cmath> | ||
+ | <cmath>= 4(8i - 19)(14i - 73)</cmath> | ||
+ | <cmath>= 4(1275 - 850i)</cmath> | ||
+ | <cmath>-10\cot \tan^{-1}-\frac{850}{1275} = 10\cot \tan^{-1}\frac{850}{1275}</cmath> | ||
+ | <cmath> = 10\frac{1275}{850}</cmath> | ||
+ | <cmath> = 10\frac{3}{2} = 15.</cmath> | ||
== See also == | == See also == |
Revision as of 12:50, 20 April 2014
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 |