Difference between revisions of "1984 AIME Problems/Problem 13"
(→Solution 1) |
(→Solution 3) |
||
Line 29: | Line 29: | ||
=== Solution 3 === | === 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 | + | 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=== | ===Solution 4=== |
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 |