Difference between revisions of "1984 AIME Problems/Problem 13"
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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. | 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. | ||
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+ | === Solution 3 === | ||
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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}(\angleA_2OA_1)=3</math>, <math>\cot^{-1}(\angleB_2OB_1)=7</math>, <math>\cot^{-1}(\angleC_2OC_1)=13</math>, and <math>\cot^{-1}(\angleD_2OD_1)=21</math>. The sum of these four angles forms the angle of trangle <math>OD_2H</math>, which has a cotangent of <math>\frac{5100}{3400}=\{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*(\frac{3}{2})=\boxed{015}</math>. | ||
== See also == | == See also == |
Revision as of 22:52, 27 May 2013
Problem
Find the value of
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 $\cot^{-1}(\angleA_2OA_1)=3$ (Error compiling LaTeX. Unknown error_msg), $\cot^{-1}(\angleB_2OB_1)=7$ (Error compiling LaTeX. Unknown error_msg), $\cot^{-1}(\angleC_2OC_1)=13$ (Error compiling LaTeX. Unknown error_msg), and $\cot^{-1}(\angleD_2OD_1)=21$ (Error compiling LaTeX. Unknown error_msg). The sum of these four angles forms the angle of trangle , which has a cotangent of $\frac{5100}{3400}=\{3}{2}$ (Error compiling LaTeX. Unknown error_msg), which must mean that . So the answer is .
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 |